Solutions 2r1l2c

Solutions to Stat 641 Homework Assignments Chapter 2 Exercise 2.1 - Specific to the student Exercise 2.3 - For experiment #2 complete steps a-c on the checklist a) Define the objectives of the experiment. It is believed that the boiling point of water may be affected by the concentration of salt in the water. The purpose of this experiment is to determine the extent to which the boiling point of water changes with varying levels of salt concentrations in the water. b) Identify all sources of variation. a. treatments - various concentrations of salt. Four (4) equally spaced concentration levels (C1 , C2, C3, and C4) will be used in the experiment. b. experimental units - Twenty (20) glass beakers containing constant, known volumes of distilled water will be made available for the study. c. blocking factors - days on which the experiment is executed could be considered as blocking factors due to differences in relative humidity and atmospheric pressure. These tests, however, will be executed in an environmentally controlled laboratory so blocking will not be necessary. c) Choose a rule by which to assign the experimental units to the levels of the treatment factors. a. A completely randomized design will be employed. Each of the 20 beakers will be randomly assigned a unique identification number between 1 and 20. A computer program has been used to generate a randomized sequence of the integers from 1 to 20, and the assignment of concentration levels to the experimental units is presented below. The experiment will be run in the original order of experimental unit numbers. CONC C1

C1

C1

C1

C1

C2

C2

C2

C2

C2

C3

C3

C3

C3

C3

C4

C4

C4

C4

C4

9

7

14

19

4

18

12

3

2

16

1

8

17

5

11

13

10

6

15

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

EU 20 Run order 1

1 C:\OSU\Courses\Stat 641 - DOE\Homework Assignments\Solutions\Bishop Solutions\Solutions to Stat 641 Homework Assignmentschap1-6.doc

Exercise 2.7 - For experiment #8 write down all the possible sources of variation. a) b) c) d) e) f) g) h)

course (confounding effects like professor and time of day) class topic (English, history, mathematics, statistics, etc) Students length of exam type of exam - multiple choice, written, true false, problem solving, etc color of exam paper unknown unexplainable, random effects

This design should probably be run as some type of block design. The blocking variables might be some combination of course, topic and length of the exam. Exercise 2.8 - For experiment #8 write down all the possible sources of variation


Chapter 3 Exercise 3.3 TABLE 3.3.1 TREATMENT DEFINITION

Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

TREATMENT 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

TREAMENT CODE 111 112 113 121 122 123 131 132 133 211 212 213 221 222 223 231 232 233

A 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2

B 1 1 1 2 2 2 3 3 3 1 1 1 2 2 2 3 3 3

C 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

TABLE 3.3.2 EXPERIMENTAL ASSIGNMENT AND RUN ORDER

Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

TREATMENT 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18

TREAMENT CODE 111 111 112 112 113 113 121 121 122 122 123 123 131 131 132 132 133 133 211 211 212 212 213 213 221 221 222 222 223 223 231 231 232 232 233 233

A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

B 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3

C 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

EU 15 29 22 8 2 21 5 36 13 20 4 6 25 27 3 33 35 32 12 14 7 17 31 30 19 11 34 10 23 9 28 1 26 18 16 24

RUN ORDER 22 25 31 13 5 1 34 2 4 20 29 12 6 33 35 9 3 10 8 27 24 7 23 36 11 18 32 16 17 19 15 14 26 21 28 30


Exercise 3.14 - Pedestrian light experiment

Scatterplot of TIME vs PUSHES 38.4

38.3

TIME

38.2

38.1

38.0

37.9 0.0

0.5

1.0

1.5 PUSHES

2.0

2.5

3.0

a) This plot indicates that there is no evidence that the number of pushes affects the waiting time. b) ANOVA table. We fail to reject the hypothesis that there is no effect of pushes on wait time. ( p-value > .05) One-way ANOVA: TIME versus PUSHES Source PUSHES Error Total

DF 3 28 31

S = 0.1045

Level 0 1 2 3

N 7 10 10 5

SS 0.0080 0.3060 0.3140

MS 0.0027 0.0109

R-Sq = 2.56%

Mean 38.207 38.171 38.194 38.212

StDev 0.068 0.116 0.100 0.130

F 0.25

P 0.864

R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on Pooled StDev ---------+---------+---------+---------+ (-------------*------------) (----------*----------) (-----------*----------) (---------------*---------------) ---------+---------+---------+---------+ 38.160 38.220 38.280 38.340

Pooled StDev = 0.105

c) Estimates of the mean waiting time and standard deviation by number of pushes.


Data Display

Row 1 2 3 4

Mean Wait Time 38.2071 38.1710 38.1940 38.2120

StDev 0.068243 0.116089 0.099577 0.129885

Sample Size 7 10 10 5

d) contrast estimate: θˆ = 0.0148 e) Var(θˆ ) = σ 2 [(1 / 7) ] + (σ 2 / 9)[(1 / 10) + (1 / 10) + (1 / 5)] = .1873σ 2 So the variance for the estimator of the contrast is approximately 1/5 the size of the variance associates with the random error in the model.


Exercise 3.15 - Trout experiment Scatterplot of HEMOGLOBIN(GMS PER 100 ML) vs SULFAMERAZINE - gms

HEMOGLOBIN(GMS PER 100 ML)

12 11 10 9 8 7 6 5 0

2

4

6 8 10 SULFAMERAZINE - gms

12

14

16

a) There appears to be an effect on the hemoglobin response due to the level of sulfamerazine in the food. The variance of the unexplained variation around the mean appears to be comparable across all levels of sulfamerazine. b) Yit = µ it + ε it with the usual assumptions on the error . c) Least squares estimates Descriptive Statistics: HEMOGLOBIN SULFAMERAZINE 0 5 10 15

N 10 10 10 10

Mean 7.200 9.330 9.030 8.690

StDev 1.019 1.717 1.135 1.000

The effect may be non-linear. It appears as though hemoglobin is significantly increased when any level of sulfamerzine is added to the food, but that the amount is increased the hemoglobin level may decrease.

d) The hypothesis of no effect of sulfamerazine on hemoglobin is rejected at the p < .003 level of significance.


Scatterplot of HEMOGLOBIN vs SULFAMERAZINE with GROUP MEANS 12

Variable HEMOGLOBIN * SULFAMERA ZINE Mean1 * By Var1

11

Hemoglobin

10 9 8 7 6 5 0

2

4

6 8 10 Sulfamerizine

12

14

16

One-way ANOVA: HEMOGLOBIN versus SULFAMERAZINE Source SULFAMERAZINE Error Total S = 1.252

Level 0 5 10 15

N 10 10 10 10

DF 3 36 39

SS 26.80 56.47 83.27

R-Sq = 32.19%

Mean 7.200 9.330 9.030 8.690

StDev 1.019 1.717 1.135 1.000

MS 8.93 1.57

F 5.70

P 0.003

R-Sq(adj) = 26.54%

Individual 95% CIs For Mean Based on Pooled StDev ------+---------+---------+---------+--(-------*-------) (-------*-------) (-------*-------) (-------*-------) ------+---------+---------+---------+--7.0 8.0 9.0 10.0


e)

2 2 χ 36 ,.95 = 23.3, SSE = 5647, so a 95% confidence bound for σ is 2.42 = 56.47 / 23.3.


Exercise 3.17 - Soap experiment Power and Sample Size One-way ANOVA Alpha = 0.05

SS Means 0.03125

Assumed standard deviation = 0.2828

Sample Size 4

Power 0.144533

Number of Levels = 3

Maximum Difference 0.25

The sample size is for each level.


SS Means 0.00005

Assumed standard deviation = 0.03

Sample Size 342

Target Power 0.98

Actual Power 0.980229




SS Means 10.125

Assumed standard deviation = 3

Sample Size 15

Target Power 0.9

Actual Power 0.907403



The sample size is for each level.


Exercise 4.3 - Pedestrian experiment

1) contrast estimate: θˆ = 0.0148 2) Var(θˆ ) = σ 2 [(1 / 7) ] + (σ 2 / 9)[(1 / 10) + (1 / 10) + (1 / 5)] = .1873σ 2 3) 4) 5) 6) 7)

MSE = .0109 based on 28 degrees of freedom critical t -distribution value for 28 df and alpha = .05: t = 1.70 test statistic: T = θˆ / MSEx (.1873) = .327 reject the null hypothesis iff T > 1.70 fail to reject the null hypothesis

One-way ANOVA: TIME versus PUSHES Source PUSHES Error Total

DF 3 28 31

S = 0.1045

Level 0 1 2 3

N 7 10 10 5

SS 0.0080 0.3060 0.3140

MS 0.0027 0.0109

R-Sq = 2.56%

Mean 38.207 38.171 38.194 38.212

StDev 0.068 0.116 0.100 0.130

F 0.25

P 0.864

R-Sq(adj) = 0.00%

Individual 95% CIs For Mean Based on Pooled StDev ---------+---------+---------+---------+ (-------------*------------) (----------*----------) (-----------*----------) (---------------*---------------) ---------+---------+---------+---------+ 38.160 38.220 38.280 38.340


Dunnett's comparisons with a control Family error rate = 0.05 Individual error rate = 0.0196 Critical value = 2.48 Control = level (0) of PUSHES Intervals for treatment mean minus control mean Level 1 2 3

Lower -0.1637 -0.1407 -0.1467

Center -0.0361 -0.0131 0.0049

Upper 0.0914 0.1144 0.1564

------+---------+---------+---------+--(-----------*------------) (------------*-----------) (--------------*---------------) ------+---------+---------+---------+---0.10 0.00 0.10 0.20

Exercise 4.5 - Trout experiment


One-way ANOVA: HEMOGLOBIN(GMS PER 100 ML) versus SULFAMERAZINE - gms Source SULFAMERAZINE Error Total S = 12.52

Level 0 5 10 15

N 10 10 10 10

DF 3 36 39

SS 2680 5647 8327

R-Sq = 32.19%

MS 893 157

Mean 72.00 93.30 90.30 86.90

F 5.70

P 0.003

R-Sq(adj) = 26.54%

StDev 10.19 17.17 11.35 10.00

Individual 99% CIs For Mean Based on Pooled StDev ---------+---------+---------+---------+ (--------*--------) (--------*--------) (--------*--------) (--------*--------) ---------+---------+---------+---------+ 72 84 96 108

Pooled StDev = 12.52 Tukey 99% Simultaneous Confidence Intervals All Pairwise Comparisons among Levels of SULFAMERAZINE - gms Individual confidence level = 99.81% SULFAMERAZINE - gms = SULFAMERAZINE - gms 5 10 15

Lower 2.57 -0.43 -3.83

0 subtracted from: Center 21.30 18.30 14.90

Upper 40.03 37.03 33.63

---+---------+---------+---------+-----(---------*--------) (--------*---------) (--------*---------) ---+---------+---------+---------+------20 0 20 40

SULFAMERAZINE - gms =

5 subtracted from:

SULFAMERAZINE - gms 10 15

Center -3.00 -6.40

Lower -21.73 -25.13

Upper 15.73 12.33

---+---------+---------+---------+-----(---------*--------) (---------*--------) ---+---------+---------+---------+------20 0 20 40

SULFAMERAZINE - gms = 10 subtracted from: SULFAMERAZINE - gms 15

Lower -22.13

Center -3.40

Upper 15.33

---+---------+---------+---------+-----(--------*---------) ---+---------+---------+---------+------20 0 20 40


Exercise 5.1 - Pedestrian light experiment Residual Plots for TIME Residuals Versus the Fitted Values 0.2

90

0.1 Residual

Percent

Normal Probability Plot of the Residuals 99

50 10 1

-0.2

-0.1

0.0 Residual

0.1

0.2

38.17

38.18

38.19 38.20 Fitted Value

38.21

Residuals Versus the Order of the Data 0.2

8

0.1

6

Residual

Frequency

-0.1 -0.2

Histogram of the Residuals

4 2 0

0.0

0.0 -0.1 -0.2

-0.20 -0.15 -0.10 -0.05 0.00

Residual

0.05

0.10

0.15

2

4

6

8 10 12 14 16 18 20 22 24 26 28 30 32

Observation Order

There is no evidence of model inadequacy.


Exercise 5.4 - Reaction time experiment One-way ANOVA: REACTION TIME versus TC Source TC Error Total

DF 5 12 17

SS 0.025549 0.003472 0.029021

S = 0.01701

Level 11 12 13 21 22 23

N 3 3 3 3 3 3

MS 0.005110 0.000289

R-Sq = 88.04%

Mean 0.18500 0.17867 0.21200 0.26833 0.25933 0.26500

StDev 0.01735 0.01041 0.02088 0.01102 0.02401 0.01389

F 17.66

P 0.000

R-Sq(adj) = 83.05%

Individual 95% CIs For Mean Based on Pooled StDev -----+---------+---------+---------+---(-----*-----) (-----*-----) (------*-----) (-----*-----) (-----*-----) (-----*-----) -----+---------+---------+---------+---0.175 0.210 0.245 0.280


Residual Plots for REACTION TIME Normal Probability Plot of the Residuals

Residuals Versus the Fitted Values

99 0.02 Residual

Percent

90 50 10

0.01 0.00 -0.01 -0.02

1 -0.04

-0.02

0.00 Residual

0.02

0.04

0.18


0.20


0.26

Residuals Versus the Order of the Data

8

Residual

Frequency

0.02 6 4 2 0

0.01 0.00 -0.01 -0.02

-0.02

-0.01

0.00 Residual

0.01

0.02

2

4

6 8 10 12 14 Observation Order

16

18


Residuals Versus ORDER (response is REACTION TIME) 0.03 0.02

Residual

0.01 0.00 -0.01 -0.02 -0.03 0

5

10 ORDER

15

20


Chapter 6

Exercise 6.1

The main effects model is given by Yijt = µ + α i + β j + ε ijt ε ijt ~ N(0, σ 2 ) ε ijt ' s are mutually independent i = 1, 2, ..., a j = 1,2, ..., b t = 1,2, ..., rij The two-way complete model is given by

Yijt = µ + α i + β j + αβ ij + ε ijt ε ijt ~ N(0, σ 2 ) ε ijt ' s are mutually independent i =1, 2, ..., a j = 1,2, ..., b t =1,2, ..., rij The two-way main effects model should be used when the effect of each factor on the response does not depend upon the level of the other factor. In that case we say the two factors do not "interact" with each other. In this case their effects on the response are "additive". The significance of this case is that the effects of the two factors can be described separately by merely analyzing the factor level means or the factor main effects. The main effects in this case directly describe the effect of changing each factor across their various levels. Under these circumstances it is also of benefit to use the two-way main effects model because the contrasts or degrees of freedom for the interaction are estimating error. Using this model these degrees of freedom go into the error term resulting in a better estimate for the error variance σ2. If the two factors do interact, then the two-way complete model must be used. In this case the main effects must be treated with caution. In this case for example the main effect for factor A is the effect of factor A averaged over the levels of factor B. This makes the interpretation of the main effects difficult and in many cases not even meaningful. In fact under some interaction conditions, the main effects can be zero while both factor A and factor B have a significant impact on the response variable.


Exercise 6.7 - Weld Strength Experiment

The cell means model Yijt = µ + τ ij + ε ijt ε ijt ~ N(0, σ 2 ) ε ijt ' s are mutually independent i = 1, 2, ..., a j = 1,2, ..., b t = 1,2, ..., rij

6.7 (a) - Results for: WELD_STRENGTH.MTW MTB > Oneway 'STRNTH' 'TRTMT'. One-way ANOVA: STRNTH versus TRTMT Source TRTMT Error Total

DF 14 15 29

S = 3.307

Level 11 12 13 14 15 21 22 23 24 25 31 32 33 34 35

N 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

SS 1261.2 164.0 1425.2

MS 90.1 10.9

F 8.24

R-Sq = 88.49%

Mean 11.000 15.000 25.500 17.000 19.000 17.000 13.000 34.000 13.000 13.000 9.000 10.500 7.500 14.500 15.000

StDev 1.414 2.828 6.364 1.414 2.828 2.828 1.414 5.657 2.828 1.414 1.414 2.121 3.536 0.707 5.657

P 0.000

R-Sq(adj) = 77.75%

Individual 95% CIs For Mean Based on Pooled StDev -------+---------+---------+---------+-(----*----) (----*----) (----*---) (----*----) (----*----) (----*----) (----*----) (----*----) (----*----) (----*----) (----*----) (----*---) (----*---) (----*---) (----*----) -------+---------+---------+---------+-10 20 30 40


Since the p-value is less than .001, we would reject the null hypothesis and claim that there is a statistically significant difference in the average weld strength across the treatments.


6.7(b) -

For the cell means model the pairwise comparisons are of the form τ ij − τ sh The contrast coefficients are all either {0,-1,+1} where the +1's are in the ij positions, the -1 in the sh positions and 0 elsewhere. The contrast comparing gage bar setting 3 to the average of the other two settings is given by τ3. − [ τ1. + τ 2. ] / 2 There are several possible strategies for intervals. The most sensible is probably to use Tukey's method at level 99% for pairwise comparisons, and a t interval at level 99% for the difference of averages contrasts. The overall level will then be at least 98%. 6.7(c) -

For the Tukey 99% simultaneous intervals for τ ij − τ sh , the formulae are y ij. − y sh . ±

q 155,15,.01 2

2MSE 2

The interval for τ13 − τ15 is 6.500 ±

6.927 2

10.933 = (−9.695, 22.695)

This interval tells us that, with overall 98%confidence, the difference in weld strength between the third and fifth time of welding for the first gage bar setting is somewhere between -9.695 and 22.695 units. The interval for τ3. − [ τ1. + τ 2. ] / 2 is given by τ3. − [ τ1. + τ 2. ] / 2 ± t 15,.005 (1 / 2 + 1 / 8 + 1 / 8) xMSE 17.5 − [18.0 + 11.3] / 2 ± 2.95 .75x10.9

17.5 −14.65 ± 8.43

2.85 ± 8.43 (-5.58,11.28)


6.7(d) -

The analysis of variance table gives SSE = 164.0. The chi-squared value at 15 degrees of freedom with p = .90 in the right hand tail is 8.547. Using formula (3.4.10), we have a 90% confidence bound for σ2 as σ 2 < 164 / 8.547 = 19.188 6.7(e) -

The formula for the confidence intervals for pairwise comparisons is y ij. − y sh . ±

q 155,15,.01 2

2MSE r

so we need q 15,15( r −1),.01

MSE <4 r

The upper bound for σ2 is 19.188 which use for MSE so we need q 15,15( r −1),.01

19.188 < 4 or (q 15,15( r −1),.01 ) 2 < .0834r r

We need to solve for r which we must do in an iterative fashion. r

15(r-1)

(q 15,15( r −1),.01 ) 2

0.834r

7 20 25 35 36

90 285 360 510 525

32.5 29.7 29.7 29.7 29.7

5.84 16.68 20.85 29.19 30.02

Action Increase r Increase r Increase r Increase r

So r = 36 is just about right which requires 540 total observations.


Exercise 6.8 - Weld Strength Experiment Continued

6.8(a) Minitab Project Report

————— 4/16/2007 9:13:59 AM ———————————————————— Welcome to Minitab, press F1 for help. MTB > WOpen "C:\OSU\COURSES\STAT64~1\DATA\MINITA~1\WELD_STRENGTH.MTW". Retrieving worksheet from file: 'C:\OSU\COURSES\STAT64~1\DATA\MINITA~1\WELD_STRENGTH.MTW' Worksheet was saved on Fri Dec 17 2004

Results for: WELD_STRENGTH.MTW MTB > GLM 'STRNTH' = GAGE| TIME; SUBC> Brief 1 ; SUBC> GFourpack; SUBC> RType 1 . General Linear Model: STRNTH versus GAGE, TIME Factor GAGE TIME

Type fixed fixed

Levels 3 5

Values 1, 2, 3 1, 2, 3, 4, 5

Analysis of Variance for STRNTH, using Adjusted SS for Tests Source GAGE TIME GAGE*TIME Error Total

DF 2 4 8 15 29

S = 3.30656

Seq SS 278.60 385.53 597.07 164.00 1425.20

Adj SS 278.60 385.53 597.07 164.00

R-Sq = 88.49%

Adj MS 139.30 96.38 74.63 10.93

F 12.74 8.82 6.83

P 0.001 0.001 0.001

R-Sq(adj) = 77.75%

The p-value for the test of no interaction effect is less than .05 so we reject that hypothesis in favor of the alternative hypothesis that there is an interaction effect. Therefore the effects on weld strength between different gage bar settings depend upon the level of time used.

6.8(b) -

The interaction plot does our conclusion in 6.8(a). It indicates that the effect of time is different for gage 3 compared to the other two gages. 6.8(c) -

The differences in gage bar settings are comparisons of averages taken over the levels of time. Given the significant interaction effect, these comparisons are probably not meaningful. In and industrial setting it is probably the case that both gage and time can be set independently and so the combination that provided the best mean strength would be selected and main effects would not be of interest. 18 C:\OSU\Courses\Stat 641 - DOE\Homework Assignments\Solutions\Bishop Solutions\Solutions to Stat 641 Homework Assignmentschap1-6.doc

Residual Plots for STRNTH Residuals Versus the Fitted Values

99

5.0

90

2.5 Residual

Percent

Normal Probability Plot of the Residuals

50 10 1

-5.0

-2.5

0.0 Residual

2.5

0.0 -2.5 -5.0

5.0

6.0 4.5

2.5

3.0 1.5 0.0

-4

-2

0 Residual

2

20 Fitted Value

30


Residual

Frequency


10

0.0 -2.5 -5.0

4

2

4

6

8 10 12 14 16 18 20 22 24 26 28 30

Observation Order

Main Effects for Weld Strength GAGE

TIME

22

Mean of STRNTH

20 18 16 14 12 10 1

2

3

1

2

3

4

5


Interaction Chart for Weld Strength 35

GA GE 1 2 3

30

Mean

25 20 15 10

1

2

3 TIME

4

5


Exercise 6.9 - Sample Size Calculation

For a design with factor A at 3 levels and factor B at 4 levels there are v=12 treatments. Therefore there are 12(r-1) degrees of freedom for error. The half-width of the Tukey 99% simultaneous pairwise intervals is q

12 ,12 ( r −1),.01

15 r

so r must be selected so that q 12,12 ( r −1),.01

15 <5 r

(q 12,12 ( r −1),.01 ) 2

15 < 25 r

or

or r > .6(q 12,12 ( r −1),.01 ) 2

r

12(r-1)

(q 12,12 ( r −1),.01 ) 2

.6(q 12,12 ( r −1),.01 ) 2

Action

5 10 15 18

48 108 168 204

33.2 30.25 29.59 27.98

19.9 18.5 17.7 16.8

Increase r Increase r Increase r

So it appears that 17 to 18 observations should be adequate.


Exercise 6.16 - Survival Experiment Minitab Project Report

————— 4/16/2007 9:26:51 AM ———————————————————— Welcome to Minitab, press F1 for help. MTB > WOpen "C:\OSU\Courses\Stat 641 - DOE\Data\Minitab Files\survival.MTW". Retrieving worksheet from file: 'C:\OSU\Courses\Stat 641 - DOE\Data\Minitab Files\survival.MTW' Worksheet was saved on Fri Dec 17 2004

Results for: survival.MTW MTB > GLM 'TIME' = POISON| TRTMT; SUBC> Brief 1 ; SUBC> GFourpack; SUBC> RType 1 . General Linear Model: TIME versus POISON, TRTMT Factor POISON TRTMT

Type fixed fixed

Levels 3 4

Values 1, 2, 3 1, 2, 3, 4

Analysis of Variance for TIME, using Adjusted SS for Tests Source POISON TRTMT POISON*TRTMT Error Total

S = 0.149139

DF 2 3 6 36 47

Seq SS 1.03301 0.92121 0.25014 0.80073 3.00508

Adj SS 1.03301 0.92121 0.25014 0.80073

R-Sq = 73.35%

Adj MS 0.51651 0.30707 0.04169 0.02224

F 23.22 13.81 1.87

P 0.000 0.000 0.112

R-Sq(adj) = 65.21%


Residual Plots for TIME Normal Probability Plot of the Residuals


99

0.4 Residual

Percent

90 50

0.0 -0.2

10 1 -0.4

0.2

-0.2

0.0 Residual

0.2

-0.4

0.4


0.6 Fitted Value

0.8

0.4

18

Residual

Frequency

0.4


24

12

0.2 0.0 -0.2

6 0

0.2

-0.3 -0.2 -0.1

0.0 0.1 Residual

0.2

0.3

-0.4

0.4

1

5

10


40

45

MTB > let c4 = 1/c3 MTB > GLM '1/TIME' = POISON| TRTMT; SUBC> Brief 1 ; SUBC> GFourpack; SUBC> RType 1 . General Linear Model: 1/TIME versus POISON, TRTMT Factor POISON TRTMT

Type fixed fixed

Levels 3 4

Values 1, 2, 3 1, 2, 3, 4

Analysis of Variance for 1/TIME, using Adjusted SS for Tests Source POISON TRTMT POISON*TRTMT Error Total

S = 0.489985

DF 2 3 6 36 47

Seq SS 34.8771 20.4143 1.5708 8.6431 65.5053

Adj SS 34.8771 20.4143 1.5708 8.6431

R-Sq = 86.81%

Adj MS 17.4386 6.8048 0.2618 0.2401

F 72.63 28.34 1.09

P 0.000 0.000 0.387

R-Sq(adj) = 82.77%


Residual Plots for 1/TIME Normal Probability Plot of the Residuals


99

1.0

50

0.0 -0.5

10 1

0.5

Residual

Percent

90

-1.0

-0.5

0.0 Residual

0.5

-1.0

1.0


3 Fitted Value

4

5

1.0

12

0.5

Residual

Frequency

2


16

8 4 0

1

0.0 -0.5

-0.75 -0.50 -0.25 0.00

0.25

Residual

0.50

0.75

1.00

-1.0

1

5

10


40

45

MTB > Interact 'POISON' 'TRTMT'; SUBC> Response 'TIME' '1/TIME'.

Interaction Plot (data means) for TIME 0.9

POISON 1 2 3

0.8

Mean

0.7 0.6 0.5 0.4 0.3 0.2 1

2

3

4

TRTMT


Interaction Plot (data means) for 1/TIME 5

POISON 1 2 3

Mean

4

3

2

1 1

2

3

4

TRTMT

16 (a) -

The equal variance assumption is clearly violated using the response data on the original scale. The residual variance increases with increasing values for survival times. The normality assumption is also violated. 16 (b) -

The model assumptions look much better using the transformed response variable. Based on this response variable there appears to be no interaction between poison and treatment and significant main effects for both factors. The rate of dying is significantly higher for poison #3 no matter which treatment is used. The rate of dying appears to be the lowest for poison #1 independent of which treatment is used. 16 (c) -

Based on the interaction graphs it appears that the average survival time for poison #3 is significantly lower across all treatments compared to poisons #1 and #2, and that the differences in the effects of the treatments are much smaller for poison #3 compared to poisons #1 and #2 which is creating the interaction effect.


Chapter 9 -Homework Solutions Results for: papertoweltable9-8.MTW

Scatterplot of ABSORBANCY vs DROP RATE 0.8

ABSORBANCY

0.7 0.6 0.5 0.4 0.3 0.2 1.8

1.9

2.0 DROP RATE

2.1

2.2

2.3

Scatterplot of ABSORBANCY vs DROP RATE 0.8

TRTMT 1 2 3 4 5 6

ABSORBANCY

0.7 0.6 0.5 0.4 0.3 0.2 1.8

1.9

2.0 2.1 DROP RATE

2.2

2.3

1 C:\OSU\Courses\Stat 641 - DOE\Homework Assignments\Solutions\Bishop Solutions\Solutions to Stat 641 Homework Assignmentschap9.doc

General Linear Model: ABSORBANCY versus TRTMT Factor TRTMT

Type fixed

Levels 6

Values 1, 2, 3, 4, 5, 6

Analysis of Variance for ABSORBANCY, using Adjusted SS for Tests Source DROP RATE TRTMT Error Total

DF 1 5 5 11

Seq SS 0.114712 0.241986 0.027941 0.384640

S = 0.0747546

Term Constant DROP RATE

Adj SS 0.040553 0.241986 0.027941

R-Sq = 92.74%

Coef 1.4951 -0.5076

SE Coef 0.3824 0.1884

Adj MS 0.040553 0.048397 0.005588

F 7.26 8.66

P 0.043 0.017

R-Sq(adj) = 84.02%

T 3.91 -2.69

P 0.011 0.043

Unusual Observations for ABSORBANCY Obs 1 3

ABSORBANCY 0.735500 0.388400

Fit 0.652311 0.471589

SE Fit 0.062604 0.062604

Residual 0.083189 -0.083189

St Resid 2.04 R -2.04 R

R denotes an observation with a large standardized residual.

Residual Plots for ABSORBANCY Residuals Versus the Fitted Values 0.10

90

0.05 Residual

Percent


50 10 1

-0.10

-0.05

0.00 Residual

0.05

0.00 -0.05 -0.10

0.10


0.6

0.7

0.10

2

Residual

Frequency



3

1 0

0.3

-0.075

-0.050

-0.025

0.000

0.025

0.050

0.075

0.05 0.00 -0.05 -0.10

Residual

1

2

3

4 5 6 7 8 9 Observation Order

10 11 12


Residuals Versus RUN (response is ABSORBANCY) 0.10

Residual

0.05

0.00

-0.05

-0.10 0

2

4

6 RUN

8

10

12

Residuals Versus DROP RATE (response is ABSORBANCY) 0.10

Residual

0.05

0.00

-0.05

-0.10 1.8

1.9

2.0 2.1 DROP RATE

2.2

2.3


General Linear Model: ABSORBANCY versus BRAND, PATTERN Factor BRAND PATTERN

Type fixed fixed

Levels 3 2

Values BRAND-1, BRAND-2, BRAND-3 PRINTED, WHITE

Analysis of Variance for ABSORBANCY, using Adjusted SS for Tests Source DROP RATE BRAND PATTERN BRAND*PATTERN Error Total

DF 1 2 1 2 5 11

S = 0.0747546

Term Constant DROP RATE

Seq SS 0.114712 0.142377 0.006369 0.093241 0.027941 0.384640

Adj SS 0.040553 0.104557 0.001698 0.093241 0.027941

R-Sq = 92.74%

Coef 1.4951 -0.5076

SE Coef 0.3824 0.1884

Adj MS 0.040553 0.052278 0.001698 0.046620 0.005588

F 7.26 9.36 0.30 8.34

P 0.043 0.020 0.605 0.026

R-Sq(adj) = 84.02%

T 3.91 -2.69

P 0.011 0.043

Unusual Observations for ABSORBANCY Obs 1 3

ABSORBANCY 0.735500 0.388400

Fit 0.652311 0.471589

SE Fit 0.062604 0.062604

Residual 0.083189 -0.083189

St Resid 2.04 R -2.04 R

R denotes an observation with a large standardized residual.

Residual Plots for ABSORBANCY Residuals Versus the Fitted Values 0.10

90

0.05 Residual

Percent


50 10 1

-0.10

-0.05

0.00 Residual

0.05

0.00 -0.05 -0.10

0.10


0.6

0.7


2

Residual

Frequency


0.10

3

1 0

0.3

-0.075

-0.050

-0.025

0.000

0.025

0.050

0.075

0.05 0.00 -0.05 -0.10

Residual

1

2

3

4 5 6 7 8 9 Observation Order

10 11 12


Residuals Versus RUN (response is ABSORBANCY) 0.10

Residual

0.05

0.00

-0.05

-0.10 0

2

4

6 RUN

8

10

12

Residuals Versus DROP RATE (response is ABSORBANCY) 0.10

Residual

0.05

0.00

-0.05

-0.10 1.8

1.9

2.0 2.1 DROP RATE

2.2

2.3


Main Effects Plot (fitted means) for ABSORBANCY BRAND

0.60

PATTERN

Mean of ABSORBANCY

0.55 0.50 0.45 0.40 0.35 0.30 BRAND-1

BRAND-2

BRAND-3

PRINTED

WHITE

Interaction Plot (fitted means) for ABSORBANCY 0.7

BRA ND BRA ND-1 BRA ND-2 BRA ND-3

0.6

Mean

0.5

0.4

0.3

0.2 PRINTED

WHITE PATTERN


Fractional Factorial Designs Homework Assignment Problem #1 1. Create a Minitab database for the single replication of the 24 experiment below that was conducted in a completely randomized design. Factor A = Formulation (1 or 2) Factor B = Cycle Time (15 sec or 25 sec) Factor C = Pressure (300 psi or 375 psi) Factor D = Temperature (110 degrees F or 130 degrees F) Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

A 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

B 15 15 25 25 15 15 25 25 15 15 25 25 15 15 25 25

C 300 300 300 300 375 375 375 375 300 300 300 300 375 375 375 375

D 110 110 110 110 110 110 110 110 130 130 130 130 130 130 130 130

Y 71 73 74 75 77 77 78 80 71 72 74 74 77 77 76 78

2. Add columns to the Minitab database corresponding to the contrasts presented in Table 7.6 on page 26 of the class notes. 3 Analyze these data by fitting a model that only includes the main effects and two-factor interactions. 4. Use the BCD interaction as the generating contrast and the +1 values to construct the corresponding 24-1 fractional factorial design matrix. 5. Write down the confounding pattern. Give a reason why this design might be of interest to an experimenter. 6. Try analyzing the data for this fractional design by fitting a model that only includes the main effects and two-factor interactions. What happened and why? 7. Try analyzing the fractional design fitting only the main effects. Compare the results to those obtained in part #3.

1 C:\OSU\Courses\Stat 641 - DOE\Homework Assignments\Fractional Factorial HW.doc

Solutions to Part 3. Results for: fractional factorial HW.MTW MTB > GLM 'RESPONSE - Y' = SUBC> Brief 1 ; SUBC> GFourpack; SUBC> RType 1 .

A B C D A*B A*C A*D B*C B*D C*D;

General Linear Model: RESPONSE - Y versus A, B, C, D Factor A B C D

Type fixed fixed fixed fixed

Levels 2 2 2 2

Values -1, 1 -1, 1 -1, 1 -1, 1

Analysis of Variance for RESPONSE - Y, using Adjusted SS for Tests Source A B C D A*B A*C A*D B*C B*D C*D Error Total

DF 1 1 1 1 1 1 1 1 1 1 5 15

Seq SS 4.000 12.250 81.000 2.250 0.250 0.000 0.250 2.250 1.000 0.250 3.500 107.000

S = 0.836660

Adj SS 4.000 12.250 81.000 2.250 0.250 0.000 0.250 2.250 1.000 0.250 3.500

Adj MS 4.000 12.250 81.000 2.250 0.250 0.000 0.250 2.250 1.000 0.250 0.700

R-Sq = 96.73%

F 5.71 17.50 115.71 3.21 0.36 0.00 0.36 3.21 1.43 0.36

P 0.062 0.009 0.000 0.133 0.576 1.000 0.576 0.133 0.286 0.576

R-Sq(adj) = 90.19%

Residual Plots for RESPONSE - Y Residuals Versus the Fitted Values 0.8

90

0.4 Residual

Percent


50 10 1

-1.0

-0.5

0.0 Residual

0.5

-0.4 -0.8

1.0

Histogram of the Residuals 4 3

0.4

2 1 0

-0.8

-0.4

0.0 Residual

0.4

72

74 76 Fitted Value

78

80


Residual

Frequency

0.0

0.0 -0.4 -0.8

0.8

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15 16

Observation Order


Main Effects Plot (fitted means) for RESPONSE - Y A

B

77

Mean of RESPONSE - Y

76 75 74 73 -1

1

-1

1

C

D

77 76 75 74 73 -1

1

-1

1

Interaction Plot (fitted means) for RESPONSE - Y -1

1

-1

1

-1

1 78

75

A

72 78

75

B

72 78

75

C

72

D


A -1 1

B -1 1

C -1 1

Solutions to Part 5. A = ABCD B = CD C = BD D = BC This design might be of interest if factor A were the primary factor of interest. It is not confounded with two- or three-factor interactions. Solutions to Part 6. It will not fit the model because the main effects and two-factor interactions are confounded. General Linear Model: RESPONSE - Y versus A, B, C, D Factor A B C D


Levels 2 2 2 2

Values -1, 1 -1, 1 -1, 1 -1, 1

Analysis of Variance for RESPONSE - Y, using Adjusted SS for Tests

Source A B C D A*B A*C A*D B*C B*D C*D Error Total

Model DF 1 1 1 1 1 1 1 1 1 1 -3 7

Reduced DF 1 1 1 1 1 1 1 0+ 0+ 0+ 0 7

Seq SS 2.0000 4.5000 32.0000 4.5000 0.5000 0.0000 0.5000 0.0000 0.0000 0.0000 0.0000 44.0000

+ Rank deficiency due to empty cells, unbalanced nesting, collinearity, or an undeclared covariate. No storage of results or further analysis will be done.

S = *


Solutions to Part 7. The results are basically the same except that the main effect for factor D is now significant. Factor D, however, is confounded with the BC interaction. In looking at the original ANOVA table these two factors had p-values of .133 and which were the smallest non-significant pvalues. So the main effect for D and the BC interaction are probably not zero and we are probably picking up the combined effect of these two factors.

General Linear Model: RESPONSE - Y versus A, B, C, D Factor A B C D


Levels 2 2 2 2

Values -1, 1 -1, 1 -1, 1 -1, 1

Analysis of Variance for RESPONSE - Y, using Adjusted SS for Tests Source A B C D Error Total

DF 1 1 1 1 3 7

Seq SS 2.000 4.500 32.000 4.500 1.000 44.000

S = 0.577350

Adj SS 2.000 4.500 32.000 4.500 1.000

Adj MS 2.000 4.500 32.000 4.500 0.333

R-Sq = 97.73%

F 6.00 13.50 96.00 13.50

P 0.092 0.035 0.002 0.035

R-Sq(adj) = 94.70%


Residual Plots for RESPONSE - Y Normal Probability Plot of the Residuals


99

0.50 Residual

Percent

90 50 10

0.25 0.00 -0.25 -0.50

1 -1.0

-0.5

0.0 Residual

0.5

1.0

72.0

0.50

3

0.25

2 1

76.5

0.00 -0.25 -0.50

0

-0.4

-0.2

0.0 Residual

0.2

0.4

1

2

3 4 5 6 Observation Order

Main Effects Plot (fitted means) for RESPONSE - Y A

B

77 76

Mean of RESPONSE - Y

78.0


4 Residual

Frequency



75 74 73 -1

1

-1

C

1 D

77 76 75 74 73 -1

1

-1

1


7

8

Problem #2 1. Suppose that you are faced with an experiment that involves 7 factors each at two levels. List the available fractional factorial designs. 2. Select a 1/2 fraction design using the seven factor interaction as the generating contrast. List the alias structure. Explain why this would be a good design for assessing main effects and twofactor interactions. 3. Select a 1/16th fraction and list the alias structure. Explain what this design might be useful for and why.


Solutions to Part 1

Solutions to Part 2

Fractional Factorial Design Factors: Runs: Blocks:

7 64 1

Base Design: Replicates: Center pts (total):

7, 64 1 0

Resolution: Fraction:

VII 1/2

Design Generators: G = ABCDEF

Alias Structure I + ABCDEFG A + BCDEFG B + ACDEFG C + ABDEFG D + ABCEFG E + ABCDFG F + ABCDEG G + ABCDEF AB + CDEFG AC + BDEFG AD + BCEFG


AE + BCDFG AF + BCDEG AG + BCDEF BC + ADEFG BD + ACEFG BE + ACDFG BF + ACDEG BG + ACDEF CD + ABEFG CE + ABDFG CF + ABDEG CG + ABDEF DE + ABCFG DF + ABCEG DG + ABCEF EF + ABCDG EG + ABCDF FG + ABCDE ABC + DEFG ABD + CEFG ABE + CDFG ABF + CDEG ABG + CDEF ACD + BEFG ACE + BDFG ACF + BDEG ACG + BDEF ADE + BCFG ADF + BCEG ADG + BCEF AEF + BCDG AEG + BCDF AFG + BCDE BCD + AEFG BCE + ADFG BCF + ADEG BCG + ADEF BDE + ACFG BDF + ACEG BDG + ACEF BEF + ACDG BEG + ACDF BFG + ACDE CDE + ABFG CDF + ABEG CDG + ABEF CEF + ABDG CEG + ABDF CFG + ABDE DEF + ABCG DEG + ABCF DFG + ABCE EFG + ABCD


Solutions to Part 3 Fractional Factorial Design Factors: Runs: Blocks:

7 8 1

Base Design: Replicates: Center pts (total):

7, 8 1 0

Resolution: Fraction:

III 1/16

* NOTE * Some main effects are confounded with two-way interactions. Design Generators: D = AB, E = AC, F = BC, G = ABC

Alias Structure I + ABD + ACE + AFG + BCF + BEG + CDG + DEF + ABCG + ABEF + ACDF + ADEG + BCDE + BDFG + CEFG + ABCDEFG A + BD BCDEFG B + AD ACDEFG C + AE ABDEFG D + AB ABCEFG E + AC ABCDFG F + AG ABCDEG G + AF ABCDEF

+ CE + FG + BCG + BEF + CDF + DEG + ABCF + ABEG + ACDG + ADEF + ABCDE + ABDFG + ACEFG + + CF + EG + ACG + AEF + CDE + DFG + ABCE + ABFG + BCDG + BDEF + ABCDF + ABDEG + BCEFG + + BF + DG + ABG + ADF + BDE + EFG + ABCD + ACFG + BCEG + CDEF + ABCEF + ACDEG + BCDFG + + CG + EF + ACF + AEG + BCE + BFG + ACDE + ADFG + BCDF + BDEG + ABCDG + ABDEF + CDEFG + + BG + DF + ABF + ADG + BCD + CFG + ABDE + AEFG + BCEF + CDEG + ABCEG + ACDEF + BDEFG + + BC + DE + ABE + ACD + BDG + CEG + ABDF + ACEF + BEFG + CDFG + ABCFG + ADEFG + BCDEF + + BE + CD + ABC + ADE + BDF + CEF + ABDG + ACEG + BCFG + DEFG + ABEFG + ACDFG + BCDEG +

This design will probably only be useful as a pilot or screening design in cases where the main effects are thought to be the dominant effects because all the main effects are heavily confounded with two-factor and higher order interaction .


Design and Analysis of Experiments Angela Dean and Daniel Voss Solutions to Chapter 6 12 February 2003 Solutions available in this file c

2003 Angela Dean. All rights reserved. No part of this work may be displayed on the web. No part of this work may be reproduced in any form without the written permission of Angela Dean, The Ohio State University.

CHAPTER 6.2 CHAPTER 6.7 — weld strength experiment CHAPTER 6.8 — weld strength experiment CHAPTER 6.15 — ink experiment CHAPTER 6.16 —- survival experiment CHAPTER 6.17 —estimability CHAPTER 6.21 —water boiling

1

Solution to Question 6.2 that (τij − τ i. − τ .j + τ .. ) is an interaction contrast for the two-way complete model. Write down the list of contrast coefficients in of the τij ’s as a = 3 levels and factor B has b = 4 levels. We can this by substituting the equivalent two-way complete model notation with τij = αi + βj + (αβ)ij . This gives the contrast (αβ)ij − (αβ)i. − (αβ).j + line(αβ).. which is a function of interaction parameters only. For simplicity, let i = j = 1, then we want the contrast coefficients for the contrast τ11 − τ 1. − τ .1 + τ .. = τ11 −

1 b

Pb

j=1 τ1j

−

1 a

Pa

j=1 τi1

−

1 ab

Pa

i=1

Pb

j=1 τij

with Now let a = 3 and b = 4. The, the coefficients must be as in the following table. τij τ11 τ12 τ13 τ14

term 1 1 14

term 2

term 3

1 3 1 4 1 4 1 4

1 12

1 3

τ21 τ22 τ23 τ24

1 3

τ31 τ32 τ33 τ34

2

term4

T otal 20 12

1 12 1 12 1 12 1 12 1 12 1 12 1 12 1 12 1 12 1 12 1 12

4 12 4 12 4 12 4 12 1 12 1 12 1 12 5 12 1 12 1 12 1 12

Solution to Question 6.7 — weld strength experiment a)

Using the cell-means model (6.2.1) for these data, test the hypothesis that there is no difference in the effects of the treatment combinations on weld strength against the alternative hypothesis that at least two treatment combinations have different effects. The analysis of variance for the cell means model is as follows: General Linear Models Procedure Dependent Variable: STRNTH Source Model Error Corrected Total

DF 14 15 29

Sum of Squares 1261.20000000 164.00000000 1425.20000000

F Value 8.24

Pr > F 0.0001

Source TC

DF 14

Type III SS 1261.20000000

F Value 8.24

Pr > F 0.0001

Since the p-value is less than 0.0001, for most reasonable choices of significance level, we would reject the hypothesis of no effect of the treatment combinations and conclude that the different combinations of gage bar setting and time of welding do have different effects on the strength of the weld. b)

Suppose the experimenters had planned to calculate confidence intervals for all pairwise comparisons between the treatment combinations, and also to look at the confidence interval for the difference between gage bar setting 3 and the average of the other two. Show what these contrasts look like in of the parameters τij of the cell-means model, and suggest a strategy for calculating all intervals at overall level “at least 98%.” In of the parameters τij of the cell means model, the pairwise comparisons of interest are of the form τij − τsh which has coefficient list [000...010.....0 − 10...0] where the 1 and -1 are in positions ij and sh respectively. The other contrast is of the form τ 3. − [τ 1. + τ 2. ]/2 which has coefficient list [−0.5 − 0.5....... − 0.51.0.......1.0] where there are 10 occurrences of -0.5 and five occurrences of 1.0 There are several possible strategies for intervals. The most sensible is probably to use Tukey’s method at level 99% for the pairwise comparisons, and a t interval at level 99% for the difference of averages contrast. The overall level will then be at least 98%.

c) Give formulae for the intervals in part (b). As an example, calculate the actual interval for τ13 −τ15 (the difference in the true mean strengths at the 3rd and 5th times of welding for the first gage bar setting). Explain what this interval tells you. For the Tukey 99% simulataneous intervals for τij − τsh , the formulae are y ij. − y sh

q15,15,.01 √ ± 2 3

r

2 msE . 2

The interval for τ13 − τ15 from SAS is

√ √ 6.500 ± (6.927/ 2) × 10.9333 = (−9.695, 22.695) .

This interval tells us that, with overall 98% confidence, the difference in weld strength between the third and fifth time of welding for the first gage bar setting is somewhere between -9.695 and 22.695 units. The interval for τ 3. − [τ 1. + τ 2. ]/2, is s y 3.. − (y 1.. + y 1.. )/2 ± t15,.005

1 1 1 + + 2 8 8

msE

=

d)

Calculate an upper 90% confidence limit for σ 2 . The analysis of variance table gives ssE = 164.0. The chi-squared value with 15 degrees of freedom and prob .9 in the right hand tail is 8.547. Using formula (3.4.10), we have a 90% confidence bound for σ 2 as σ 2 ≤ 164/8.547 = 19.188.

e)

If the experimenters were to repeat this experiment and needed the pairwise comparison intervals in (b) to be of width at most 8, how many observations should they take on each treatment combination? How many observations is this in total? The formula for the confidence intervals for pairwise comparisons in b) is √ p (y ij. − y sh. ) ± q15,15r−15,.01 / 2 ∗ msE (2/r) So we need q15,15r−15,.01

p

msE/r ≤ 4

2

using the upper bound for σ of 19.188 in place of msE, we need p q15,15r−15,.01 19.188/r ≤ 4 that is 2 q15,15r−15,.01 ≤ 0.834r

We now need to solve for r . r 7 20 25 35 36

15(r − 1) 90 285 360 510 525

2 q15,1(5r−1),0.01 5.72 = 32.49 5.452 = 29.7 5.452 = 29.7 5.452 = 29.7 5.452 = 29.7

0.834r 5.84 16.68 20.85 29.19 30.02

Action Increase Increase Increase Increase

r r r r

r = 36 is about right which requires 540 observations in total. If you use msE=10.9333, you will require fewer observations. But you need to justify why you think the msE would be about the same value in the next experiment.

4

Solution to Question 6.8— weld strength a) Test the hypothesis of no interaction between gage bar setting and time of weld and state your conclusion. The analysis of variance table for the two-way complete model is as follows: General Linear Models Procedure Dependent Variable: STRNTH Source Model Error Corrected Total

DF 14 15 29

Sum of Squares 1261.20000000 164.00000000 1425.20000000

F Value 8.24

Pr > F 0.0001

Source GAGE TIME GAGE*TIME

DF 2 4 8

Type III SS 278.60000000 385.53333333 597.06666667

F Value 12.74 8.82 6.83

Pr > F 0.0006 0.0007 0.0008

We are doing three hypothesis tests. If we choose an overall level of, say, 0.06, we can do each test at level 0.02. Since the p-value for the test of no interaction is less than 0.02, we reject the hypothesis of negligible interaction and conclude that the effects on the weld strength between differnt gage bar settings are not the same at the different times of weld. b) Draw an interaction plot for the two factors Gage bar setting and Time of welding. Does your interaction plot the conclusion of your hypothesis test? Explain. The interaction plot below shows that some of the times of weld are similar, but that the interaction of gage bar setting with time 3, in particular is quite different. This agrees with the hypothesis test that there is an interaction between gage bar setting and weld time. Plot of AVSTR*TIME. Symbol is value of GAGE. AVSTR | 40 + | | 2 | 30 + | | 1 | 20 + 1 |2 1 | 1 3 3 | 2 2 2 10 +1 3 | 3 | | 0 + | -+-----------+-----------+-----------+-----------+1 2 3 4 5 TIME NOTE: 1 obs hidden.

5

c) In view of your answer to part (b), is it sensible to investigate the differences between the effects of gage bar setting? Why or why not? Indicate on your plot what would be compared. The differences between gage bar settings are comparisons of averages (averaging over the levels of time, and hence over the interaction). This is probably not of interest since the time of weld can presumably be set in the industrial process. d) Regardless of your answer to (c), suppose the experimenters had decided to look at the linear trend in the effect of gage bar settings. Test the hypothesis that the linear trend in gage setting is negligible (against the alternative hypothesis that it is not negligible). From SAS, a test of the hypothsis of no linear trend in the weld strength due to time of weld gives a sum of squares (y 3.. − y 1.. )2 /(2/10) = (11.3 − 17.5)2 /(2/10) = 192.2 General Linear Models Procedure Dependent Variable: STRNTH Contrast LIN GAGE

DF 1

Contrast SS 192.20000000

F Value 17.58

Pr > F 0.0008

The p-value is 0.0008, so for most choices of type I error probability (significance level) the hypothesis of no linear trend would be rejected. (Notice that the hypothesis test gives no indication of direction of trend, since the alternative hypothesis is two-sided.)

6

Solution to Question 6.15 — ink experiment a) Obvious difficulties in running the experiment include: • Getting the same amount of stain on each piece of cloth • Washing in water of exactly the same temperature • Reading the 19-point scale Obvious difficulties in analysing the experiment include: • The possibility of non-constant variance due to inaccurate reading of the scale and different propoerties of the cloth. • The possibility of non-normal data due to a discrete scale and difficulties reading it. —Ways to reduce the difficulties in reading the scale is to have the same person do all the reading and have him/her have plenty of practice and training beforehand. —Controlling water temperature in the washing machine is difficult unless the cold setting is used. Cold tap water should be less variable. —Some careful measuring of the amount of stain used and istering it to a constant spot would be needed. —Non-constant variance would need a transformation or Satterthwaite’s method. Non-normal data with equal variances could be analysed with nonparametric methods. b) The water temperature may vary from wash to wash. If some pieces of cloth are washed in the same wash and some in different, the error variability will no tbe constant. if all pieces of cloth are washed together then the natrual variability that occurs between washes will not be seen in the experiment. c)

Plot of Y*CLOTH. Symbol is value of METHOD. Y | | 11 + 1 10 + 9 +1 1 8 +1 2 7 + 6 +2 2 5 +2 4 + 1 3 + 1 2 + 1 + 2,2 | -+-----------------------+-----------------------+1 2 3 CLOTH

7

Plot of Y*METHOD.

Symbol is value of CLOTH.

Y | | | | | | | 11 + 2 10 + 9 + 1,2 8 + 1 2 7 + 6 + 1,2 5 + 1 4 + 3 3 + 3 2 + 1 + 3,3 | --+--------------------------------+--------------1 2 METHOD If points for corresponding treatment levels are ed up on the two plots, the lines are remarkably parallel. This suggests that the two-way main effects model would be suitable for this experiment. Since the stain remover is presumably designed to work on the stain itself rather than to interact with the cloth, perhaps this could have been anticipated. d)

The GLM Procedure Dependent Variable: Y Sum of Source DF Squares Mean Square Model

3

109.2500000

36.4166667

Error

8

5.6666667

0.7083333

11

114.9166667

Corrected Total

F Value

Pr > F

51.41

<.0001

The analysis of this pilot data gives ssE = 5.666 based on df = 8 error degrees of freedom. A 90% unpper bound for sigma2 is 5.666 ssE = = 1.624. σ2 ≤ 2 χ8,.9 3.49 For pairwise differences, we require that each interval in a set of simultaneous 95% confidence intervals (for each factor separately), using Tukey’s method, satisfies msd ≤ 1. For the two-way main effects model, the error degrees of freedom are df = n − a − b + 1 = 6r − 3 − 2 + 1 = 6r − 4. So, for A, we require qa,6r−4,.05 √ 2

r

2 that is, q3,6r−4,.05 ≤ 1.2315r.

8

2 (1.624) ≤ 1 rb

r 10 8 7

6r − 4 56 44 38

2 q3,6r−4,.05 3.42 = 11.56 3.442 = 11.83 3.422 = 11.70

1.2315r 12,31 9.85 11.08

Action Decrease r Increase r Increase r

Taking r = 10 observations on each of the six treatment combinations is likely produce confidence intervals of the required length. Now we need to do a similar calculations for the single interval for B. So, for B, we require r t6r−4,.05 2 √ (1.624) ≤ 1 ar 2 that is, t26r−4,.05 ≤ 0.9236r. If we take r = 10 observations per treatment combination, then t256,.05 = 1.6732 = 2.799 which is less than 0.9236r = 9.236 so r = 10 observations is plenty to satisfy both requirements.

9

Solution to Question 6.16— survival experiment a. EQUAL VARIANCE ASSUMPTIONS Plot of Z*YHAT. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | A | | A | 2 + A | A | A | | C A A | A D A A A 0 + B B CA A A | A A A A | A C B A | A AB A | A A A | A -2 + --+--------+--------+--------+--------+--------+--0.0 0.2 0.4 0.6 0.8 1.0 YHAT The equal variance assumption is definitely violated, with variance increasing for large yhats. There is no need to check for any other assumptions, but if we do we would see normalisty violated also: Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | | | A | | A | 2 + A | A | A | | ABB | BCBA 0 + BDCA | CA | BBC | ABB | AAA | A -2 + -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 Rank for Variable Z 10

Independence cannot be checked since we don’t have the observation orders. b. TRANSFORMED DATA Plot of Z*YHAT. Legend: A = 1 obs, B = 2 obs, etc. Z | | | | A | 2 + A A | A | A | A A A A |A A A A A |B A A B A 0 + A A A A A |A A A A A | AA A A A | A A A B A A | A A | A A A -2 + -+-----------+-----------+-----------+-----------+1 2 3 4 5 YHAT Plot of Z*TRTMT. Legend: A = 1 obs, B = 2 obs, etc. Z | | | A | 2 + A A | A | A | A B A | A B A A | C C A 0 + A A C | A A B A | B A A A | C B A A | A A | A A A -2 + --+------------+------------+------------+--------1 2 3 4 TRTMT

11

Plot of Z*POISON. Legend: A = 1 obs, B = 2 obs, etc. Z | | | A | 2 +A A | A | A |A A B |B B A |E A A 0 +A A C |A A C |A A C |C C A | A A |A B -2 + -+-----------------------+-----------------------+1 2 3 POISON The equal variance assumption looks more likely to be satisfied now. The two residual vs factor plots do not show any obvious differences in variance between factor levels. Obs 1 2 3 4 5 6 7 8 9 10 11 12

TC 11 12 13 14 21 22 23 24 31 32 33 34

AVTY 2.48688 1.16346 1.86272 1.68968 3.26847 1.39339 2.71392 1.70153 4.80269 3.02897 4.26499 3.09181

VTY 0.24667 0.03980 0.23949 0.13302 0.67622 0.30602 0.17432 0.49267 0.28051 0.17761 0.05514 0.05956

We can see that, there is still some imbalance in the variances in the cells. Ratio 0.6762/0.0398. So the transformation has not worked completely. The original ratio of variances was 0.1131/0.0007 !! With the transformed data Normality appears approximately satisfied as the Normal Prob. plot looks fairly straight. (see next plot) Independence cannot be checked since we don’t have the observation orders.

12

Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | | | | A | 2 + AA | A | A | BB | ABB | BCB 0 + DA | BC | ACA | BBBA | AA | A AA -2 + -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 Rank for Variable Z c. INTERACTION PLOTS For the original data, there is some evidence of interaction as survival times for treatments 1-3 are decreasing for poison 1-3, but for treatment 4, the survival time is higher for poison 2. Plot of MN_Y*POISON. Symbol is value of TRTMT. MN_Y | 1.00 + | |2 | 2 0.75 + | 4 |4 |3 0.50 + |1 | 3 | 1 2 0.25 + 3 | 1 | | 0.00 + | -+-----------------------+-----------------------+1 2 3 POISON NOTE: 1 obs hidden. The interaction plots for the death rates also show some interaction, only now, the pattern is reversed, i.e. death rates increases for treatments 1-3 from poison 1-3, but for treatment two the death rate is lowest for poison 3. 13

Plot of MN_INV*POISON. Symbol is value of TRTMT. MN_INV | | 5 + | 1 | | | 3 | 4 + | | | | 1 | 4 3 + 2 | | 3 |1 | | 2 + |3 |4 4 | | 2 |2 1 + | -+-----------------------+-----------------------+1 2 3 POISON ** survival experiment; OPTIONS LINESIZE=75; DATA SVL; INPUT POISON TRTMT TIME; LINES; 1 1 0.31 1 1 0.45 1 1 0.46 1 1 0.43 1 2 0.82 1 2 1.10 1 2 0.88 1 2 0.72 1 3 0.43 1 3 0.45 1 3 0.63 1 3 0.76 1 4 0.45 1 4 0.71 1 4 0.66

** chapter 6;

14

1 4 2 1 2 1 2 1 2 1 2 2 2 2 2 2 2 2 2 3 2 3 2 3 2 3 2 4 2 4 2 4 2 4 3 1 3 1 3 1 3 1 3 2 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 4 3 4 3 4 3 4 ; RUN;

0.62 0.36 0.29 0.40 0.23 0.92 0.61 0.49 1.24 0.44 0.35 0.31 0.40 0.56 1.02 0.71 0.38 0.22 0.21 0.18 0.23 0.30 0.37 0.38 0.29 0.23 0.25 0.24 0.22 0.30 0.36 0.31 0.33

PROC GLM; CLASSES POISON TRTMT; MODEL TIME=POISON TRTMT; OUTPUT OUT=DATA2 PREDICTED=YHAT RESIDUAL=Z; PROC STANDARD STD=1.0; VAR Z; PROC RANK NORMAL=BLOM; VAR Z; RANKS NSCORE; PROC PLOT; PLOT Z*POISON Z*TRTMT Z*YHAT Z*NSCORE/VPOS=20 HPOS=50; RUN; QUIT; ******************************************************* * TRANSFORM DATA *

15

******************************************************* DATA INV_SVL; SET SVL; INV_TIME=1/TIME; RUN; PROC PRINT; RUN; PROC GLM; CLASSES POISON TRTMT; MODEL INV_TIME=POISON TRTMT; OUTPUT OUT=DATA3 PREDICTED=YHAT RESIDUAL=Z; PROC STANDARD STD=1.0; VAR Z; PROC RANK NORMAL=BLOM; VAR Z; RANKS NSCORE; PROC PLOT; PLOT Z*POISON Z*TRTMT Z*YHAT Z*NSCORE/VPOS=20 HPOS=50; RUN; QUIT; ******************************************************* * INTERACTION PLOTS * ******************************************************* PROC SORT DATA=DATA2; BY POISON TRTMT; PROC MEANS DATA=DATA2 NOPRINT MEAN VAR; VAR TIME; BY POISON TRTMT; OUTPUT OUT=INT_Y MEAN=MN_Y VAR=VAR_Y; PROC PLOT; PLOT MN_Y*POISON=TRTMT / VPOS=20 HPOS=50; PLOT MN_Y*TRTMT=POISON / VPOS=20 HPOS=50; RUN; QUIT; PROC SORT DATA=DATA3; BY POISON TRTMT; PROC MEANS DATA=DATA3 NOPRINT MEAN VAR; VAR INV_TIME; BY POISON TRTMT; OUTPUT OUT=INT_INV MEAN=MN_INV VAR=VAR_INV; PROC PLOT; PLOT MN_INV*TRTMT=POISON / VPOS=20 HPOS=50; PLOT MN_INV*POISON=TRTMT / VPOS=20 HPOS=50; RUN; QUIT; *************************************************** *AVERAGES * *************************************************** ;

16

DATA data8; SET SVL; PROC GLM ; CLASS TC; MODEL TIME=TC; PROC MEANS NOPRINT MEAN VAR; VAR TIME; BY TC; OUTPUT OUT=OUT8 MEAN=AVTY VAR=VTY; PROC PRINT; VAR TC AVTY VTY;

DATA data9; SET INV_SVL; PROC GLM data=data3; CLASS TC; MODEL INV_TIME=TC; PROC MEANS NOPRINT MEAN VAR; VAR INV_TIME; BY TC; OUTPUT OUT=OUT8 MEAN=AVTY VAR=VTY; PROC PRINT; VAR TC AVTY VTY; RUN; QUIT;

17

Solution to Question 6.17 — estimability a) For the two-way main effects model, all combinations of µ + αi + βj are estimable; that is, all functions of the form Σi Σj dij (µ + αi + βj ). (i) Setting d12 = 1 and all other dij = 0 gives the required function, so it is estimable. (ii) Setting d11 = d12 = 0.5 and all other dij = 0 gives the required function, so it is estimable. (iii) There is no way to choose the dij so that the αi and the µ cancel. So the function is not estimable —it is not a contrast in the βj . b) For equal sample sizes, E[Y i.. ]

= =

1 E[Yi11 + . . . + Yi1r + Yi21 + . . . + Yi2r + Yi31 + . . . + Yi3r ] 3r 1 [r(µ + αi + β1 ) + r(µ + αi + β2 ) + r(µ + αi + β3 )] 3r

= µ + αi + β . Similarly, E[Y .j. ]

= µ + α. + βj

E[Y ... ]

= µ + α. + β.

and the result follows. It is easiest to calculate the variance if we write the expression in of the individual random variables (otherwsie we have to worry about covariances between ). Y i.. + Y .j. − Y ...

=

[Yi11 + Yi12 + . . . + Yibr ]/br +

[Y1j1 + Y1j2 + . . . + Yajr ]/ar

− [Y111 + Y112 + . . . + Yabr ]/abr

So , if we collect up the , [Yij1 + Yij2 + . . . + Yijr ] has multiplier

a+b−1 abr

b−1 for each s 6= i abr a−1 [Yih1 + Yih2 + . . . + Yihr ] has multiplier for each h 6= j abr −1 [Ysh1 + Ysh2 + . . . + Yshr ] has multiplier for each s 6= i andh 6= j abr [Ysj1 + Ysj2 + . . . + Ysjr ] has multiplier

Since V ar[Yijt ] = σ 2 , it follows that V ar[Y i.. + Y .j. − Y ... ] = σ2 r

(a + b − 1) abr

2

+ σ 2 (a − 1)r

2

+ σ (b − 1)r

18

(a − 1) abr

2

(b − 1) abr 2

2

+ σ (a − 1)(b − 1)r

1 abr

2

=

σ2 [(a + b − 1)2 + (a − 1)((b − 1)2 a2 b2 r + (a − 1)2 (b − 1) + (a − 1)(b − 1)]

= = =

σ2 a2 b2 r

[(a + b − 1)2 + (a − 1)(b − 1)(a + b − 1)]

σ2 [(a + b − 1)(a + b − 1 + ab − a − b + 1)] a2 b2 r σ2 a2 b2 r

[(a + b − 1)(ab)] =

(a + b − 1)σ 2 . abr

c) From part a), E[Y i.. ] = µ + αi + β . . So E[Σci Y i.. ] = Σci µ + Σci αi + Σci β . , and since Σci = 0, it follows that E[Σci Y i.. ] = Σci αi .

19

Solution to Question 6.21— water boiling experiment a) Check the assumptions on the two-way main-effects model. The plot of residuals versus predicted values is as follows Plot of Z*PREDY. Legend: A = 1 obs, B = 2 obs, etc. 2 + | A A A | B | BA | A B B | E B D 0 + BC BC | | C C Z | B A | A A | A -2 + A | | | | | A -4 + -+-----------+-----------+-----------+-----------+2 4 6 8 10 PREDY There appears to be an extreme outlier. This corresponds to observation burner 1 (right back), salt 2 teaspoons, order=13, y121 = 4. We can see the outlier in the normal prob plot also, which appears reasonably like a straight line without the outlier. Similiarly there are no problems with independence apart from the outlier. The analysis without the outlier is done in part d). If we include the outlier the analysis is as in parts (b) and (c). b) Calculate a 99% set of Tukey confidence intervals for pairwise differences between the levels of salt, and calculate separately a 99% set of intervals for pairwise differences between the levels of burner. Dependent Variable: TIME Source Model Error Corrected Total

DF 6 41 47

Sum of Squares 117.333333 19.333333 136.666667

Mean Square 19.555556 0.471545

F Value 41.47

Pr > F 0.0001

Source BURNER SALT

DF 3 3

Type III SS 110.500000 6.833333

Mean Square 36.833333 2.277778

F Value 78.11 4.83

Pr > F 0.0001 0.0057

With the outlier, the Tukey intervals at oveall confidence level 99% show all burners except 1 and 3 to be different. The biggest difference is between burners 2 and 4 with burner 2 boiling the water between 3.3208 and 5.179 minutes faster. We are unable to see any differences between the effects of the salt levels:

20

The GLM Procedure Tukey’s Studentized Range (HSD) Test for TIME NOTE: This test controls the type I experimentwise error rate. Alpha 0.01 Error Degrees of Freedom 41 Error Mean Square 0.471545 Critical Value of Studentized Range 4.68765 Minimum Significant Difference 0.9292 Comparisons significant at the 0.01 level are indicated by ***.

BURNER Comparison 4 - 1 4 - 3 4 - 2 1 - 3 1 - 2 3 - 2

Difference Between Means 1.7500 2.3333 4.2500 0.5833 2.5000 1.9167

Simultaneous 99% Confidence Interval 0.8208 2.6792 1.4041 3.2626 3.3208 5.1792 -0.3459 1.5126 1.5708 3.4292 0.9874 2.8459

*** *** *** *** ***

Tukey’s Studentized Range (HSD) Test for TIME NOTE: This test controls the type I experimentwise error rate. Alpha 0.01 Error Degrees of Freedom 41 Error Mean Square 0.471545 Critical Value of Studentized Range 4.68765 Minimum Significant Difference 0.9292 Comparisons significant at the 0.01 level are indicated by ***.

SALT Comparison 4 - 0 4 - 2 4 - 6 0 - 2 0 - 6 2 - 6


Simultaneous 99% Confidence Interval -0.8459 1.0126 -0.1792 1.6792 -0.0959 1.7626 -0.2626 1.5959 -0.1792 1.6792 -0.8459 1.0126

c) Test a hypothesis that there is no linear trend in the time to boil water due to the level of salt. Do a similar test for a quadratic trend. The tests for the hypotheses H0 :{ no linear trend due to salt} and H0 : {no quadratic trend due to salt} have p-values 0.0982 and 0.6764 respectively. So the null hypotheses would not be rejected at most usual choices of significance level. The GLM Procedure Dependent Variable: TIME Contrast LIN SALT QUAD SALT

DF 1 1

Contrast SS 1.35000000 0.08333333

Mean Square 1.35000000 0.08333333

21

F Value 2.86 0.18

Pr > F 0.0982 0.6764

d) The experimenter believed that observation number 13 was an outlier, since it has a large standardized residual and it was an observation taken late on a Friday evening. Repeat the analysis in (b) and (c) removing this observation. Which analysis do you prefer? Why? If we remove the outlier, some of the residual plots are: Plot of Z*PREDY. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | A 2 + A | A | A AD | C B B CA 0 + B B B | B BC C | E A A | A A -2 + A | A | | -4 + | -+-----------+-----------+-----------+-----------+2 4 6 8 10 PREDY Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | A 2 + A | A | ABB A | ABCCB 0 + DB | BCCB | AB D | AA -2 + A | A | | -4 + | -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 RANK FOR VARIABLE Z

22

Plot of Z*ORDER. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | A 2 + A | A | A A A A A A | A AA AA AA AAA A 0 + A A A AAA | A A AAA A A A A A | A A A A A A A | A A -2 + A | A | | -4 + | -+-------+-------+-------+-------+-------+-------+0 8 16 24 32 40 48 ORDER The equal variance assumption per cell looks fairly well satisfied. here are a lot of ties in the data making it difficult to gauge accurately. The normality assumption is questionable, possibly due to the ties. The plot of residuals against order of observation shows no patterns. The analysis of variance given by SAS is: Dependent Variable: TIME Source Model Error Corrected Total

DF 6 40 46

Sum of Squares 118.147033 12.959350 131.106383

Mean Square 19.691172 0.323984

F Value 60.78

Pr > F 0.0001

Source BURNER SALT

DF 3 3

Type III SS 112.510347 5.449741

Mean Square 37.503449 1.816580

F Value 115.76 5.61

Pr > F 0.0001 0.0026

At an overall alpha level of at most 0.01 (each test at .05), the hypotheses of no effect of burner and no effect of salt would both be rejected.

i

Least Squares Means for Effect BURNER Difference Simultaneous 99% Between Confidence Interval for j Means LSMean(i)-LSMean(j)

1 1 1 2 2 3

2 3 4 3 4 4

2.727642 0.810976 -1.522358 -1.916667 -4.250000 -2.333333

1.937580 0.020913 -2.312420 -2.688137 -5.021470 -3.104803

23

3.517705 1.601038 -0.732295 -1.145197 -3.478530 -1.561863

i 1 1 1 2 2 3

j 2 3 4 3 4 4

Least Squares Means for Effect SALT Difference Simultaneous 99% Between Confidence Interval for Means LSMean(i)-LSMean(j) 0.439024 -0.351038 1.229087 -0.083333 -0.854803 0.688137 0.750000 -0.021470 1.521470 -0.522358 -1.312420 0.267705 0.310976 -0.479087 1.101038 0.833333 0.061863 1.604803

We can see the Tukey 99% confidence intervals show all burners to be different and, again the largest difference is between burners 2 and 4 with a similar difference as the analysis in b). The 99% intervals for salt show small difference in time between salt levels 3 and 4. The test for the trends in salt requires coefficients to be calculated since the levels no longer have the same number of observations. The formula for the linear trend coefficients is given on Page 71 as ci = ri (47xi − 118) where xi are the coded level 1, 2, 3, 4. These coefficients work out to be -852 -264 276 840. We can divide by 12 and use the coefficients -71 -22 23 70. (Notice these are just a little different from the equally spaced level coefficients of -3 -1 1 3.) The GLM Procedure Dependent Variable: TIME Contrast DF Contrast SS Mean Square LIN SALT 1 1.82689454 1.82689454

F Value 5.64

Pr > F 0.0225

The p-value is 0.0225 instead of 0.0982 that was obtained when the outlier was included. This would now be significant if an individual significance level of over 0.023 had been selected. Otherwise, the results are surprisingly close. Here the experimenter truely believed the observation was incorrect. In this the case, the analysis without the outlier might be preferred. Otherwise, since they are so similar, the analysis with the outlier might be preferable.

24

Design and Analysis of Experiments Angela Dean and Daniel Voss Solutions to Chapter 3 23 August 2006 Solutions available in this file c

2006 Angela Dean. All rights reserved. No part of this work may be displayed on the web without permission. No part of this work may be reproduced in any form without the written permission of Angela Dean, The Ohio State University.

Question 3.2 — Randomization Question 3.3 — Randomization Question 3.4 — Estimation Question 3.5 — Estimation Question 3.12 —Balloon experiment Question 3.13 — Heart-lung pump experiment, continued Question 3.15 —Trout experiment Question 3.16 — Trout experiment, sample sizes Question 3.19 — Sample size calculation

1

Solution to Question 3.2 Suppose that you are planning to run an experiment with one treatment factor having three levels and no blocking factor. It has been determined that r1 = 3, r2 = r3 = 5. Assign at random 13 experimental units to the v = 3 treatments so that the first treatment is assigned 3 units and the other two treatments are each assigned 5 units. SAS commands as given in Section 3.8.1 with treatment 1 listed 3 times and treatments 2 and 3 each listed 5 times are given below. Alternatively, one can assign the random numbers by hand using Table A.1 and following Section 3.2. data design; input treat @@; ranno=ranuni(0); lines; 1 1 1 2 2 2 2 2 3 3 3 3 3 ; proc sort; by ranno; proc print; run;

SAS PRINTOUT Obs 1 2 3 4 5 6 7 8 9 10 11 12 13

treat 2 3 2 2 3 2 3 1 3 2 3 1 1

ranno 0.17343 0.26774 0.39478 0.43455 0.52564 0.60382 0.72904 0.77178 0.79472 0.80521 0.90078 0.91355 0.99821

We then assign the experimental units in order to the randomly ordered list of treatments. Since this is a random ordering, no two solutions are likely to be identical.

2

Solution to Question 3.3 — Randomization Suppose that you are planning to run an experiment with three treatment factors, where the first factor has two levels and the other factors have three levels each. Write out the coded form of the 18 treatment combinations. Assign 36 experimental units at random to the treatment combinations so that each treatment combination is assigned two units. SAS commands as given in Section 3.8.1 with each of the 18 treatment combinations listed twice. data design; input treat @@; ranno=ranuni(0); lines; 111 111 112 112 113 113 121 121 122 122 123 123 131 131 132 132 133 133 211 211 212 212 213 213 221 221 222 222 223 223 231 231 232 232 233 233 ; proc sort; by ranno; proc print; run; SAS PRINTOUT OBS 1 2 3 4 5 6 : : 33 34 35 36

TREAT 121 113 212 122 222 113 : : 231 131 131 223

RANNO 0.00247 0.05134 0.05963 0.08012 0.08204 0.11382 : : 0.88784 0.92804 0.96630 0.98881

We then assign the experimental units in order to the randomly ordered list of treatments. Since this is a random ordering, no two solutions are likely to be identical.

3

Solution to Question 3.4 — Estimation For the one-way analysis of variance model (3.3.1), page 36, the solution to the normal equations ˆ = y v. . used by the SAS software is τî = y i. − y v. (i = 1, . . . , v) and µ (a) Is τi estimable? Explain. τi is not estimable since it cannot be written in the form Σvi=1 bi (µ + τi ). Any choice of the bi would result in a function which includes µ and/or at least one of the other τj . (b) Calculate the expected value of the least squares estimator for τ1 − τ2 corresponding to the above solution. Is τ1 − τ2 estimable? Explain. The above solution gives least squares estimate of τ1 −τ2 as y 1. −y 2. , so the least squares estimator of τ1 − τ2 is Y 1. − Y 2. and " # ri 1 X E Y i. = E Yit ri t=1 =

ri 1 X (µ + τi ) ri t=1

= µ + τi So, E Y 1. − Y 2. = (µ + τ1 ) − (µ + τ2 ) = τ1 − τ2 , so τ1 − τ2 estimable. Solution to Question 3.5 — Estimation Consider a completely randomized with observations on three treatments (coded 1, 2, 3). For the one-way analysis of variance model (3.3.1), page 36, determine which of the following are estimable. For those that are estimable, state the least squares estimator. (a) Yes, τ1 +τ2 −2τ3 is estimable. It has the form Σci τi with Σi ci = 0, since c1 = c2 = 1, c3 = −2 and all other ci zero. This is a contrast. All contrasts are estimable (page 37). The least squares estimator of τ1 + τ2 − 2τ3 is given by Y¯1. + Y¯2. − 2Y¯3. . P (b) Yes, µ + τ3 is estimable. It has the form i bi (µ + ti ) with b3 = 1 and b1 = b2 = 0. The least squares estimator of µ + τ3 is given by Y¯3. . (c) No, τ1 − τ2 − τ3 is not written as Σbi (µ + τi ) for any selection of Pestimable. it cannot be P constants bi . In order to have ci τi estimable, we need ci = 0. P (d) Yes, µ + (τ1 + τ2 + τ3 )/3 is estimable. It has the form i bi (µ + ti ) with b1 = b2 = b3 = 1/3. The least squares estimator of µ + (τ1 + τ2 + τ3 )/3 is given by (Y¯1. + Y¯2. + Y¯3. )/3.

4

Solution to Question 3.12 —Balloon experiment a). Plot inflation time versus color and comment on the results. From the plot, it appears as though balloon inflation time does depend on balloon color. In particular, it looks as ithough colors 1 and 4 are easier to blow up than colors 2 and 3. Plot of TIME*COLOR.

Legend: A = 1 obs, B = 2 obs, etc.

TIME | | 30 + | A | A | | | A 25 + A | B A | | A *A* B | *** | 20 + B A B A | A A A | *** B *A* | A A | A A | B A A 15 + | A ---+----------------+----------------+----------------+-1 2 3 4 COLOR b). Estimate the mean inflation time for each balloon color, and add these estimates to the plot from part (a). SAS was used to calculate the means using the SAS program listed after the solution to part e). On the above plot, the means are indicated by the asterisks. c). Construct an analysis of variance table and test the hypothesis that color has no effect on inflation time. From SAS PROC GLM: Dependent Variable: TIME Source

DF

Sum of Squares

Mean Square

Model Error Corrected Total

3 28 31

126.15125 305.64750 431.79875

42.05042 10.91598

F Value

Pr > F

3.85

0.0200

If we select a significance level of α = 0.05, the p-value of 0.02 leads to a rejection of the hypothesis that that the color of the balloon has no effect on the inflation time and we would conclude that the mean inflation times for these 4 colors of balloons are not the same. However, if we select a 5

significance level of α = 0.01, then we would not have sufficient evidence to reject the null hypothesis of no difference in inflation times of the 4 colors. d). Plot the data for each color in the order that it was collected. Are you concerned that the assumptions on the model are not satisfied? If so, why? If not, why not? From the plots shown at the end of this question, it is evident that for all four colors of balloons, the inflation time tended to decrease the later the balloons were inflated in the experiment. Perhaps the assistant improved at blowing up the balloons. In the analysis of variance, we are assuming that the the eight observations for each treatment are independent and identically distributed. The plots suggest that this is not the case: it appears that the means are decreasing as the run order increases. e). Is the analysis conducted in part (c) satisfactory? Since the assumptions for the analysis of variance do not appear to be satisfied, the stated p-value in part c) is not correct. We cannot predict whether it is too high or too low. Plot of TIME*ORDER. Legend: A = 1 obs, B = 2 obs, etc., COLOR 1 | | A 22 + | TIME | | | A 20 + | A | A | | 18 + | A | | | A 16 + A A | ---+--------+--------+--------+--------+--------+--------+-------+ORDER 0 5 10 15 20 25 30 35

6

Plot of TIME*ORDER. Legend: A = 1 obs, B = 2 obs, etc. COLOR 2 | A | A | 27.5 + | TIME | | | A 25.0 + | | | | 22.5 + | A | | | 20.0 + A | | A | A | A 17.5 + ---+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+-6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 ORDER

7

Plot of TIME*ORDER. Legend: A = 1 obs, B = 2 obs, etc. COLOR 3 | A | A A 24 + | TIME | | A | A 22 + | | | | A A 20 + | | | | 18 + | | | | 16 + A ---+--------+--------+--------+--------+--------+--------+--------+- ORDER 0 5 10 15 20 25 30 35

8

Plot of TIME*ORDER. Legend: A = 1 obs, B = 2 obs, etc. COLOR 4 24 + A | | | | 22 + | TIME | | | 20 +A | | A | A | 18 + | | A | A | 16 + A | | | | 14 + A -+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 ORDER

9

Solution to Question 3.13 — Heart-lung pump experiment, continued. a). Calculate an analysis of variance table and test the null hypothesis that the number of revolutions per minute has no effect on the fluid flow rate. Using the one-way analysis of variance model, the analysis of variance table obtained from PROC GLM is as follows:

The GLM Procedure Dependent Variable: FLOW

Source

DF

Sum of Squares

Model Error Corrected Total

4 15 19

16.12551 0.02084 16.14635280

Mean Square 4.03138 0.00139

F Value

Pr > F

2901.61

<.0001

Since the p-value is less than most usual choices of significance level, we reject the null hypothesis H0 : {τ1 = τ2 = τ3 = τ4 = τ5 }, and conclude there is an effect of the rpm on the fluid flow rate. b).

Are you happy with your conclusion? Why or why not?

We need to examine the assumptions of the model. The plot in Figure 3.1 on Page 40 suggests that the variability of the data may increase as the RPM increase. This would violate the assumption that the error varibles in the one way analysis of variance models are identically distributed with common variance σ 2 . If the model assupmptions are violated then the stated significant levels and confidence levels are not correct (see Chapter 5). c. Calculate a 90% upper confidence limit for the error variance σ 2 . χ215,0.9 = 8.547 so a 90% upper confidence limit for the error variance is ssE/8.547 = .0024. However, this interval also requires that the assumptions of the model are satisfied.

10

Solution to Question 3.15 — Trout experiment (a) Plot the data and comment on the results. There is a good deal of overlap between the data obtained at each level of sulfamerazine, so it is possible that there is not much difference in the hemoglobin content obtained from the different levels of sulfamerazine (but see answer to part c). The variability of the data looks about the same at each level, although a little smaller for level 3. Plot of Hem*Sulf.


Hem | 12 + A | | | C C 10 + A A | A D | -B__ | B B A B__ 8 + B D A | A | B__ A B | A A 6 + A | A | | 4 + | --+------------+------------+------------+--------0 1 2 3 SULF b) Write down the suitable model for this experiment. A suitable model would be the one-way analysis of variance model:

Yit = µ + τi + it it ∼ N (0, σ 2 ) it 0 s are mutually independent t = 1, . . . , 10, i = 1, . . . , 4,

(1)

where Yit is the hemoglobin content from a sample of blood from the tth trout fed level i of sulfamerazine, µ is a constant, τi is the effect on the hemoglobin content of level i of sulfamerazine, and it is a random error variable.

11

(c)Calculate the least squares estimate of the mean response for each treatment. Show these estimates on the plot in (a). Can you draw any conclusion from these estimates? SAS PRINTOUT from the MEANS statement Level of Sulf

N

1 2 3 4

10 10 10 10

The The The The

least least least least

squares squares squares squares

-------------Hem------------Mean Std Dev 7.20000000 9.33000000 9.03000000 8.69000000

estimate estimate estimate estimate

of of of of

µ + τ1 µ + τ2 µ + τ3 µ + τ4

is is is is

y¯1 y¯2 y¯3 y¯4

= = = =

1.01871379 1.71661812 1.13534136 1.00049988 7.2 9.33 9.03 8.69

The means for sulfamerazine levels 2, 3, and 4 are very close together and each one falls within the range of the data for the other two levels. The mean for sulfamerazine level 1 is outside the range of the data from level 3, so it is possible that the effects of these two levels are different. (d) Test the hypothesis that sulfamerazine has no effect on the hemoglobin content of trout blood. SAS PRINTOUT from PROC GLM

Dependent Variable: Hem Source DF Model 3 Error 36 Corrected Total 39 Source Sulf

DF 3

Sum of Squares 26.80275 56.47100 83.27375

Type III SS 26.80275

Mean Square 8.9342 1.5686

Mean Square 8.93425

F Value 5.70

F Value 5.70

Pr > F 0.0027

Pr > F 0.0027

If we select a significance level of 0.01. Then, since F > F(3,36,0.95) = 2.87, we reject H0 (or, equivalently, since P-value=0.0027 < 0.01, we reject H0 ). So we conclude (at significance level 0.05) that the level of sulfamerazine does affect the average hemoglobin content in the blood of brown trout. (e) Calculate a 95% upper confidence limit for σ 2 . σ2 ≤

ssE 56.471 = = 2.51 χ36,0.95 22.47

12

Solution to Question 3.16 — Trout experiment, continued (a) For calculating the number of observations needed on each treatment, what would you use as a guess for σ 2 ? I would use the 95% confidence bound found in problem 3.15(e), (b) Calculate the sample size needed for an analysis of variance test with α = 0.05 to have power 0.95 if (i) ∆ = 1.5. (ii) ∆ = 1.0. (iii) ∆ = 2.0. (i) For v = 4, ∆ = 1.5, P = 0.95, α = 0.05, r 47 42

ν2 = 4(r − 1) 1000 184 164

r = 8.629φ2 46.846 41.764 41.764

Action Round up to r = 47 Round up to r = 42 Round up to r = 42

r = 18.416φ2 105.408 93.97 93.97


φ 2.33 2.20 2.20

Take 42 observations on each treatment. (ii) For v = 4, ∆ = 1.0, P = 0.95, α = 0.05, r 106 94

ν2 = 4(r − 1) 1000 420 372

φ 2.33 2.20 2.20

Take 94 observations on each treatment. (iii) For v = 4, ∆ = 2.0, P = 0.95, α = 0.05, r 27 24

ν2 = 4(r − 1) 1000 104 92

φ 2.33 2.20 2.20

r = 18.416φ2 26.35 23.49 23.49

Take 24 observations on each treatment.

13


Solution to Question 3.19 — Sample size calculation There are v = 5 levels of lighting and we wish to test the null hypothesis that all levels are similar in of their effect on performance. We require α = 0.05

∆ = 4.5 seconds

σ ≤ 3 seconds

Π(∆) = 0.9

Now

(2)(5)(32 )φ2 2vσ 2 φ2 = = 4.444φ2 . 2 ∆ 4.52 We also have degrees of freedom ν1 = 4 and ν2 = v(r − 1) = 5(r − 1). Using the top table on page 715, we have r=

r 15 16

ν2 = 5(r − 1) 1000 70 75

φ approx 1.8 approx 1.85 1.85

r = 4.444φ2 14.4 15.2

Action round up to 15 round up to 16

The table on page 715 has very few values, so we are unable to get an accurate answer. However it looks as though 16 observations per treatment (80 in total) should be enough to achieve the required power.

14

Design and Analysis of Experiments Angela Dean and Daniel Voss Solutions to Chapter 4 31 January 2003 Solutions available in this file c


SOLUTIONS AVAILABLE IN THIS FILE 4.3

Pedestrian light experiment

4.4

Reaction time experiment

4.5

Trout experiment, continued

4.6– Battery experiment, continued 4.7– Soap experiment, continued 4.8


4.9– Battery experiment, continued 4.10


4.11

Pedestrian light experiment, continued

1

Design and Analysis of Experiments Dean and Voss Solutions to Chapters 4 Solution to Question 4.3– Pedestrian Light Experiment a) We wish to test the null hypothesis that the contrast (τ1 + τ2 + τ3 )/3 − τ0 is zero against the alternative hypothesis that it is less than zero. [Note: If you did not the corrections from the website, then you will be testing that this contrast is greater than zero]. From computer output, msE = 0.01093. The least squares estimate of the contrast is (y 1. + y 2. + y 3. )/3 − y 0 = −0.0148095 . and the corresponding standard error is s 1 1 1 1 + + + msE 10 × 9 10 × 9 5 × 9 7

= 0.04524.

So, the test statistic for the one sided test is least squares estimate standard error

−0.0148095 = − 0.32735 0.04524

=

Since the value of the test statistic is not less than −t28,0.05/2 = −2.048 (or, for any other reasonable choice of α, we have no evidence to suggest that pushing the pedestrian light button shortens the waiting time at this particular light! b) Using SAS, we can obtain a set of simulataneous 95% confidence intervals for the treatment versus control contrasts using Dunnett’s method. We cannot use Dunnett’s method from the tables at the back of the book since we do not have equal sample sizes. The SAS output is as follows: The GLM Procedure Dunnett’s t Tests for TIME NOTE: This test controls the Type I experimentwise error for comparisons of all treatments against a control. Alpha 0.05 Error Degrees of Freedom 28 Error Mean Square 0.010927 Critical Value of Dunnett’s t 2.47622 Comparisons significant at the 0.05 level are indicated by ***.

NPUSH Comparison 3 - 0 2 - 0 1 - 0

Difference Between Means 0.00486 -0.01314 -0.03614

Simultaneous 95% Confidence Limits -0.14671 0.15642 -0.14070 0.11442 -0.16370 0.09142

Conclusion: Each of these intervals includes zero, so again there is no evidence to suggest that either one, two or three pushes is better than no pushes for this pedestrian light.

2

If you do not have access to SAS and decide to use Tukey’s method instead, the intervals will be slightly wider, and will still include zero. The GLM Procedure Tukey’s Studentized Range (HSD) Test for TIME NOTE: This test controls the Type I experimentwise error rate. Alpha 0.05 Error Degrees of Freedom 28 Error Mean Square 0.010927 Critical Value of Studentized Range 3.86125 Comparisons significant at the 0.05 level are indicated by ***.

NPUSH Comparison 3 - 0 2 - 0 1 - 0

Difference Between Means 0.00486 -0.01314 -0.03614

Simultaneous 95% Confidence Limits -0.16226 0.17197 -0.15379 0.12751 -0.17679 0.10451

Solution to Question 4.4 – Reaction time experiment Solution to Question 4.4 – Reaction Time Experiment a) Identify a set of contrasts that you would find particularly interesting in this experiment. treatment combinations are coded as: 1= auditory, 5 seconds4= visual, 5 seconds 2= auditory, 10 seconds 5= visual, 10 seconds 3= auditory, 15 seconds6= visual, 15 seconds Contrasts of interest 1) c1= 1/3, c2 = 1/3, c3= 1/3, c4= -1/3, c5= -1/3, c6= -1/3 Test for differences between the responses due to visual and auditory cues. 2) c1=1/2 , c2 = -1/4, c3= -1/4, c4= 1/2, c5= -1/4, c6= -1/4 Test for differences between the response times after 5 seconds vs. 10 or 15 seconds. 3) c1=1/4 , c2 = 1/4, c3= -1/2, c4= 1/4, c5= 1/4, c6= -1/2 Test for differences between the response times after 5 or 10 seconds vs. 15 seconds. b) Plot the data. What does the plot suggest about the treatments? The plot suggests that the subjects responded faster with the auditory cue than the visual cue. Also, as the elapsed time between the cue and stimulus increased, there was an increase in the amount of time that the subjects took to respond. c) Test the hypothesis that the treatments do not have different effects on the reaction time against the alternative hypothesis that they do have different effects. Ho: (1 = (2 = (3 = (4 = (5 = (6 H1: at least two of the treatments means differ We find that the F* critical value is 17.66 with a p-value of ¡.001 so we reject Ho and conclude at the alpha =0.05 level that at least two of the means differ. (The mean response time differs for at least two of the treatments). One-way ANOVA: time versus treatment Analysis of Variance for time Source DF SS MS F P treatmen 5 0.025549 0.005110 17.66 0.000 Error 12 0.003472 0.000289 Total 17 0.029021 d) Calculate a set of simultaneous 90 simultaneous 90(ci ( tn-(, (/2m tn-(, (/2 = t18-6, .01/2*3 = t12, .00167 = 3.6381 treatment means: =.185 =.179 =.212 =.268 =.259 =.265 = = .0080 Confidence interval: 1. (ci= .573/3 - .792/3 = -.073 (-.073 ( .0291) (-.102, -.044) We conclude at the 90 2. (ci= .185/2 - .179/4 - .212/4 + .268/2 - .259/4 - .265/4 = -.002

3

(-.002 ( 3.6381*sqrt(.000289*.25)) (-.002 ( 3.6381*.0085) (-.002 ( .0309) (-.0329, .0289) Since the value zero is contained in the 90we conclude at the 90between the response time after an elapsed time of 5 seconds vs. 10 or 15 seconds. 3. (ci= .185/4 + .179/4 - .212/2 + .268/4 + .259/4 - .265/2 = -.01575 (-.01575 ( 3.6381(sqrt(.000289*.25)) (-.01575 (0.0309) (-.04665, .01515) Since the value zero is contained in the 90we conclude at the 90difference between the response time after an elapsed time of 5 or 10 seconds vs. 15 seconds. e) Ignoring the previous parts of the exercise, use Hsu’s method of multiple comparisons with the best to determine the best/worst treatment or treatments. Define the “best” to be the treatment that produces the quickest response (that is, the smallest value of the response variable). One-way ANOVA: time versus treatment Analysis of Variance for time Source DF SS MS F P treatmen 5 0.025549 0.005110 17.66 0.000 Error 12 0.003472 0.000289 Total 17 0.029021 Individual 95Based on Pooled StDev Level N Mean StDev -+——— +———+———+—– 1 3 0.18500 0.01735 (—-*—–) 2 3 0.17867 0.01041 (—–*—-) 3 3 0.21200 0.02088 (—-*—-) 4 3 0.26833 0.01102 (—-*—-) 5 3 0.25933 0.02401 (—–*—-) 6 3 0.26500 0.01389 (—-*—–) -+———+———+———+—– Pooled StDev = 0.01701 0.160 0.200 0.240 0.280 Hsu’s MCB (Multiple Comparisons with the Best) Family error rate = 0.0500 Critical value = 2.50 Intervals for level mean minus largest of other level means Level Lower Center Upper —–+———+———+———+– 1 -0.11809 -0.08333 0.00000 (——*—— ———-) 2 -0.12442 -0.08967 0.00000 (——*—————–) 3 -0.09109 -0.05633 0.00000 (——*———-) 4 -0.03142 0.00333 0.03809 (——*——) 5 -0.04375 -0.00900 0.02575 (——*——) 6 -0.03809 -0.00333 0.03142 (——*——) —–+———+———+———+– -0.100 -0.050 -0.000 0.050 Hsu’s MCB (Multiple Comparisons with the Best) Family error rate = 0.0500 Critical value = 2.50 Intervals for level mean minus smallest of other level means Level Lower Center Upper ———+———+———+——– 1 -0.02842 0.00633 0.04109 (——*——) 2 -0.04109 -0.00633 0.02842 (——*——) 3 -0.00142 0.03333 0.06809 (——*——) 4 0.00000 0.08967 0.12442 (—————–*——) 5 0.00000 0.08067 0.11542 (—————*——) 6 0.00000 0.08633 0.12109 (———— —-*——) ———+———+———+——– 0.000 0.050 0.100 The first and second treatments (auditory, 5 and 10 seconds) yield the lowest sample means. This implies that these are the best because they have the shortest sample mean response time. Solution to Question 4.5 – Trout experiment, continued. SAS Commands: data one; infile ’DV3-15.dat’; input Hem Sulf; Proc glm; class Sulf; model Hem=Sulf; means Sulf/TUKEY cldiff alpha=0.01; contrast ’1 - (2+3+4)/3’ Sulf 3 -1 -1 -1/divisor=3; run; (a) Compare the four treatment using Tukey’s method of pairwise comparisons and a 99% overall confidence level. SAS PRINTOUT The GLM Procedure Tukey’s Studentized Range (HSD) Test for Hem NOTE: This test controls the Type I experimentwise error rate. Alpha 0.01 4

Error Degrees of Freedom 36 Error Mean Square 1.568639 Critical Value of Studentized Range 4.72945 Minimum Significant Difference 1.8731 Comparisons significant at the 0.01 level are indicated by ***. Difference Between Means 0.3000 0.6400 2.1300 -0.3000 0.3400 1.8300 -0.6400 -0.3400 1.4900 -2.1300 -1.8300 -1.4900

Sulf Comparison 2 - 3 2 - 4 2 - 1 3 - 2 3 - 4 3 - 1 4 - 2 4 - 3 4 - 1 1 - 2 1 - 3 1 - 4

Simultaneous 99% Confidence Limits -1.5731 2.1731 -1.2331 2.5131 0.2569 4.0031 -2.1731 1.5731 -1.5331 2.2131 -0.0431 3.7031 -2.5131 1.2331 -2.2131 1.5331 -0.3831 3.3631 -4.0031 -0.2569 -3.7031 0.0431 -3.3631 0.3831

***

***

The only pair of treatment means that are significantly different at α = 0.01 are treatments 1 and 2. Therefore, deifference between adding 5 grams of Sulfamerazine and not adding any Sulfamerazine produces an average difference of between 2.1300 and 4.0031 gm of Hemoglobin per 100ml of brown trout blood. However, adding 10 or 15 grams of Sulfamerazine does not lead to a difference in Hemoglobin levels. (b) Our new confidence interval must have confidence level of 99% if the overall confidence level is to be at least 98%. To do the actual comparison we will need to use the contrast (1, −1/3, −1/3, −1/3) because it will give the estimate for the effect of treatment 1 minus the average effect of the other 3 treatments. Our least squares estimate for τ1 − (τ2 + τ3 + τ4 )/3 is y¯1. − 1/3(¯ y2. + y¯3. + y¯4. ) = 7.2 − (9.33 + 9.03 + 8.69) = −1.817. The standard deviation is given by

q P4 msE i=1

c2i ri

and there are 36 degrees of freedom for error.

From SAS, we know msE =1.568639, and 4 X c2 i

i=1

ri

Thus our standard error is

4

=

1 X 2 1 2 1 1 1 ci = (1 + (− )2 + (− )2 + (− )2 ) = 0.133. 10 i=1 10 3 3 3 v u 4 u X tmsE i=1

c2i . ri = 0.457

Also, t36,( 0.01 = 2.719. Thus our 99% Confidence Interval for τ1 − (τ2 + τ3 + τ4 )/3 is 2 ) −1.817 ± 2.719 × 0.457 = (−3.0604, −0.5730). Since 0 is not in this interval we may conclude thatthe average effect of no sulfermerazine is significantly different form the average effect of the other 3 levels of sulfermerazine at the α = 0.01 level, producing between 0.573 and 3.06 gm less hemoglobin per 100 ml of brown trout blood. Alternative Method: 5

SAS PRINTOUT

The GLM Procedure Dependent Variable: Hem Parameter 1 - (2+3+4)/3

Estimate -1.81666667

Standard Error 0.45733123

t Value -3.97

Pr > |t| 0.0003

Solution to Question 4.6 – Battery experiment, continued that Tukey’s method gives shorter confidence intervals than would either of the Bonferroni or Scheffé methods (for v = 4 and r = 4). In this experiment, v = 4,r = 4, n = v ∗ r = 16 and α = 0.05. And only the 6 pairwise comparisons τi − τs , i 6= s, are of interest. We can compare the critical coefficients for these methods, Bonferroni: Scheffé:

wB = t12, .025 = 3.15, 6 p wS = 3F3,12,.05 = 3.24,

Tukey:

wT =

√1 q4,12,.05 2

= 2.97.

Since wT is less than wB , which is less than wS , the Tukey intervals will be shorter than either of the Bonferroni or Scheffé intervals.

6

Solution to Question 4.7– Soap experiment, continued a) Suppose that the experimenter had been interested only in the contrast τ1 − 12 (τ2 +τ3 ), which compares the weight loss for the regular soap with the average weight loss for the other two soaps. Calculate a confidence interval for this single contrast. The SAS ESTIMATE statement gives Contrast T1-(T2+T3)/2

DF 1

Parameter T1-(T2+T3)/2

Estimate -2.38125000

Contrast SS 15.12093750 T for H0: Parameter=0 -14.00

F Value 195.93 Pr > |T| 0.0001

Pr > F 0.0001

Std Error of Estimate 0.17011944

Since t9,0.025 = 2.262, a 95% confidence interval for the contrast τ1 − 21 (τ2 + τ3 ) is −2.38125 ± 2.262 ∗ 0.17 = −2.38125 ± 0.385 = (−2.766, −1.9964) b. Test the hypothesis that the regular soap has the same average weight loss as the average of the other two soaps. Do this via your confidence interval in part (a) and also via (4.3.13) and (4.3.15). Since the 95% confidence interval does not include 0, we would reject, at level 0.05, the null hypothesis the regular soap has the same average weight loss as the other soaps. By (4.3.13), | − 2.38125/.17| = 14 which corresponds to a p-value of the t9 distribution of less than .0001. By (4.3.15), ssc/msE = 195.93 corresponds to a p-value from the F1,9 distribution of less than .0001 also. Hence we would reject the null hypothesis for any reasonable choice of significance level α. c. In Example 4.4.5 (page 89), Dunnett’s method was used for simultaneous 99% confidence intervals for two preplanned treatment-versus-control contrasts. Would either or both of the Bonferroni and Tukey methods have given shorter intervals? The critical coefficient for Bonferroni method for two preplanned treatment versus control intervals √ √ is wB = t9,(0.01)/4 = 3.69 and the critical coefficient for Tukey’s method is wT = q3,9,.01 / 2 = 5.43/ 2 = 3.84, as compared with Dunnett’s two-sided critical coefficient of wD2 = 3.63. Since the standard error does not depend on the method used, the confidence intervals for the treatment versus control contrasts sre shorter using Dunnett’s method. d. Which method would be the best if all pairwise differences are required? Calculate a set of simultaneous 99% confidence intervals for all of the pairwise differences. Why are the intervals longer than those in part (c)? There are three pairwise comparisons. The critical coefficient for Bonferroni’s method is t(9, .01/(2 ∗ 3)) = 3.95422 compared with Tukey’s method which has critical coefficient wT = 3.84 as in part (c). Consequently, Tukey’s method gives shorter intervals than Bonferroni’s method. Dunnett’s method is not considered since it is not valid for all pairwise comparisons. The Bonferroni intervals are longer than those in part (c) since the critical value takes of the number of intervals and t9,.01/(2∗3) > t9,.01/(2∗2) .

7

Solution to Question 4.8 – Trout experiment, continued SAS Command: data one; input Hem Sulf; lines; 1 6.7 1 7.8 1 5.5 : : : : 4 7.2 proc print; Proc glm; class Sulf; model Hem=Sulf; means Sulf/dunnett(’1’) cldiff ALPHA=0.01; contrast ’linear’ Sulf l -3 -1 1 3; contrast ’quadratic’ Sulf 1 -1 -1 1; run; (a) For the trout experiment in Exercise 15 of Chapter 3, test the hypothesis that the linear and quadratic trends in hemoglobin content of trout blood due to the amount of sulfamerazine added to the diet is negligible. State the overall significance level of your tests. There are two tests to be done. If we do each test at significance level of α = 0.025, the overall level will be at most 0.05. Linear Trend: The contrast coefficients for the test H0 : Linear Trend is negligible vs HA : Linear Trend is NOT negligible are given in Table A2 (page 702) as (−3, −1, 1, 3). Using the CONTRAST statement in SAS, we get a p-value of 0.0241, so that we would reject the null hypothesis at the α = 0.025 level. Therefore, we conclude that there is a linear trend in hemoglobin levels due to increasing sulfamerazine. The contrast sum of squares is ssc =

[4.17]2 [−3(7.2) − 9.33 + 9.03 + 3(8.69)]2 = = 8.6944 1 1 9 9 2.0 + 10 + 10 + 10 ] [ 10

SAS PRINTOUT

Dependent Variable: Hem Contrast DF Contrast SS linear 1 8.69445000 quadratic 1 15.25225000

Mean Square 8.69445000 15.25225000

F Value 5.54 9.72

Pr > F 0.0241 0.0036

Quadratic Trent: The contrast coefficients for the test H0 : Quadratic Trend is negligible vs HA : Quadratic Trend is NOT negligible are given in Table A2 as (1, −1, −1, 1). Using the CONTRAST statement in SAS, we get a p-value of 0.0036 and we reject H0 at the α = 0.025 level. Therefore, we conclude that there is a quadratic trend in hemoglobin levels due to invcreasing sulfamerazine. 8

(b) Regarding the absence of sulfamerazine in the diet as the control treatment, calculate simultaneous 99% confidence intervals for the three treatment-versus-control comparisons. Which method did you use and why? Using Dunnett’s method, since it is specifically designed for treatments vs control experiments and therefore gives the smallest confidence intervals, using SAS, we get SAS PRINTOUT Dunnett’s t Tests for Hem NOTE: This test controls the Type I experimentwise error for comparisons of all treatments against a control. Alpha 0.01 Error Degrees of Freedom 36 Error Mean Square 1.568639 Critical Value of Dunnett’s t 3.11170 Minimum Significant Difference 1.7429 Comparisons significant at the 0.01 level are indicated by ***. Difference Sul Between Simultaneous 99% Comparison Means Confidence Limits 2 - 1 2.1300 0.3871 3.8729 *** 3 - 1 1.8300 0.0871 3.5729 *** 4 - 1 1.4900 -0.2529 3.2329 (c) What is the overall confidence level of the intervals in part (b) together with those in Exercise 5? Is there a better strategy than using three different procedure for the three sets of intervals? Explain. The overall confidence level for the intervals in 5(a) was 99%, the interval in 5(b) had confidence level 99%, and the intervals in 8(b) had overall confidence level 0.99. Thus the combined overall confidence level is 100(1 − (0.01 + 0.01 + 0.01)0% = 97% . The method used to answer each question was the method that would yield the narrowest confidence intervals. For instance, Dunnett’s method gives the narrowest confidence intervals for treatment vs control experiments. Using a different method in each case would widen the confidence intervals unnecessarily. However, one might want to compare the Scheffe’s method since that method allows you to make an infinite number of statements with a specified overall confidence level, and the overall α would not need to be split into three parts.

9

Solution to Question 4.9– Battery experiment, continued a) Suppose the battery experiment of Section 2.5.2 (page 26) is to be repeated. The experiment involved four treatments, and the error standard deviation is estimated from that experiment to be about 48.66 minutes per dollar (minutes/dollar). Calculate a 90% upper confidence limit for the error variance σ2 . Since msE = 48.662 , then ssE=(16 − 4) × 48.662 = 28, 413.5. So, a 90% upper confidence bound for σ is 28, 413.55 ssE = = 4507.23 . σ2 ≤ 2 χ12,0.9 6.304 2

(b). How large should the sample sizes be in the new experiment if Tukey’s method of pairwise comparisons is to be used and it is desired to obtain a set of 95% simultaneous confidence intervals of length at most 100 minutes per dollar? We need r wT

2 1 msE = √ q4,4r−4,.05 r 2

s

2 4507.23 ≤ 50 . r

This uses the upper bound for σ 2 obtained in part (a), which is the largest likely value of msE for the repeat experiment. So we need 2 q4,4r−4,.05 ≤

r 11 29 20 24 25

4r − 4 40 120 76 92 96

502 r = 0.5547r . 4507.23

2 q4,4r−4,0.05 3.792 = 14.36 3.682 = 13.54 3.72 = 13.69 3.72 = 13.69 3.72 = 13.69

0.5546r 6.10 66.56 11.09 13.31 13.69

Action Increase r Decrease r Increase r Increase r Increase r

So we need approximately 25 observations per battery type, which is a total of 100 observations. c. How large should the sample sizes be in the new experiment if Scheffé’s method is to be used to obtain a set of 95% simultaneous confidence intervals for various contrasts and if the confidence interval for the duty contrast is to be of length at most 100 minutes per dollar? The confidence interval for the duty contrast needs to have at most 100 minutes per dollar, as part of a set of 95% simultaneous Scheffé intervals. The duty contrast (page 71) is τ1 + τ2 τ3 + τ4 − 2 2 with least squares estimate 0.5(y 1. + y 2. − y 3. − y 4. ). The corresponding variance is 0.25(4σ 2 /r), so using the upper bound for σ 2 from part (a) as the largest likely value of msE, we require p p 3F3,4r−4,.05 4507.23/r ≤ 50 ; that is F3,4r−4,.05 ≤ 0.1848r. r 26 16 15

4r − 4 100 60 56

F3,4r−4,.05 2.70 2.76 2.77

0.1848r 4.80 2.96 2.77

Action Decrease r Decrease r

About 15 observations per treatment would meet this requirement.

10

Solution to Question 4.10 Trout experiment, continued (a) Suppose the experiment were to be repeated. Suggest the largest likely value for the error mean square MSE. From section 3.46, a 95% upper confidence limit for σ 2 is given by σ≤

ssE . χ2df,0.95

From problem 3.15, we have ssE = 56.471. And χ236,0.95 = 23.26861. Thus, a good largest value for σ 2 56.471 is given by σ 2 = 23.26861 = 2.427. (b) How many observations should be taken in each treatment so that the length of each interval in a set of simultaneous 95% confidence intervals for pairwise comparison should be at most 2g per 100ml. Since the length of each interval must be less than 2 it follows that the msd for each interval must be less than 1. For pairwise comparisons, Tukey’s method or Bonferroni’s method will lead to the smaller numbert of observations for given msd. (i) Using Bonferonni’s Method The q radius of a Bonferroni interval for 6 pairwise comparisons and 4

α M SE( 2r ) , where r is the number of samples taken at each level treatment levels is given by tn−4, 12 and n is the total number of samples taken (n = 4r). To be conservative,qI will use the ”largest” value 4.8538 α of σ 2 supplied by part (a) of this problem. Thus we desire msd = tn−4, 12 ≤ 1. r

r 10 20 30 35 34

Degreeof f reedom 36 76 116 136 132

t4r−4, 0.05 12 2.7920 2.7091 2.6843 2.6774 2.6786

msd 1.8452 1.3346 1.0797 0.9971 1.0121

By using the Bonferroni’s method, we need 35 samples at each level in order to get the desired length. By using the Scheffe’s method, we need 39 samples at each level in order to get the desired length. q q 2M SE (ii) Using Tukey’s Method We need msd = µ,n−µ,α ≤ 1. For this particular problem, we 2 r q want msd = q4,4r−4,0.05 2.4269 ≤ 1. r

r 30 31 32 33

Degreeof f reedom 116 120 124 128

q4,4r−4,0.05 3.69 3.68 3.68 3.68

msd 1.05 1.03 1.01 0.998

By using the Tukey’s method, we need 33 samples at each level in order to get the desired length.

11

Solution to Question 4.11 – pedestrian light experiment (a) Suppose that you are planning to repeat the pedestrian light experiment at a pedestrian crossing of your choosing. Select v = 4 levels for the treatment factor ”number of pushes” including the level ”no pushes.” Give reasons for your selection. Level 1 2

Number of pushes No pushes 1 push

3

5 pushes

4

Continuous pushing

Reasons For control purposes This is how the device is supposed to work, you push it once and then the signal changes (in theory) 5 seems like a nice round number that the average person might push Continuous pushing until the light changes

(b) Using ”no pushes” as the control treatment, write down the formula for a set 95% simultaneous confidence intervals for treatment-versus-control contrasts. Dunnett’s method gives shorter intervals than Tukey’s method and Scheffé’s method for treatment versus control contrasts. It may or may not give shorter intervals than Bonferroni method for preplanned comparisons. The formula for a set of simultaneous confidence intervals for the 3 treatment versus control contrasts using Dunnett’s method and equal sample sizes is p y¯i. − y¯1. ± wD2 msE(2/r) for i = 2, 3, 4 n where wD2 is based on α = 0.05 and degrees of freedom v − 1 = 3 and n − v = 4(r − 1). (c) How many observation would you need to ensure your treatment-versus-control confidence intervals are of length less than 0.1 seconds? What value are you going to use for msE and why? One could use the upper 95% confidence limit for σ 2 generated by the data in Table 3.12, however, this data is for 0, 1, 2, and 4 pushes and not for 0, 1, 5, and ”continuous” pushes. Since we are assuming that all of our variances are equal and the two experiments share levels 1 and 2 and Table 3.12 is all we have to make any kind of estimate, we will proceed with the upper 95% confidence limit for σ 2 based upon the data in Table 3.12. Using SAS and the data from Table 3.12, we get SSE = 0.3059529. Also, we have χ232−4,0.95 ≈ 16.9279. Thus, our confidence upper bound for σ 2 is 0.01807. Now, if our intervals are to have length less than 0.1 seconds then they must have a minimum significant difference less than 0.05 seconds. From part (b), we know our msd is given by p msd = wD2 msE(2/r), 2 so we require wD2 ≥ 0.052 r/2msE = 0.06917r. If we use Table A.10 with α = 0.05 and degrees of freedom equal to ν1 = v − 1 = 3 and ν2 = 4(r − 1), we obtain the following:

r 10 100 80

4(r − 1) 36 396 316

2 wD2 2.472 = 6.1 2.352 = 5.52 2.352 = 5.52

0.06917r 0.69 6.92 5.53

Action Increase r Decrease r About right

Thus, r = 80 observations per treatment should generate confidence intervals for treatment versus control with length of less than 0.1. This we require a total of 320 observations. (d) If you had selected v=6 instead of v=4 would you have required more observation per treatment, or fewer, or the same. 12

Our new msd would be given by msd = wD2

p msE(2/r),

with α = 0.05 and degrees of freedom equal to ν1 = v − 1 = 5 and ν2 = 6(r − 1), and we still require 2 wD2 ≥ 0.052 r/2msE = 0.06917r. it appears that the value of wD2 increases as v − 1 increases and decreases as n − v increases making it difficult to judge what will happen. If we calculate the msd with r = 80, we obtain r 80

6(r − 1) 474

2 wD2 2.51

msd 0.053

Based upon this, it looks as though would need slightly more observations per treatment to achieve a msd of 0.05 if v = 6.

13



SOLUTIONS AVAILABLE IN THIS FILE 5.3 Margerine experiment 5.4a – Reaction time experiment 5.5 – Catalyst experiment – needs rewriting 5.6 – Bicycle experiment 5.7 – Dessert experiment 5.9 – Spaghetti sauce experiment

1

Solution to Question 5.3 (a). A plot of the standardized residuals vs. the predicted values indicates unequal variance. This is also apparent from the standard deviations given in the problem. Before proceeding, a variance-stabilizing transformation is needed. Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. Z | 2 + A A | | A | A B | A | A | B | A A | C B 0 +-------C--A------------------B-----------------A-| A | C A | A | A B | A A A | A | A | -2 + A A -+-----------+-----------+-----------+-----------+160 180 200 220 240 YPRED If we plot log(var)vs. log(mean), where var and mean are the variances and means of the responses for the four levels of treatment, we get

2

Plot of LN_VAR*LN_AVG. Legend: A = 1 obs, B = 2 obs, etc. LN_VAR | 4.5 + | A | A | 4.0 + | | | 3.5 + | A | | 3.0 + | A | | 2.5 + -+-----------+-----------+-----------+-----------+5.1 5.2 5.3 5.4 5.5 LN_AVG This plot has a slope of approximately 4, so by Eqn 5.5.3 of the text, the transformation 1/y is suggested. The analysis of the data was then carried out using 1/y as the response. We’ll now check the assumptions on the model again, using the transformed data: From the plot of standardized residuals vs. predicted, we see no apparent outliers, and they show no non-random pattern about zero. The run order of the experiment is not given, so independence of error cannot be checked with a plot of residuals vs. order. The variances appear similar from the plot, and this is checked by calculating the variances with SAS (see below). The normal probability plot looks linear, (apart from possibly one observation) indicating that the residuals approximately follow a normal distribution. The assumptions appear to be approximately satisfied.

3

Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. Z | 2 + A A | | A A | A | A A | A A | B A | B | A A 0 +-----A-------------B--------------------A---B----| A A A | B C | | B | B | A A A | B | -2 + A -+-----------+-----------+-----------+-----------+4.0 4.5 5.0 5.5 6.0 YPRED OBS 1 2 3 4

BRAND 1 2 3 4

AVGTIME .0056740 .0041961 .0058376 .0047940

VARTIME .000000032103 .000000023033 .000000021217 .000000037671

4

Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | 2 + A A | | AA | A | B | B | C | B | B 0 +-----------------------ABC-----------------------| AB | ACA | | B | B | AAA | AA | -2 + A -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 RANK FOR VARIABLE Z (b). Using the SAS ESTIMATE statement to calculate a 95% C.I. for the contrast t4 −(t1 +t2 +t3 )/3, we get −0.00044184 ± t(0.025,36) × (0.00006165) = (−0.000567, −0.000317) This does not include zero, so we conclude Ha , at the 0.05 level, that the average melting time of the margarines is less than that of butter. (c). Using Satterthwaite’s approximation for unequal variances, Here c1 = c2 = c3 = −1/3, c4 = 1 s1 , s2 , s3 , s4 are given in the problem r1 = r2 = r3 = r4 = 10.df = 12.6 from Eqn. (5.5.4) — (round down to 12 to be conservative) ˆ = 8.523 from Eqn (5.5.4) Var From (5.5.5), and getting the point estimate for the contrast from the SAS ESTIMATE statement, the 95% C.I. for the contrast τ4 −(τ1 +τ2 +τ3 )/3 is (7.08, 19.79). This does not include zero, so the conclusion is the same as in part (b). (d). The interpretation of the results using Satterthwaite’s approximation is simpler: the confidence interval for the butter-avg(margarine) contrast in part (c), indicates that butter takes longer to melt than the average of the three margarines (with confidence level 95%) by between 7 and 20 seconds, approximately. The conclusion using the transformed data, also suggests that butter takes longer to melt, but here the results are measured in seconds−1 . Thus, Satterthwaite’s approximation may be preferred. SAS Program options linesize=75; filename marg ’margerine.data’; data marg1; 5

infile marg; input brand time; proc glm; class brand; model time=brand; estimate ’btr_marg’ brand -1 -1 -1 3 / divisor=3; output out=marg2 predicted=ypred residual=z; proc standard std=1.0; var z; proc rank normal=blom; var z; ranks nscore; proc plot ; plot time*brand z*brand z*ypred z*nscore / vref=0 vpos=20 hpos=50; data; set marg1; proc sort; by brand; proc means noprint mean var; var time; by brand; output out=marg3 mean=avgtime var=vartime; data; set marg3; ln_avg=log(avgtime); ln_var=log(vartime); proc print; var brand avgtime vartime ln_avg ln_var; proc plot; plot ln_var*ln_avg/vpos=20 hpos=50; data marg10; set marg1; invtime=1/time; proc glm; class brand; model invtime=brand; estimate ’btr_marg’ brand -1 -1 -1 3 / divisor=3; output out=marg11 predicted=ypred residual=z; proc standard std=1.0; var z; proc rank normal=blom; var z; ranks nscore; proc plot ; plot invtime*brand z*brand z*ypred z*nscore / vref=0 vpos=20 hpos=50; data; set marg10; proc sort; by brand; proc means noprint mean var; var invtime; by brand; output out=marg13 mean=avgtime var=vartime; proc print; var brand avgtime vartime ; run;

6

Solution to Question 5.4 – reaction time experiment (a) A plot of the standardized residuals from the one-way analysis of variance model against the predicted values yît show no obvious outliers, since all the standardized residuals are within the ±2 bands. However, we may be a little concerned about the equality of variances. Plot of Z*PDY. Legend: A = 1 obs, B = 2 obs, etc. 2 + | A | A | A Z | A | | A A | | A 0 +----------------------------------------A----A---| A | A | A A | A A A | A | | | A -2 + -+-----------+-----------+-----------+-----------+0.175 0.200 0.225 0.250 0.275 PDY Obs 1 2 3 4 5 6

TC 1 2 3 4 5 6

AVRTM 0.18500 0.17867 0.21200 0.26833 0.25933 0.26500

VARRTM .000301000 .000108333 .000436000 .000121333 .000576333 .000193000

LN_AV -1.68740 -1.72223 -1.55117 -1.31553 -1.34964 -1.32803

LN_VAR -8.10840 -9.13030 -7.73787 -9.01697 -7.45882 -8.55282

The ratio of the maximum to the minimum variance (treatment combinations 5 to 1) is 5.32 which is above our “rule of thumb” threshhold. However, the plot of ln(s2 ) against ln(¯ y ) shows that no simple transformation will equalize the variances. This means that Satterthwaite’s approximations should be used. If the ususal analysis is done, the stated alpha levels will not nexessarily be very close to the actual values. Plot of LN_VAR*LN_AV. | | | -8 + | A | LN_VAR | | | -9 +

Legend: A = 1 obs, B = 2 obs, etc. A A

A

A 7

| A | --+--------+--------+--------+--------+--------+---1.8 -1.7 -1.6 -1.5 -1.4 -1.3 LN_AV The experimenters were particularly concerned about the fatigue of the subject. this can be checked by plotting the standardized residuals against the order of collection of the observations: Plot of Z*ORDER. Legend: A = 1 obs, B = 2 obs, etc. 2 + | A | A | A Z | A | | A A | | A 0 +-----------------------A-------A-----------------| A | A | A A | A A A | A | | | A -2 + --+---------+---------+---------+---------+-------0 5 10 15 20 ORDER There is no obvious pattern in this plot, and so the subject does not appear to have tired significantly (unless a learning effect cancelled out a tiring effect). A plot of the standardized residuals against normal scores is as follows. Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. 2 + | | | A | | A | | A Z | | A | | | | A A | | | | A 0 +------------------------+-B----------------------| A| | A | | A A | | A A A | | A | | | | | 8

| A | -2 + | -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 Rank for Variable Z There are three observations on the low end and one on the high end which might give cause for concern about the tails of the distribution. However, this effect may also be caused by the unequal variacnes, which appears to be the greater worry.

9

Solution to Question 5.5 – catalyst experiment i.) A residual plot of residuals vs. fitted values will help check the constant variance assumption. Here we see fairly random scatter about 0, suggesting constant variance. Plot of Z*YPRED.


Z | 2.077 + A | 1.384 + A A | 0.692 + B B A B | 0.000 +-----------------------------B---B-----| -0.692 + B B A B | -1.384 + A A | -2.077 + A | --+---+---+---+---+---+---+---+---+---+-5 6 7 8 9 10 11 12 13 14 YPRED ii.) To check more formally the constant variance assumption, we see the ratio max(s2i )/min(s2i )= 9 > 3, suggesting heteroscedacity. However, note there are only 2 observations per treatment and some of the variances are zero, so it is a little difficult to tell. OBS 1 2 3 4 5 6 7 8 9 10 11 12

TREAT 1 2 3 4 5 6 7 8 9 10 11 12

AVY 5 9 7 5 14 8 14 12 13 12 13 8

VARY 2 8 8 2 2 2 2 18 0 0 2 2

LN_AV 1.60944 2.19722 1.94591 1.60944 2.63906 2.07944 2.63906 2.48491 2.56495 2.48491 2.56495 2.07944

LN_VAR 0.69315 2.07944 2.07944 0.69315 0.69315 0.69315 0.69315 2.89037 . . 0.69315 0.69315

There are no apparent trends that suggests any transformations, but we will plot log(s2i ) vs. log(y i ) and determine the slope just to see. The scatterplot does not show a linear relationship which suggests that no simple transformation exists that will equalize the variances. If there is non-constant variance, we can make make formal inferences by using Satterthwaite’s approximation. Plot of LN_VAR*LN_AV.


LN_VAR | 3 + | | |

A

10

2 + A A | | | 1 + | B B A B | | 0 + | -+------------+------------+------------+ 1.5 2.0 2.5 3.0 LN_AV NOTE: 2 obs had missing values. iii) To check the independence assumption, we plot the residuals vs. order. The increasing trend clearly indicates the independence assumption is violated. Plot of Z*ORDER. Legend: A = 1 obs, B = 2 obs, etc. Z | 2.077 + A | 1.384 + A A | 0.692 + A AAA AAA | 0.000 +----A----A-A-A-------------------------| -0.692 + AAA A A A A | -1.384 + A A | -2.077 + A | --+---------+---------+---------+-------0 10 20 30 ORDER

11

Solution to Queuestion 5.6—- Bicycle experiment (a) Plot the standardized residuals against ybit , compare the sample variances, and evaluate equality of the error variances for the treatments. Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. 2 + | | A | A Z | | A | | A | A B B 0 +-------------------------------------------------| | B | A A | | | | | A -2 + A -+-----------+-----------+-----------+-----------+0 20 40 60 80 YPRED Notice that the spread of the observations decreases when the predicted values of crank-rate increase. So it is unlikely that the error variances for the different treatments (speeds)are equal. we can investigate this using our “rule of thumb”. The mean (AVRATE) and variance (VARRATE) for each treatmentis as follows OBS 1 2 3 4 5

TREAT 1 2 3 4 5

AVRATE 18.6667 31.0000 45.0000 60.3333 74.3333

VARRATE 12.3333 13.0000 3.0000 1.3333 1.3333

LN_AV 2.92674 3.43399 3.80666 4.09988 4.30856

LN_VAR 2.51231 2.56495 1.09861 0.28768 0.28768

Since the ratio of the maximum to the minimum variance is much greater than 3.0, a transformation of the data should be sought. (b) Choose the best transformation of the data of the form (5.6.3), and test the hypotheses that the linear and quadratic trends in crank rates due to the different speeds are negligible, using an overall significance level of 0.01. First, we plot log(s2i ) against logy i. .

12

Plot of LN_VAR*LN_AV.


3 + | | | A A | | 2 + | | LN_VAR | | | A 1 + | | | | A A | 0 + -+-----------+-----------+-----------+-----------+2.5 3.0 3.5 4.0 4.5 LN_AV The relationship is fairly linear, so either the SAS PROC REG procedure can be used to evaluate the slope, or a line can be drawn by hand and slope = rise/run. The SAS PROC REG output is as follows: Model: MODEL1 Dependent Variable: LN_VAR

Variable

DF

INTERCEP LN_AV

1 1

Parameter Estimates Parameter Standard T for H0: Estimate Error Parameter=0 8.478647 -1.918730

1.64116538 0.43792761

5.166 -4.381

Prob > |T| 0.0141 0.0220

Now the slope is q = −1.92, so y 1.96 which is approximately y 2 should be used for the data transformation. At this point we need to check that all of the assumptions on the model are fairly well satisfied with the transformed data. With the transformed (squared data) we have. OBS 1 2 3 4 5

TREAT 1 2 3 4 5

AVSRATE 356.67 969.67 2027.00 3641.00 5526.33

VARSRATE 16784.33 47796.33 24843.00 19200.00 29205.33

Max/Min =2.84, so the variances are now much closer, and within our rule of thumb. The plot shows that the spreads look much closer.

13

Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. Z | 2 + | | | A A 1 + A | B | A B | 0 +---A---------------------------------------------| | | B -1 + A | A | A | A -2 + | --+------------+------------+------------+--------0 2000 4000 6000 YPRED We should check normailty with the transformed data. We cannot check independence as we do not have the information about the order of the observations. Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | 2 + | | | A A 1 + A | B | A B | 0 +----------------------A--------------------------| | | B -1 + A | A | A | A -2 + | -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 RANK FOR VARIABLE Z The normal scores do not form a very straight line. Sometimes, a transformation can ruin the normality. If we were to proceed, ignoring worry about the normality, the two contrasts can be tested approximately using the CONTRAST statement with the transformed data:

14

General Linear Models Procedure Dependent Variable: RATE Contrast LINEAR QUADRATIC

DF 1 1

Contrast SS 50783234.1 2061057.5

Mean Square 50783234.1 2061057.5

F Value 1842.26 74.77

Pr > F 0.0001 0.0001

With an overall level of at most 0.01, we can do each test at level 0.005. Since the p-values for both the LINEAR and QUADRATIC contrasts are less than 0.005, the hypothesis of negligible linear trend is rejected, and so is the hypothesis of negligible quadratic trend. We conclude that the linear and quadratic trends in crank rates due to the different speeds are significantly different from zero. ( that the normality assumption is questionable. However, our p-values are nowhere near the significance levels, so the decision on whether to reject or not is fairly clear-cut). (c) Repeat part (b), using the untransformed data and Satterthwaite’s approximation for unequal variances. Returning to the untransformed data, as above we have OBS 1 2 3 4 5

TREAT 1 2 3 4 5

AVRATE 18.6667 31.0000 45.0000 60.3333 74.3333

VARRATE 12.3333 13.0000 3.0000 1.3333 1.3333

The sample sizes are all r = 3. For the linear trend, the contrast coefficient list is (-2 -1 0 1 2) and the least squares estimate is −2y 1. − y 2. + y 4. + 2y 5. = 140.6666 . Using (5.6.4), we obtain d Var(Σc î ) = iτ

X c2 i

r

s2i = 23

(Σc2i s2i /r)2 and df = P (c2 s2 /r) . 2 i i = 3.62 (r−1)

We round the degrees of freedom down to 3. Using the Bonferroni method, at overall level at least 99%, (i.e. each interval at 99.5%), a confidence interval for the linear trend is given by √ l.s.e. ± wB [ est var] =

√ 140.6667 ± 5.481[ 23]

=

[114.38, 166.95].

Since the interval does not contain 0, the hypothesis of no linear trend is rejected. For the quadratic trend, the contrast is given by (2 -1 -2 -1 2) with a least squares estimate of 4.6667. Using (5.6.4) again, we have d Var(Σc î ) = iτ

X c2 i

r

s2i = 27

(Σc2i s2i /r)2 and df = P (c2 s2 /r) , 2 i i = 4.72 (r−1)

—round down to 4. So the 99.5% confidence interval for the quadratic trend is given by √ 4.6667 ± 4.604[ 27] = [−19.26, 28.59] . 15

This does contain 0 and, therefore, the hypothesis of negligible quadratic trend is not rejected. (d)

Discuss the relative merits of the methods applied in parts (b) and (c).

In of (untransformed) crank rate, there is no quadratic trend, but in of squared (transformed) crank rate, there is one. Which result is more appealing to you? The transformed analysis is simpler to do, but we did question the normality assumption. The untransformed analysis is simpler to interpret and the normality assumption is perhaps not quite so questionable. It is possible that by squaring the data, we induced a quadratic trend. (For example, the values 1, 2, 3, 4 are evenly spaced but the values 1, 4, 9, 16 are not). Plot of Z*NSCORE.


2 + | | | Z |

A A

| A | | A | A D 0 +-------------------------------------------------| | B | B | | | | | A -2 + A -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 Rank for Variable Z

16

Solution to Question 5.7 – dessert experiment a) A plot of melting time vs treatment combination suggests that there are differences in the effects of the different treatments. for example, it appears cubes with treatment 2 melt more slowly than those with treatment 3 and 6. Plot of PMLT*TC. Legend: A = 1 obs, B = 2 obs, etc. 15 + | | | A | A | A 10 + A A A | B A | A A A A PMLT | A A | A | A 5 + | A | | | | 0 + --+------+------+------+------+------+--1 2 3 4 5 6 TC b) From the plot of the standardized residuals versus the predicted values, there are indications of non-constant variance as variability of the residuals alternate between “large” and “small”. The ratio max(s2i )/min(2i )=18.07 suggests non-constant variance also. OBS 1 2 3 4 5 6

TC 1 2 3 4 5 6

_TYPE_ 0 0 0 0 0 0

_FREQ_ 3 3 3 3 3 3

APMLT 8.9600 10.8800 7.1333 8.0967 10.1500 7.5633

VARPMLT 0.36480 1.24680 6.59613 4.39043 6.10930 0.81923

However, we cannot simply make transformations of the data to cure this type of heteroscedacity. The plot of log(s2i ) vs. log(y i ) shows no clear linear trend, so a simple transformation does not exist. A possible solution for making formal formal statistical inferences on the treatment effects, is to use Satterthwaite’s approximation. Note there are only 3 observations per treatment, which makes it difficult to tell whether we have outliers or non-constant variance. Plot of Z*YPRED. 5 + | | Z | | A | | A


A A A A

A 17

0 +----------A---A----A------------A------| A A A A | | A A | A | | -5 + -+------------+------------+------------+ 6 8 10 12 YPRED Plot of LN_VAR*LN_AV. Legend: A = 1 obs, B = 2 obs, etc. 2 + | A A | | A | | 1 + | | LN_VAR | | | A 0 + | A | | | | -1 + A -+------------+------------+------------+ 1.8 2.0 2.2 2.4 LN_AV A plot of residuals vs position shows a possible “V” pattern indicating independence assumptions may be violated. Plot of Z*POS. Legend: A = 1 obs, B = 2 obs, etc. 5 + | | Z | A | A A | A | A A A 0 +-----A--A---A--A-----------------------| A A A A | | A A | A | | -5 + 18

--+---------+---------+-----------------0 10 20 POS c) A comparison of tc2 to tc6 (that with the largest mean and that with the smallest) has standard error p S. E. = 1.247/3 + .819/3 = .829, and Satterthwaite’s method give approximate degrees of freedom df =

.8292 = 3.835 . (1.247/3)2 + (.819/3)2

If we round down to 3√degrees of freedom, we have q6,3,0.05 = 8.04 and y¯2 − y¯6 = 3.32. Since msd=(8.04/ 2) ∗ .829 = 4.713 > 3.32, treatments 2 and 6 are statistically insignificant if we consider the 15 possible pairs with 95% simultaneous confidence level, i.e. the confidence interval is (-1.393, 8.033). Similar calculations for treatments 2 and 3 gives msd=9.19 and y¯2 − y¯3 = 3.747. d.) Based only on the above two comparisons, it seem there is no statistically significant difference between the treatments. Based on the sample statistics however, perhaps mixes with 1/2 cup sugar (treatments 2 and 5) will stay frozen longest. Another possible factor to consider is the proportion of orange juice to water. We might also be concerned in the analysis to have small variance as well as large mean.

19

Solution to Question 5.9 – spaghetti sauce experiment a) Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | A 2 + | A | A A A | A 0 +--------B--A---------------------------| AA B | AA A A | A -2 + | --+--------+--------+--------+--------+-0 20 40 60 80 YPRED The residual plot shows larger spread for some treatment combinations than others indicating possibly unequal variance. The ratio max(vari )/min(vari )=49/2.33=21 s this conclusion. Plot of Z*ORDER. Legend: A = 1 obs, B = 2 obs, etc. 4 + | | | | | A 2 + | | A Z | AA | A | A A 0 +----------A---A------------------------| A A A | A | A AAA | A | -2 + --+---------+---------+-----------------0 10 20 ORDER This plot of residuals vs. order is hard to interpret. Though there may be a slight increasing pattern for observations 6-14 and then a decreasing pattern for observations 15-18, without access to the original experimenters’ reports it is hard to know whetehr there is a reason. We proceed with the analysis as though thiese trends occured by pure chance, but if they didn’t our analysis will not be correct.

20

Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | | | A 2 + | | A | A A | A | A A 0 +--------------------AA-----------------| B A | A | A A B | A | -2 + -+------------------+------------------+-2 0 2 Rank for Variable Z The normal prob. plot shows a slight departure from linearity at the bottom end. So we should interpret our confidence levels and significance levels with a little caution in case normality of the error tems is not quite satisfied. b) The GLM Procedure Dependent Variable: WT Source TC Error Corrected Total Obs 1 2 3 4 5 6

TRTMT 1 2 3 4 5 6

DF 5 12 17

Sum of Squares 7976.444444 194.000000 8170.444444

_TYPE_ 0 0 0 0 0 0

_FREQ_ 3 3 3 3 3 3

Mean Square 1595.288889 16.166667

MNWT 58.0000 65.6667 17.0000 23.0000 15.3333 15.6667

F Value 98.68

Pr > F <.0001

VARWT 49.0000 8.3333 19.0000 9.0000 9.3333 2.3333

We will calculate Bonferroni Intervals. As an example the CI for (τ1 − τ2 ) is calculated as follows: sd(contrast)=sqrt(49/3+8.3/3)=4.370355 df=(3 − 1) ∗ (49 + 8.3)2 /(492 + 8.32 ) = 2.658653 2 and w=t(2,.03/6)=t(2,.005)=9.925 So the CI for (τ1 − τ2 ) is (58-65.677) ± 5.841*4.37 = (-33.2, 17.87). The 94% simultaneous CIs are: Contrast CI ----------------------------------

21

tau1-tau2 tau3-tau4 tau5-tau6 tau1-tau5 tau1-tau3 tau3-tau5

(-51.05, (-23.84, (-19.91, ( -1.1, ( -6.25, (-16.28,

35.72) 11.84) 19.24) 86.43) 88.25) 19.62)

c) The plot of Residuals vs WT does not offer any clear suggestions of what type of transformation to use. A plot of log(VARi ) vs. log(aveWT) and corresponding regression also does not indicate any of the transformations in (5.6.3) are suitable sincve no straight line is apparent. Plot of LN_VAR*LN_AV. Legend: A = 1 obs, B = 2 obs, etc. LN_VAR | 4 + A | | | 3 + A | | | A A 2 + A | | | 1 + | A | | 0 + | --+--------+--------+--------+--------+-2.5 3.0 3.5 4.0 4.5 LN_AV Using a regression routine from SAS, the best fitting straight line has q=0.72 which indicates that √ 1/ y would be a reasonable transformation. However, we already know that this won’t be very successful. Variable Intercept LN_AV

DF 1 1

Parameter Estimate 0.00241 0.72201

Standard Error 2.21152 0.66239

22

t Value 0.00 1.09

Pr > |t| 0.9992 0.3370

Design and Analysis of Experiments Angela Dean and Daniel Voss Solutions to Chapter 7 March 1st 2003 Solutions available in this file c


Question 7.6 — Weathering experiment, continued Question 7.8 — Paper towel strength experiment Question 7.9 — Rocket experiment Question 7.12 — Washing power experiment Question 7.14 — Popcorn-robust experiment Question 7.15 — Steel.bar

Solution to Question 7.6 Weathering experiment, continued First, we need to recreate the anlaysis of variance table. The sums of squares for the treatment factors can be calculated by hand using the rules in Chapter 7. Alternatively, the averages can be entered into a computer p[ackage, such as SAS PROC GLM and the resulting sums of squares and mean squares multiplied by r. In this experiment, r = 2. The error mean square can only be obtained from the raw data. According to the information in the paper, msE = 6.598, with 36 degrees of freedom. In the analysis of variance table shown below, the mean squares are those obtained from the computer program using the averages. The F-value is calculated from the corrected mean square (multiplied by r = 2) and divided by the error mean square. The GLM Procedure Dependent Variable: Y Source E A D F E*A E*D E*F A*D A*F D*F E*A*D E*A*F E*D*F A*D*F E*A*D*F

DF 1 2 1 2 2 1 2 2 4 2 2 4 2 4 4

Type III SS 260.82250 519.57389 1.03361 51423.42389 13.56167 1.03361 40.68500 6.06056 198.21778 240.67722 8.67722 26.90333 149.92722 43.53111 6.73444

Mean Square 260.82250 259.78694 1.03361 25711.71194 6.78083 1.03361 20.34250 3.03028 49.55444 120.33861 4.33861 6.72583 74.96361 10.88278 1.68361

F value 79.061 78.747 0.313 7793.790 2.055 0.313 6.166 0.918 15.021 36.477 1.315 2.039 22.723 3.299 0.510

If we test each hypothesis of negligible main effect or interaction at individual level α∗ = 0.001, the overall level will be at most α = 0.015. From the F -tables we have F1,36,0.001 = 12.9 F2,36,0.001 = 8.47

F4,36,0.001 = 5.88

The reults of the hypothesis tests are that, at overall level at most α = 0.015, the interactions EDF , AF are non-negligble. Averaged over the the third factor, the interactions DF and possibly EF are non-negligible. Averaged over these interactions, there is a significant difference in the levels of E, A and F . In order to be able to draw conclusions about the factors, it would be necessary to examine interaction plots. b). To test the null hypothesis that the interaction F A is negligible, against the alternative hypothesis that it is not negligible, we obtain the formula from rules 1-8 of Section 7.3. First, write the factors in alphabetical order (A, D, E, F ). AF has (a − 1)(f − 1) = af − a − f + 1 degrees of freedom. So the corresponding sum of squares is ssFA = edr

f a X X

(¯ yi..l. − y¯i.... − y¯...l. + y¯..... )2 .

i=1 l=1

Using the cell means model or the equivalent four-way complete model, the error sum of squares is ssE = (yijklt − y¯ijkl. )2 . We reject the null hypothesis of no AF interaction if ssFA/4 > F4,36,α . ssE/36

c) In order to answer this question, we need to examone the interaction plots. We see that there is a clear difference in the fabrics and the interaction, in comparison, is relatively small (– see also the analysis of variance table). Consequently, it may well be of interest to calculate confidence intervals for the differences in the fabrics averaged over the various weather conditions to which the fabric may be exposed. A formula for such confidence intervals using Bonferroni method of mutiple comparisons is p y¯...1. − y¯...2. ± t36,0.01/6 (2/12)msE Plot of AVY*ED. Symbol is value of F. 50 + | | | | | 2 0 + 2 | 2 2 | AVY | | | -50 + 1 1 1 1 | | | | 3 | 3 3 -100 + 3 --+------------+------------+------------+--------1 2 3 4 ED Plot of AVY*E. Symbol is value of F. 50 + | | | | | 0 + 2 | 2 | AVY | | | -50 + 1 1 | | | | | 3 -100 + 3 --+--------------------------------+--------------1 2 E

Plot of AVY*D. Symbol is value of F. 50 + | | | | | 0 + 2 | 2 | AVY | | | -50 + 1 1 | | | | | 3 -100 + 3 --+--------------------------------+--------------1 2 D Plot of AVY*A.

Symbol is value of F.

0 +2 2 2 | | | AVY | | | | |1 -50 + 1 | 1 | | | | |3 | | 3 -100 + 3 -+-----------------------+-----------------------+1 2 3 A d) We can see from the D ∗F interaction plot, that averaging over the exposure fabrics 2 and 3 reacted differently to direction of the pull. One had larger breaking strength in direction 1 while the oterh had the larger strength in direction 2. From the ED*F interaction plot, we can see this difference was more marked at exposure 1 than exposure 2. these differences are significantly larger than the experimental error. Therefore, it makes no sense to say that direction has no effect on the breaking strength – although one can argue that the effect is small as compared with the fabric differences.

Solution to Question 7.8 — Paper towel strength experiment a) The assumed model is

Yijkt = µ + αi + βj + γk + (αβ)ij + ijkt ijkt ∼ N (0, σ 2 ) ijkt ’s mutually independent t = 1, 2, 3; i = 1, 2; j = 1, 2; k = 1, 2 .

(1)

b) All factors have two levels, so the experimenters were problably interested in γ1 − γ2 (since C is not involved in any interactions), plus the interaction contrast (αβ)11 − (αβ)12 − (αβ)21 + (αβ)22 . If the interaction appears to be negligible, then they would be interested in the two main effects contrasts α1∗ − α2∗ and β1∗ − β2∗ , where αi∗ = αi + (αβ)i. and βj∗ = βj + (αβ).j . On the other hand, if there is a non-negligible interaction, then the experimenters may be interested in pairwise differences in the combinations of the different levels of A and B averaged over C. Thus we might plan to look at 10 contrasts in total. c) The AC and BC interaction plots are shown below. Neither of these plots exhibits interaction and the assumptions of no AC and BC interactions look reasonable. The AB interaction plot (not shown) suggests that the AB interaction is also neglgible.

Obs 1 2 3 4

AMOUNT 1 1 2 2

The SAS System LIQUID avy 1 3302.33 2 2903.68 1 3035.00 2 2677.10

vary 465084.23 139966.03 361395.28 248273.80

Plot of avy*AMOUNT. Symbol is value of LIQUID. avy | 3400 + | | 1 | 3200 + | | | 1 3000 + | | 2 | 2800 + | | 2 | --+--------------------------------+--------------1 2 AMOUNT

Obs 1 2 3 4

BRAND 1 1 2 2

LIQUID 1 2 1 2

avy 3635.45 3161.20 2701.88 2419.58

vary 277207.81 8906.13 69223.94 80140.53

Plot of avy*BRAND. Symbol is value of LIQUID. avy | | | 1 3500 + | | | 2 3000 + | | 1 | 2500 + | 2 | | 2000 + --+--------------------------------+--------------1 2 BRAND d) Some residual plots are shown below. The plot of the standardized residulas against order suggests some possible non-independence, but this is prmarily caused by the two large residuals in the center of the plot. An alternative explanation is that these two large values belong to the larger predicted responses, and so it is possible that the variance is increasing as the mean increases. Both these values belong to the low level of liquid and Brand 1. There may be a transformation that would help to equalize the variances. The normality appears to be well-satisfied. The SAS System Plot of Z*ORDER. Legend: A = 1 obs, B = 2 obs, etc. | | A 2 + A | | A Z | A | A A | A A A 0 + A A A A | A A A A | A A A A | | A | A A -2 + -+-------+-------+-------+-------+-------+-------+0 4 8 12 16 20 24 ORDER

Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. | A 2 + A | | A Z | A | A A | A A A 0 + A A A A | A A B | A A A A | | A | A A -2 + -+-----------+-----------+-----------+-----------+2000 2500 3000 3500 4000 YPRED Plot of Z*AMOUNT. Legend: A = 1 obs, B = 2 obs, etc. | A 2 + A | | A Z | A | A A | B A 0 + A C | B B | B B | | A | A A -2 + --+--------------------------------+--------------1 2 AMOUNT Plot of Z*BRAND. Legend: A = 1 obs, B = 2 obs, etc. | A 2 + A | | A Z | A | B | C 0 + A C | C A | B B | | A | A A -2 + --+--------------------------------+--------------1 2 BRAND

Plot of Z*LIQUID. Legend: A = 1 obs, B = 2 obs, etc. | | A 2 + A | | A Z | A | A A | A B 0 + B B | D | B B | | A | B -2 + --+--------------------------------+--------------1 2 LIQUID Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. | | | | A 2 + | A | | | | A Z | | A | | A A | | A AA 0 +-----------------------A+AAA---------------------| A AAA | | AA A A | | | | A | | A A | -2 + | -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 Rank for Variable Z e)

The variances are certainly not equal: Obs

TC

AVY

VARY

1 2 3 4 5 6 7 8

111 112 121 122 211 212 221 222

3817.80 3200.57 2786.87 2606.80 3453.10 3121.83 2616.90 2232.37

277021.33 4035.10 88571.60 81460.84 316243.63 13581.02 62821.75 13740.25

Without any transformation, we cannot rely on the stated p-values, significance levels and confidence levels. The analysis of variance table gives

The GLM Procedure Dependent Variable: Y Source AMOUNT BRAND LIQUID AMOUNT*BRAND Error Corrected Total

DF 1 1 1 1 19 23

Sum of Squares 365930.510 4209358.800 858551.854 3822.850 1862905.935 7300569.950

Mean Square 365930.510 4209358.800 858551.854 3822.850 98047.681

F Value 3.73 42.93 8.76 0.04

Pr> F 0.0684 <.0001 0.0081 0.8456

Thus, even if the p-values are considerably inaccurate, it is clear that there is no substantial AB interaction and that there is a substantial main effect of “Brand”. If an overall significance level of ≥ α = 0.04 is selected, so that each test is done at level 0.01, the two levels of liquid type appear to give significantly different strengths, but the amounts do not. However, if the p-values are corrected for the different variances, a different conclusion may be reached. We may note that the variances of the data at each of the two levels of each factor are not too different: Obs 1 2

AMOUNT 1 2

Obs 1 2

BRAND 1 2

Obs 1 2

LIQUID 1 2

AVY 3103.01 2856.05 AVY 3398.33 2560.73 AVY 3168.67 2790.39

VARY 318365.16 312056.60 VARY 191391.72 89627.48 VARY 395163.54 190474.47

Consequently, any adjustment that we make using Satterthwaite’s approximation for main effect comparisons will make little difference. If a transformation was to be tried, the most sensible transformation √ would be 1/ y and this would be somewhat difficult to interpret. f) We suggested a maximum of 10 contrasts of interest in part b). If we select an overall confidence level of at least 90%, then we can use Bonferroni’s method and calulate each interval at an individual 99% confidence level. It turns out that there is no substantial AB interaction, and only the three main effect contrasts would be of interest. Parameter AMOUNT DIFF BRAND DIFF LIQUID DIFF

Estimate -246.958333 -837.591667 -378.275000

Standard Error 127.833017 127.833017 127.833017

t Value -1.93 -6.55 -2.96

Pr> |t| 0.0684 <.0001 0.0081

Since t19,0.005 = 2.861, we have t19,0.005 ∗ std. err. = 2.861 ∗ 127.833 = 365.73 and α1∗ − α2∗ ∈ (−246.96 ± 365.73) = (−612.69, 118.77) β1∗ − β2∗ ∈ (−837.59 ± 365.73) = (−1203.32, −471.86) γ1 − γ2 ∈ (−378.28 ± 365.73) = (−744.01, −12.55) . Thus, at an overall 90% confidence level, the brands differ in strength by between 472 and 1203 units with brand 2 being the stronger. The amount of liquid absorbed seemed not to affect the strength substantially, and the type of liquid had a marginal effect with the towels being able to withstand water better than beer!

Solution to Question 7.9 — Rocket experiment a) State a reasonable model for this experiment, including any assumptions on the error term. Since there is only one observation per treatment combination and since the experimenters were willing to assume that the 3-factor and 4-factor interactions were negligible, a suitable model would be

Yijkl = µ + αi + βj + γk + δl + (αβ)ij + (αγ)ik + (αδ)il + (βγ)jk + (βδ)jl + (γδ)kl + ijkl , ijkl ∼ N (0, σ 2 ) , ijkl ’s mutually independent , i = 1, . . . , a; j = 1, . . . , b; k = 1, . . . , c; l = 1, . . . , d. where αi , βj , γk , δl are the effects (positive or negative) on the response of factors A, B, C, D at levels i, j, k, l, respectively, (αβ)ij , (αγ)ik , (αδ)il , (βγ)jk , (βδ)jl , (γδ)kl are the additional effects of the pairs of factors together at the specified levels. b) How would you check the assumptions on your model? Plot the standardized residuals against (i) the fitted values yîjkl and against the levels of each of the four factors to look for equal variance and outliers, (ii) the order of collection to look for independence of the error variables (–this is not possible in this experiment since we do not have the order), (iii) normal scores to check for normality of the error variables. c) Calculate an analysis of variance table and test any relevant hypotheses, stating your choice of the overall level of significance and your conclusions. There are 10 hypotheses to be tested. Suppose that we select an overall significance level of at most 0.1 so that we test the significance of each effect at level 0.01. We test the interactions first. If the interactions containing a given factor are zero, we will then test whether the corresponding main effect is zero. The ANOVA table is below. General Linear Models Procedure Dependent Variable: Y Source Model Error Corrected Total

DF 18 13 31

Sum of Squares 673.212725 9.648825 682.861550

Mean Square 37.400707 0.742217

F Value 50.39

Pr > F 0.0001

Source A B C D A*B A*C A*D B*C B*D C*D

DF 1 1 1 3 1 1 3 1 3 3

Type III SS 0.021012 0.405000 0.610513 665.516025 0.556512 0.011250 1.064613 0.427813 1.435825 3.164163

Mean Square 0.021012 0.405000 0.610513 221.838675 0.556512 0.011250 0.354871 0.427813 0.478608 1.054721

F Value 0.03 0.55 0.82 298.89 0.75 0.02 0.48 0.58 0.64 1.42

Pr > F 0.8690 0.4732 0.3809 0.0001 0.4022 0.9039 0.7030 0.4613 0.5999 0.2814

Only the main effect of D appears significantly different from zero at overall significance level 0.1.

d) Levels 0 and 1 of factor D represent temperatures −75◦ F and 170◦ F, respectively at sea level. Level 2 of D represents −75◦ F at 35,000 feet. Suppose the experimenters had been interested in two preplanned contrasts. The first compares the effects of levels 0 and 1 of D, and the second compares the effects the levels 0 and 2 of D. Using an overall level of at least 98%, give a set of simultaneous confidence intervals for these two contrasts. Assume that we are intersted in just these two pre-planned contrasts. For an overall confidence level of at least 98%, we calculate individual 99% confidence intervals for the two contrasts. Now, t13,0.005 = 3.012, and using the SAS output below, the two confidence intervals are: δ1 − δ0 : δ2 − δ0 :

− 10.18 ± 3.012(0.431)

=

(−11.48, −8.88)

− 0.99 ± 3.012(0.431)

=

(−2.29, 0.31)

At an overall confidence level of 98%, the duration of thrust is between 8.8 and 11.4 seconds longer when level 0 of D is used than when level 1 is used. There does not appear to be a significant difference between the effects of levels 0 and 2.

Parameter d1-d0 d2-d0

Estimate

T for H0: Parameter=0

-10.1762500 -0.9900000

-23.62 -2.30

Pr > |T|

0.0001 0.0388

Std Error of Estimate 0.43076017 0.43076017

e) Test the hypotheses that each contrast identified in part (d) is negligible. Be explicit about which method you are using and your choice of the overall level of significance. Using Bonferroni method at overall level at most 0.02 for the two tests, we reject H0 if √ |lse/ standard error| > t13,0.02/4 = 3.012 . The least squares estimate and standard error are given on the SAS output in the solution to part (d). For H0 : {δ1 − δ0 = 0}, the value of the test statistic is 23.62 and we reject H0 at overall level ≤ 0.02 and conclude that there is a significant difference in thrust duration between levels 0 and 1 of temperature/altitude. For H0 : {δ2 − δ0 = 0}, the value of the test statistic is 2.30, so there is not sufficient evidence to conclude a difference between levels 0 and 2 of temperature/altitude. Alternatively, we can draw the same conclusions by observing that 0 is not in the first confidence interval in part (d), but is in the second interval. f) If the contrasts in part (d) had not been preplanned, would your answer to (d) have been different? If so, give the new calculations. If these contrasts had not been pre-planned, we would need to use Scheffé confidence intervals. If only contrasts for main effects and interactions in the model are to be examined, then the numerator degrees of freedom would be the 18. p p model degrees of freedom, Now v − 1=18, so (v − 1)F0.02,v−1,13 = 18(3.15) = 7.53, so the confidence intervals are δ1 − δ0 : δ2 − δ0 :

− 10.18 ± 7.53(0.431) − 0.99 ± 7.53(0.431)

= (−13.43, −6.93) = (−4.24, 2.26)

The confidence intervals are much wider than those in part (d), so they give much less precise information. However, we still see that level 0 of D gives longer thrust duration than level 0 (—at most 14 seconds longer).

g) Although it may not be of great interest in this particular experiment, draw an interaction plot for the CD interaction and explain what it shows. Plot of avy*D. Symbol is value of C. 25 + | | | | 1 | 0 1 20 + 0 | | avy | | | 15 + | | | 0,1 | 0 | 1 10 + --+------------+------------+------------+--------0 1 2 3 D The plot shows almost parallel lines indicating almost no interaction between C and D. It also shows that levels 0 and 2 of D give a longer duration of thrust than levels 1 and 3 no matter the level of C. h) If the experimenters had included the 3-factor and 4-factor interactions in the model, how could they have decided upon the important main effects and interactions? If these interactions are included in the model, there are no degrees of freedom to estimate the error variance. A normal probability plot of the normalized contrasts or Voss-Wang method could be used to identify important effects. Those effects lying off the straight line are likely to be significantly different from zero. Note that factor D has 3 degrees of freedom, so the plot should include estimates for three orthogonal contrasts in the levels of D. (Trend contrasts would not make sense here, since D does not have 4 equally spaced levels). Similarly, each interaction involving D has three degrees of freedom.

Solution to Question 7.12— washing power experiment a) The contrast coefficients are given by the following columns:

Trtmt lA qA lC qC lAlC lAqC qAlC qAqC 111 112 113 121 122 123 131 132 133 211 212 213 221 222 223 231 232 233 311 312 313 321 322 323 331 332 333

-1 -1 -1 -1 -1 -1 -1 -1 -1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 -2 -2 -2 -2 -2 -2 -2 -2 -2 1 1 1 1 1 1 1 1 1

-1 0 1 -1 0 1 -1 0 1 -1 0 1 -1 0 1 -1 0 1 -1 0 1 -1 0 1 -1 0 1

1 -2 1 1 -2 1 1 -2 1 1 -2 1 1 -2 1 1 -2 1 1 -2 1 1 -2 1 1 -2 1

1 0 -1 1 0 -1 1 0 -1 0 0 0 0 0 0 0 0 0 -1 0 1 -1 0 1 -1 0 1

-1 2 -1 -1 2 -1 -1 2 -1 0 0 0 0 0 0 0 0 0 1 -2 1 1 -2 1 1 -2 1

-1 0 1 -1 0 1 -1 0 1 2 0 -2 2 0 -2 2 0 -2 -1 0 1 -1 0 1 -1 0 1

1 -2 1 1 -2 1 1 -2 1 -2 4 -2 -2 4 -2 -2 4 -2 1 -2 1 1 -2 1 1 -2 1

b) Using the formula for the variance of the contrasts √ in √ of coefficients for τi jk (Rule 11,sec.7.3), the divisors for Linear A and Quadratic A are 18 and 54. The LSE for the Linear A contrast (without divisors) is y3.. − y1.. = 104.796. and LSE for the Quadratic A contrast (without divisors) is [y1.. + y3.. ] − 2y2.. = −78.2. So the LSEs of the normalized contrasts are √ LSE(lA) = 104.796/( 18) = 24.70065 and

√ LSE(qA) = −78.2/( 54) = −10.64167 .

c) The contrasts [−1, 0, 1] and [.5, −1, .5] for the three levels of factor B are orthogonal. These should be interpreted as: the contrast which compares the effect of the .2% detergent with that of the .05% detergent; and the contrast which compares the effect of the .1% detergent with the average effects of the other two detergents. It probably makes more sense to take the two contrasts [-1, 1, 0] and [.5, .5, -1]. The sums of squares connected with the two orthogonal contrasts should add to ssB.

d) The least squares estimates of the normalized contrasts √ are listed under CONS. The contrast estimates with divisor 27 are listed under EST1. NORM lists 27/ 18 etc., so that CONS = EST1*NORM.

Obs

_NAME_ 1 2 3 4 5 6

lA lB lC qA qB qC

EST1

NORM

3.88148 4.60370 2.15926 -2.89630 -2.35185 -1.61852

6.36396 6.36396 6.36396 3.67423 3.67423 3.67423

CONS 24.7016 29.2978 13.7414 -10.6417 -8.6413 -5.9468

NSCORE 1.53898 1.98075 1.28155 -1.98075 -1.53898 -1.08892

. . .

Plot of CONS*NSCORE.


30 + A | | | A | | 20 + | | CONS | | A | 10 + | | | | | AA A A 0 + AAAAA A | A AAAAA | AA | A | A A | A -10 +A -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 Rank for Variable CONS The plot indicates there are several effects that are are non-neglible, in particular, the three main effects and quadratic for A.

e)

lB lA lC qA qB lAqB qC lAlB lAqC lAlCqB qAqB qAqC lBqAqC lAlC lBqA lCqA lAqBqC qAqBqC lAlBqC lBlC lCqAqB lBlCqA lBqC lCqB qBqC lAlBlC

ss 858.36055556 610.16888889 188.82722222 113.24518519 74.67129630 38.23361111 35.36462963 26.70083333 9.81777778 7.04166667 5.92675926 3.92925926 2.96055556 2.61333333 2.50694444 1.86777778 1.50222222 1.09796296 0.80666667 0.80083333 0.37555556 0.16666667 0.14694444 0.04694444 0.01564815 0.00500000

msq estimate msd .95 lb 0.9194231 29.29779097 9.396882 19.900909 0.9194231 24.70159689 9.396882 15.304715 0.9194231 13.74144178 9.396882 4.344560 0.9194231 -10.64167210 9.396882 -20.038554 0.9194231 -8.64125548 9.396882 -18.038138 0.9194231 -6.18333333 9.396882 -15.580215 0.9194231 -5.94681676 9.396882 -15.343699 0.9194231 -5.16728491 9.396882 -14.564167 0.9194231 -3.13333333 9.396882 -12.530215 0.9194231 -2.65361389 9.396882 -12.050496 0.9194231 2.43449364 9.396882 -6.962388 0.9194231 -1.98223592 9.396882 -11.379118 0.9194231 -1.72062650 9.396882 -11.117509 0.9461325 1.61658075 9.532395 -7.915815 0.9543162 -1.58333333 9.573533 -11.156866 1.0034829 1.36666667 9.817051 -8.450385 1.0316026 -1.22565175 9.953648 -11.17930 1.0626994 1.04783728 10.102557 -9.054719 1.0851068 -0.89814624 10.208509 -11.106655 1.0855556 -0.89489292 10.210620 -11.105513 1.1182692 0.61282588 10.363328 -9.750503 1.1343376 0.40824829 10.437518 -10.029270 1.1358547 -0.38333333 10.444495 -10.827829 1.1435470 0.21666667 10.479802 -10.263136 1.1459544 0.12509256 10.490828 -10.365735 1.1467735 -0.07071068 10.494576 -10.565287

.95ub 38.6946730 34.0984789 23.1383238 -1.2447901 0.7556266 3.2135487 3.4500653 4.2295971 6.2635487 6.7432682 11.8313757 7.4146461 7.6762556 11.1489762 7.9901995 11.1837181 8.727996 11.1503940 9.3103630 9.3157268 10.9761543 10.8457664 10.0611621 10.6964689 10.6159201 10.4238654

The Voss-Wang method provides results that are consistent with the Normal prob. plot. The CIs for effects B,A, C and Quadratic A don’t cover 0 and are therefore simultaneously statistically signficant.

Solution to Question 7.14 — Popcorn-robust experiment a) Analyze the experiment as a mixed array, using a three-way complete model. Draw an ABT interaction plot, similar to that of Figure 7.8, page 220. If the goal of the experiment is to find brand–oil combinations that give a high percentage of edible kernels and that are not too sensitive to the popping time, what recommendations would you make? Using a three-way complete model, the analysis of variance table is The GLM Procedure Dependent Variable: POP Source BRAND OIL TIME BRAND*OIL BRAND*TIME OIL*TIME BRAND*OIL*TIME Error Corrected Total

DF 2 1 2 2 4 2 4 18 35

Sum of Squares 562.666667 79.506944 708.791667 694.055556 1796.291667 95.263889 187.986111 1108.125000 5232.687500

Mean Square 281.333333 79.506944 354.395833 347.027778 449.072917 47.631944 46.996528 61.562500

F Value 4.57 1.29 5.76 5.64 7.29 0.77 0.76

Pr> F 0.0248 0.2707 0.0117 0.0126 0.0011 0.4760 0.5627

There are seven hypotheses to be tested. If we select an overall probability of α ≤ 0.07 of at least one Type I error, then we would use level 0.01 for each test. We would fail to reject the hypotheses of no BRAND*OIL*TIME interaction and no OIL*TIME interaction. We would reject the hypothesis of no BRAND*TIME interaction and we would want to examine this interaction. The p-value for testing no BRAND*OIL interaction is close to 0.01 so, although we cannot reject this hypothesis, it would be sensible to examine this interaction also. However, the question asks us to examine the BRAND*OIL*TIME plot first. This is given below: Obs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

TC 11 11 11 12 12 12 21 21 21 22 22 22 31 31 31 32 32 32

TIME 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

AVPC 51.25 78.50 71.25 55.50 80.25 84.50 64.50 61.50 54.25 80.00 73.25 61.75 75.50 84.75 80.00 60.00 71.25 81.75

VARPC 66.125 50.000 153.125 24.500 36.125 4.500 264.500 18.000 21.125 72.000 28.125 1.125 128.000 1.125 4.500 162.000 45.125 28.125

AVPC | 90 + | | | 31 12 | | 32 80 +22 12 31 | 11 | |31 | 22 | 32 11 70 + | | |21 | | 21 22 60 +32 | | |12 21 | |11 50 + -+-----------------------+-----------------------+1 2 3 TIME The plot shows quite a difference in the brand-oil lines across the times, However, the error variance in this experiment is sufficiently large that these differences are not significant. The combination 31 (brand 3, oil 1) looks to be a good combination with high average popping rate across all three times and fairly small variability across the times. b) Does the store brand of popcorn differ substantially in of percentage of edible kernels from the average of the name brands? Do the different types of oil differ? State your overall confidence levels or significance levels. The contrasts to be estimated are α1∗ − 0.5(α2∗ + α3∗ ) and β1∗ − β2∗ , where αi∗ is the effect of Brand i averaged over the (almost) significant brand×time and brand×oil interactions; and where βj∗ is the effect of oil j averaged over the brand×oil interaction. From SAS, the contrast estimates are as follows. Suppose that we continue to test at level 0.01 for each hypothesis. Parameter

Estimate

Standard Error

t Value

Pr > |t|

STORE-AV

-0.5000

2.7740

-0.18

0.8590

OIL 2-1

2.9722

2.6154

1.14

0.2707

At an overall type I error probability of at most 0.02, we do not have sufficient evidence to reject the null hypothesis of no difference between the store brand and the average of the other two (averaged over oil and time). Similarly, we do not have sufficient evidence to reject the null hypothesis of no difference between the oil types (averaged over brand and time). However, there is an indication of a brand×oil interaction, so the brand-oil combinations (averaged over time) should perhaps be compared (in particular, the difference between combination 31 and the others).

c) Analyze the experiment as a product array, and calculate the sample average and the log sample variance percentage of popped kernels for each brand–oil combination. Draw AB interaction plots similar to those of Figure 7.9, page 223. If the goal of the experiment is still to find brand–oil combinations that give a high percentage of edible kernels and that are not too sensitive to the popping time, what recommendations would you make? How do your recommendations compare with those that you made in part (a)? Analysed as a product array, we calculate the average o the six observations at each combination of brand and oil, and also calculate the log sample variance. This gives Obs 1 2 3 4 5 6

BRAND 1 1 2 2 3 3

OIL 1 2 1 2 1 2

AVPC 67.0000 73.4167 60.0833 71.6667 80.0833 71.0000

LNVAR 5.36223 5.34349 4.41814 4.48149 3.78058 4.95371

We see that the highest average percent popped and the lowest variance is obtained from treatment combination 31, exactly as noted in part (a). The two plots are shown below and illustrate that combination 31 is considerably better than the other possibilities in reducing variability and increasing yield. Plot of LNVAR*BRAND. Symbol is value of OIL. LNVAR | 5.5 + |1,2 | | 5.0 + 2 | | | 4.5 + 2 | 1 | | 4.0 + | | 1 | 3.5 + -+-----------------------+-----------------------+-BRAND 1 2 3

Plot of AVPC*BRAND. Symbol is value of OIL. 80 + 1 | | | AVPC | | |2 | 2 | 2 70 + | | |1 | | | | | 60 + 1 -+-----------------------+-----------------------+-BRAND 1 2 3

Solution to Question 7.15 —Steel bar experiment We will analyze the data with the outlier removed (observation y2234 = 0). i. Variance assumptions. The residual plots of residual vs predicted and vs. each factor show no indications of heteroscedacity. Plot of Z*YPRED. Legend: A = 1 obs, B = 2 obs, etc. Z | 5 + | A A A | AA A A A A | AA AA A 2 + A A A A A | AA A A B A C | A AB AAA B C C AA A |-------------BA--AA---B--C---A-----A----1 + A BA A A AA | A A AA A B A ABA A | A A A B B A | A AA A -4 + A A A | -+------------+------------+------------+ -5 0 5 10 YPRED Plot of Z*A. Legend: A = 1 obs, B = 2 obs, etc. Z | 5 + | B A | C C | C B 2 + B C | F D | H J |-C--------------------------I-----------1 + E C | H E | D D | B B -4 + B A | --+--------------------------+----------1 2 A

Plot of Z*B. Legend: A = 1 obs, B = 2 obs, etc. Z | 5 + |A B |B A C |B B A 2 +B B A |B C C B |C D D G |A------------C------------E------------C -1 +C A C A |C D D B |B B D |A B A -4 +B A | -+------------+------------+------------+ 1 2 3 4 B Plot of Z*C. Legend: A = 1 obs, B = 2 obs, etc. Z | 5 + |A B |A B C |A B B 2 +A A C |E D A |G I B |E------------------C------------------D-1 +B B D |F C D |B E A | A C -4 +A B | -+------------------+------------------+1 2 3 C

ii. Independence assumption The order of observation is not given, so we cannot check the independence assumption.

iii. Normality. The normal probability plot shows no serious departure from normality. Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | 2.438 + | AAA | BCA | CB 0.975 + DA | BFB | DJD |------------------GE--------------------0.488 + GA | FFA | DD | AC -1.950 + AAA | --+--------+--------+--------+--------+--4 -2 0 2 4 RANK FOR VARIABLE Z

General Linear Models Procedure Dependent Variable: Y Source Model Error Corrected Total

Source A B C A*B A*C B*C A*B*C

DF 23 71 94

Sum of Squares 626.58333333 395.41666667 1022.00000000

R-Square 0.613095

C.V. 58.99814

DF 1 3 2 3 2 6 6

Type III SS 82.60388128 421.46444444 22.24253941 1.94444444 0.83375563 84.01749466 21.30916132

F Value 4.89

Pr > F 0.0001

Y Mean 4.00000000 F Value 14.83 25.23 2.00 0.12 0.07 2.51 0.64

Pr > F 0.0003 0.0001 0.1433 0.9502 0.9280 0.0290 0.6996

For an OVERALL level of .05, we can conclude there is an effect due to heat (A) and machine (B) but no effects due to time (C) nor any (statistically significant) effect due to interaction. However, we note the fairly small p-value for interaction of B and C.

d.) 10 + | | AV_Y | 3 | 2 | | 1 5 + 3 2 1 | | 1 | | | 2 | 1 0 + ---+---------------+---------------+---------------+-1 2 3 4 Machine NOTE: 3 obs hidden. It appears machine 3 is most robust as the average lengths at each time (C) are closer than the average lengths at each time for the other machines. Assuming the machines are identical in make and model, a possible cause for the difference in robustness could be age of the machine or the operator. e.) Least Squares Means

B 1 2 3 4

95% Lower Confidence Limit 2.456149 5.235384 -0.085518 4.706149

Y LSMEAN 3.416667 6.222222 0.875000 5.666667

It appears that machine 4 is closest to specifications.

95% Upper Confidence Limit 4.377185 7.209061 1.835518 6.627185

Design and Analysis of Experiments Angela Dean and Daniel Voss Solutions to Chapter 10 20 August 2006 Solutions available in this file c

2006 Angela Dean. All rights reserved. No part of this work may be displayed on the web or reproduced in any form without the written permission of Angela Dean, The Ohio State University.

Question 10.6 — Candle experiment Question 10.7 — Salt water experiment Question 10.8 — Chemical experiment Question 10.10 — Length perception experiment Question 10.12 — Biscuit experiment Question 10.16 — Exam paper experiment

1

Solution to Question 10.6 — Candle experiment (a) Explain what block×treatment interaction means in the context of this experiment. Can you think of any causes that might have led to the presence of interaction in the pilot experiment? In this experiment, the treatments are the different candle colors and the blocks are the different experimenters. So a block×treatment interaction would indicate that the color of the candle affected the speed of burning differently for each of the different experimenters. For example, the red candles may have burned fastest for Tsai and Wheeler, the white fastest for Yang, and the yellow fastest for Schultz. The presence of the interaction could possibly have been caused by bias on the part of an experimenter. For example, one experimenter might have let the red candle burn past the designated mark, while blowing out the white candle prematurely. (b) Plot the data (Table 10.20) from the main experiment and interpret your plot(s). Below is a plot of the data, plotting burning time vs. experimenters (blocks) with the colors (treatments) as the labels. The plot suggests that there may exist some color (treatment) differences, as the data for color 3 (blue) are mostly below the data for the other colors for all blocks, and the data for color 1 (red) are mostly above the data for the other colors for all blocks. There probably does not exist a color×experimenter (treatment×block) interaction, since the relative position of the data for each block is similar (i.e. color 1 (red) is always near the top and color 3 (blue) is always near the bottom). Plot of AV_TIME*BLOCK. Symbol is value of COLOR. 1100 + | | | | 1 | 2 1000 + | 3 1 | 2 AV_TIME | | 4 | 1 1 900 + | 4 | 2 3 | 3 | 3 | 800 + --+------------+------------+------------+--------1 2 3 4 BLOCK

2

(c) Complete an analysis of variance table for the data using the block–treatment interaction model (??) for a general complete block design. The block-treatment interaction model is as follows:

Yhit = µ + θh + τi + (θτ )hi + hit . hit ∼ N (0, σ 2 ) , hit ’s are mutually independent , t = 1, 2, 3, 4; h = 1, 2, 3, 4; i = 1, 2, 3, 4. The anlysis of variance table is shown below. The GLM Procedure Dependent Variable: time Source Model Error Corrected Total

DF 15 48 63

Sum of Squares 227824.7500 82025.0000 309849.7500

Mean Square 15188.3167 1708.8542

F Value 8.89

Pr > F <.0001

Source Source block color block*color

DF DF 3 3 9

Type I SS Type III SS 151659.1250 60345.0000 15820.6250

Mean Square Mean Square 50553.0417 20115.0000 1757.8472

F Value F Value 29.58 11.77 1.03

Pr > F Pr > F <.0001 <.0001 0.4315

(d) Test the null hypotheses of negligible block×treatment interaction and, if appropriate, test the null hypothesis of equality of treatment effects. Using an overall significance level of at most 0.02, we test the effect on the response of the block×color interaction at individual level 0.01 and, if relevant, the effect of color at level 0.01. For the interaction, we test the hypothesis: H0θT : {(θτ )hi − (θτ )h. − (θτ ).i + (θτ ).. = 0 for all h, i} against H1θT :{ at least one interaction contrast is nonzero}. The value of the test statistic ms(θT )/msE = 1.03 with a p-value of 0.4315 > 0.01. Therefore, we fail to reject H0θT and there is not sufficent eveidence to conclude a significant block×color interaction. Since there is not a signifcant block×color interaction, we can now test for differences in average burning times between the colors; that is H0T : {τ1 = τ2 = τ3 = τ4 } against H1T : {at least two color effects differ}. The value of the test statistic msT/msE = 11.77 with a p-value of p < 0.0001 < 0.01. Therefore, we reject H0T and conclude that a statistically significant color (treatment) effect exists; that is, we conclude that at least one of the colors burns at a different speed from the others. (e) Use an appropriate multiple comparisons procedure to evaluate which color of candle is best. Interpret the results. These intervals were probably pre-planned, so we can use Bonferonni or Tukey methods to examine the m = 6 pairwise comparisons. To determine which type of multiple comparisons to compute, we first examine the critical coefficients for each type of interval. The critical coefficients for confidence level level 99% are: wB = tn−vb,α/2m = t48,0.00083 = 3.3332 and

√ √ √ wT = qv,n−vb,α / 2 = q4,48,0.01 / 2 = 4.644/ 2 = 3.283.

3

We see that wT and wB are comparable, with wT just slightly smaller. Therefore, we compute all pairwise multiple comparisons using Tukey’s method at an overall level of 99%. The following output is produced by SAS: The GLM Procedure Tukey’s Studentized Range (HSD) Test for TIME NOTE: This test controls the Type I experimentwise error rate. Alpha 0.01 Error Degrees of Freedom 48 Error Mean Square 1708.854 Critical Value of Studentized Range 4.64437 Minimum Significant Difference 47.998 Comparisons significant at the 0.01 level are indicated by ***.

COLOR Comparison 1 - 4 1 - 2 1 - 3 4 - 2 4 - 3 2 - 3


Simultaneous 99% Confidence Limits -33.75 62.25 -22.75 73.25 33.00 129.00 -37.00 59.00 18.75 114.75 7.75 103.75

*** *** ***

Based on this output, we can conclude that blue candles (color 3) burn more quickly than the other colors, but there is not a significant difference between the red (color 1), white (color 2) or yellow (color 4) candles. Therefore, if slower burning is better, one should not buy blue candles! (f) Discuss whether blocking was important in this experiment. If we compare msθ with msE, we see that msθ is about 30 times as large as msE. 830 is large. Therefore, clearly blocking was important in this experiment, and it did reduce the error variability.

4

Question 10.7— Salt water experiment a) Plot the data and interpret the results. The following plot of average temperature for each salt level for each block indicates a linear trend in boiling point as the salt level increases. There may be some interaction between block and salt contrary to the experimenters assumptions.

Plot of MN_TEMP*SALT.

Symbol is value of BLOCK.

MN_TEMP | | | 99.0 + | 1,3 | | 2 98.5 + | 1,2 | 3 | 98.0 + | | 1,3 | 97.5 +2 |1,3 3 2 | 2 | 97.0 + | 1 | | 96.5 + | -+-----------+-----------+-----------+-----------+0 8 16 24 32 SALT b) Complete an analysis of variance table for the data and test for equality of treatment effects. With no block*salt interaction in the model, we have The GLM Procedure Dependent Variable: TEMP Source Model Error Corrected Total

DF 6 38 44

Sum of Squares 15.72088889 3.31555556 19.03644444

Mean Square 2.62014815 0.08725146

F Value 30.03

Pr > F <.0001

Source BLOCK SALT

DF 2 4

Type III SS 0.14444444 15.57644444

Mean Square 0.07222222 3.89411111

F Value 0.83 44.63

Pr > F 0.4448 <.0001

5

The hypothesis of equality of treatment (salt) means would be rejected for any reasonable choice of significance level α. c) Evaluate whether blocking was worthwhile and whether the assumption of no treatment–block interaction looks reasonable. The mean square for Blocks (msθ) is less than msE, indicating blocking was not very beneficial. However, since it only s for 2 degrees of freedom and the error is based on 44 degrees of freedom, the power of the test for no treatment differences was not greatly reduced. Although the interaction plot does indicate some interaction, we note (i) the size of the differences between the blocks at each salt level is rather small and (ii) for each block, the same “story” is being told about the relationship between salt and temperature. We may conclude that the model with no treatment-block interaction is reasonable. The lack of the interaction term in the model may possibly for msE being larger than msθ. d) Compute sums of squares for orthogonal trend contrasts, and test the trends for significance, using a simultaneous error rate of 5%. Explain your results in of the effect of salt on the boiling point of water. The contrast coefficients from Table A.2 are CONTRAST CONTRAST CONTRAST CONTRAST

’LINEAR’ ’QUAD’ ’CUBIC’ ’QUARTIC’

SALT -2 -1 0 1 SALT 2 -1 -2 -1 SALT -1 2 0 -2 SALT 1 -4 6 -4

2; 2; 1; 1;

The contrast sums of squares are

SSC

Linear 12.996

Quad. 1.786

Cubic 0.784

Quartic 0.0107

and msE ∗ t38,.005 = 0.2362694. Using Table A.2, the 95% simultaneous Bonferroni intervals are

Linear Quadratic Cubic Quartic

Lower bound 2.956860 197.269050 -1.776474 -1.941851

Upper bound 4.643140 199.264284 -0.090193 2.519628

So the Linear,Quadratic and Cubic trends are simultaneously significantly diferent from zero. Alternatively, from SAS we obtain Contrast LINEAR QUAD CUBIC QUARTIC

DF 1 1 1 1

Contrast SS 12.99600000 1.78571429 0.78400000 0.01073016

Mean Square 12.99600000 1.78571429 0.78400000 0.01073016

F Value 150.24 20.64 9.06 0.12

Pr > F <.0001 <.0001 0.0045 0.7265

Selecting individual significance level 0.05/4 = 0.0125, and comparing this with the p-values, we reject the hypotheses that the linear, quadratic and cubic trends are each negligible, but fail the reject the hypothesis that the quartic trend is negligible. e) Calculate the number of observations needed per treatment if a test of equal ity of treatment effects using α = 0.05 is to have power 0.95 in detecting a difference of 1◦ C when σ = 0.5◦ C. TO BE ADDED AT A LATER DATE

6

Question 10.8 – Chemical experiment a) Draw a graph of the data and comment on it. First, recode the treatment combination labels 111, 112, 121, . . ., 122 as treatment labels 1, 2, 3, . . ., 8. A plot of the average response for each treatment in each block is shown below. Plot of Y*BLOCK. Symbol is value of TRT. 30 + | | | 6 | 5 | 4 20 + 6 | 6 4,5 8 | 5 3,7 5,6 Y | 4 3,8 4 | 3,8 2 8 | 7 2,7 1 3 10 + 2 7 | 1 2 | 1 1 | | | 0 + --+------------+------------+------------+--------1 2 3 4 BLOCK The plot indicates very little interaction between block and treatment effects, since treatments 4, 5, and 6 give consistently highest yields, while treatments 1, 2 and 7 give consistently the lowest yields. b) State a possible model for this experiment. List any assumptions you have made and discuss the circumstances in which you would expect the assumptions to be valid. A possible model for this experiment is the block-treatment model

Yhijk = µ + θh + τijk + hijk hijk ∼ N (0, σ 2 ) hijk 0 s are mutually independent h = 1, . . . , 4; i = 1, . . . , 2; j = 1, . . . , 2; k = 1, . . . , 2 ,

(1)

where µ is a constant, θh is the effect of the hth lab, τijk is the effect of the treatment combination ijk, Yhijk is the random variable representing the measurement on treatment combination ijk observed in block h, and hijk is the associated random error. The assumptions made are that the error variables are independent and normally distributed with mean zero and common variance σ 2 , and also that there is no interaction between treatments and labs. If the labs are carefully controlled and use the same experimental procedures, there is no particular reason why any one lab should favor any one treatment combination over another. So the assumption of no treatment×lab interaction is probably reasonable. (There are not sufficient degrees of freedom to be able to measure a full interaction based on 32 degrees of freedom). If the experiment is conducted carefully and the order of the treatment combinations randomized, the the error variables should be measuring only random error, in which case the error assumptions should also be valid.

7

c)

Check the assumptions on your model.

The error assumptions are checked through residual plots (shown below). There do not appear to be any serious outliers. The variability of the residuals appear to differ a little at the different levels of the treatment factors and blocks but not drastically so. One can calcvulate the s2i to check the discrepancy. The normality assumption seems fairly well satisfied. The assumption of no block-treatment interaction is checked via the plot in part a) and seems to be fairly well satisfied. Plot of Z*TRT. Legend: A = 1 obs, B = 2 obs, etc. 2 + | A | A | B A Z | A | A A | A |A A A |B B A 0 +---------------------A---------------------------A | A | | A |A A A A A | A A A | A | A A | -2 + -+------+------+------+------+------+------+------+ 1 2 3 4 5 6 7 8 TRT Plot of Z*BLOCK. Legend: A = 1 obs, B = 2 obs, etc. 2 + | A | A | A A A Z | A | A A | A | A A A | B B A 0 +--------------B----------------------------------| A | | A | A C A | A B | A | A A | -2 + --+------------+------------+------------+--------1 2 3 4 BLOCK

8

Plot of Z*NSCORE.


2 + | | | A | | A | | AAA Z | | A | | AA | | A | | AB | BBA 0 +-----------------------B+------------------------| A | | | | A | | ABB | | AB | | A | | A A | | | -2 + | -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 RANK FOR VARIABLE Z d) Analyze the data and state your conclusions. The analysis of variance table is shown below: General Linear Models Procedure Dependent Variable: Y Source Model Error Corrected Total Source BLOCK A B C A*B A*C B*C A*B*C

DF 10 21 31

DF 3 1 1 1 1 1 1 1

Sum of Squares 639.277500 9.882500 649.160000 Type III SS 182.017500 115.520000 0.911250 55.125000 284.411250 0.080000 1.201250 0.011250

Mean Square 63.927750 0.470595

Mean Square 60.672500 115.520000 0.911250 55.125000 284.411250 0.080000 1.201250 0.011250

F Value 135.84

F Value 128.93 245.48 1.94 117.14 604.36 0.17 2.55 0.02

Pr > F 0.0001

Pr > F 0.0001 0.0001 0.1786 0.0001 0.0001 0.6843 0.1251 0.8786

There appears to be a large lab difference since the block mean square is 60.6725 compared with an error mean square of only 0.470595. Thus, it was good to use blocks in our model. If we test each treatment hypothesis at level α∗ = 0.01 (for an overall level of α = 0.07), we reject the null hypotheses of negligible AB interaction and negligible A and C main effects. Let us suppose that the pre-plan was to calculate a 99% set of pairwise comparisons between the treatment combinations using Tukey’s method, to calculate 99.5% intervals for the comparisons between 9

the levels of A, B and C if these were not involved in interactions and a set of 99% intervals using Scheffé method for any other contrasts that look interesting. The overall confidence level would then be at least 96.5%. The results of the 99% Tukey intervals for pairwise comparisons of the treatment combinations are shown below: The GLM Procedure Tukey’s Studentized Range (HSD) Test for Y NOTE: This test controls the Type I experimentwise error rate. Alpha 0.01 Error Degrees of Freedom 21 Error Mean Square 0.470595 Critical Value of Studentized Range 5.79426 Minimum Significant Difference 1.9874 Comparisons significant at the 0.01 level are indicated by ***.

TRT Comparison 6 - 5 6 - 4 6 - 8 6 - 3 6 - 7 6 - 2 6 - 1 5 - 4 5 - 8 5 - 3 5 - 7 5 - 2 5 - 1 4 - 8 4 - 3 4 - 7 4 - 2 4 - 1 8 - 3 8 - 7 8 - 2 8 - 1 3 - 7 3 - 2 3 - 1 7 - 2 7 - 1 2 - 1

Difference Between Means 2.1750 2.9750 5.2750 6.1250 8.1500 9.7000 12.0000 0.8000 3.1000 3.9500 5.9750 7.5250 9.8250 2.3000 3.1500 5.1750 6.7250 9.0250 0.8500 2.8750 4.4250 6.7250 2.0250 3.5750 5.8750 1.5500 3.8500 2.3000

Simultaneous 99% Confidence Limits 0.1876 4.1624 0.9876 4.9624 3.2876 7.2624 4.1376 8.1124 6.1626 10.1374 7.7126 11.6874 10.0126 13.9874 -1.1874 2.7874 1.1126 5.0874 1.9626 5.9374 3.9876 7.9624 5.5376 9.5124 7.8376 11.8124 0.3126 4.2874 1.1626 5.1374 3.1876 7.1624 4.7376 8.7124 7.0376 11.0124 -1.1374 2.8374 0.8876 4.8624 2.4376 6.4124 4.7376 8.7124 0.0376 4.0124 1.5876 5.5624 3.8876 7.8624 -0.4374 3.5374 1.8626 5.8374 0.3126 4.2874

*** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** *** ***

The objective of the experiment was to find the combination that gives the highest yield. From the results of the Tukey confidence intervals, this would appear to be treatment 6, which corresponds to treatment combination 212. Factor C is not involved in any significant interactions. The estimate for the main effect of factor C (high−low) is positive which suggests that the higher level of C gives the higher yield averaged over A 10

and B.

Parameter c2-c1

Estimate 2.62500000

Standard Error 0.24253743

t Value 10.82

Pr > |t| <.0001

A 99.5% confidence interval for the difference between the two levels of C is 2.625 ± t21,.0025 × 0.2425 = 2.625 ± 3.135 × 0.2425 = 2.625 ± 0.760 = (1.865, 3.385) . Factors A and B are involved in the AB interaction. An interaction plot shows that the highest yield is obtained on average when A is at the high level and B at the low level. This agrees with the finding that combination 212 should give the highest yield. Plot of AVY*A. Symbol is value of B. 20 + | 1 | | | | 2 15 + | 2 | AVY | | | 10 + 1 | | | | | 5 + --+--------------------------------+--------------1 2 A

11

Question 10.10 — Length perception experiment a) Fit a block–treatment model to the data using subjects as blocks and with six treatments representing the shape–area combinations. Check the error assumptions on your model. The model for the experiment is the block-treatment model for the randomized block design is

Yhi = µ + θh + τi + hi , hi ∼ N (0, σ 2 ) , 0 hi s are mutually independent , h = 1, . . . , 14; i = 1, . . . , 6 ,

(2)

where µ is a constant, θh is the effect of the hth subject (block), τi is the effect of the ith treatment, Yhi is the random variable representing the measurement on treatment i observed in block h, and hi is the associated random error. The six treatment combinations 11, 12, 21, 22, 31, 32 are labelled 1, 2, 3, 4, 5, 6 in the plots shown below. From the plot of standardized residuals z against the predicted values yˆ, we see little evidence of variance changing with the mean, outliers or other problems, although the plot of standardized residuals versus subjects indicates a possible unequal variances from subject to subject. The ratio of maximum to minimum sample variance of the data for the subjects is 0.49069/0.03436 = 11.9. We do not have information on the variability within each subject treatment/cell, but we should treat our analysis with some caution since the equal variance assumption may not be exactly satisfied. However, the variability of the data is very similar for all treatments averaged over subjects, so Satterthwaite’s approximation will not have much effect on the treatment comparisons and we continue using the traditional analysis. Level of SHAPE 1 1 2 2 3 3

Level of AREA 1 2 1 2 1 2

N 14 14 14 14 14 14

--------------Y-------------Mean Std Dev 0.31071429 0.70142124 -0.30714286 0.47509398 0.17500000 0.69219550 -0.06785714 0.85498699 0.09285714 0.71544960 -0.13928571 0.84425173

12

Plot of Z*PRY. Legend: A = 1 obs, B = 2 obs, etc. | | A | 2 + A A | A A | A A A A A Z | AA A A A | AA BA ABB BA | B A AA A A A A A 0 + A A AA A A | A A AC AAB AA A A | A A AB B A | A AA A A AA | A A AA A A | A -2 + A A A A | | -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 PRY Plot of Z*SUBJECT. Symbol is value of TRT. Z | | | 4 2 + 6 1 1 | 3 1 3 4 4 | 3 6 2 4 2 3 | 5 1 2 5 2 6 5 2 4 3 3 5 0 +-2--5-----2--5--5--2--2-----------2--6--2--------| 1 1 1 1 1 1 1 5 2 1 | 3 1 6 4 2 3 3 1 4 6 | 4 2 5 5 -2+ 4 4 3 6 | | --+--+--+--+--+--+--+--+--+--+--+--+--+--+--------1 2 3 4 5 6 7 8 9 10 11 12 13 14 SUBJECT NOTE: 19 obs hidden.

13

Plot of Z*TRT. Legend: A = 1 obs, B = 2 obs, etc. Z | | | A 2 + B A | A B B A | B C A A | A C B C E D 0 +----------F--------A-----------------C--------C--| H A B C A | B A E B C | A A B -2 + A B A | | --+--------+--------+--------+--------+--------+--1 2 3 4 5 6 TRT The normality assumption seems to be reasonable. We cannot check independence since the order information is not available. Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. 4 + | | | | | | | | | A | | 2 + | AA | | B | | ACA Z | | CB | | EDD | |DF 0 +------------------------EA-----------------------| BFEA | ADC | | ACC | | ACB | | A | -2 + A AAA | | | -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 Rank for Variable Z b) Draw at least one graph and examine the data. One possible plot is a plot of the data against subject with the points labelled by treatment number. Although the plot gives no indication of error variability, it does indicate that there may be some subjects by treatment interaction. Note that there are a lot of hidden observations in this plot. We would need to make the plot larger to see all the data. This experiment is not sufficiently large to incorporate a subject by treatment interaction in the model, but it should be borne in mind that the block-treatment model may not be exactly correct and the error mean square may be inflated. 14

Plot of Y*SUBJECT. Symbol is value of TRT. 2 + | 3 | 1 | | 5 1 4 | 1 + 4 | 3 1 1 | 1 4 4 Y | 1 4 3 3 | 5 2 2 6 5 | 1 2 1 6 1 5 3 0 + 6 3 2 1 3 | 2 2 2 1 2 1 | 3 3 3 2 6 | 4 4 3 | 1 2 2 2 | 2 1 -1 + 6 6 | 6 | 4 2 | 4 | | -2 + --+--+--+--+--+--+--+--+--+--+--+--+--+--+--------1 2 3 4 5 6 7 8 9 10 11 12 13 14 SUBJECT NOTE: 25 obs hidden. c) Write down contrasts in the six treatment combinations representing the following comparisons: (i) differences in the effects of area for each shape separately, (ii) average difference in the effects of area, (iii) average difference in the effects of shape. In of the six treatment effects, the contrasts of interest are: (i) Area difference for each shape Square: τ1 − τ2 Circle: τ3 − τ4 Triangle: τ5 − τ6 (ii) Area difference on average (τ1 + τ3 + τ5 − τ2 − τ4 − τ6 )/3 (iii) Shape differences on average Square–circle : (τ1 + τ2 − τ3 − τ4 )/2 Square–triangle : (τ1 + τ2 − τ5 − τ6 )/2 Circle–triangle: (τ3 + τ4 − τ5 − τ6 )/2

15

d) Give a set of 99% simultaneous confidence intervals for the contrasts in c(i). State your conclusions. For the three contrasts in part c(i), at an overall level of 99%, we assume that these were preplanned and use Bonferonni’s method. From SAS: The GLM Procedure Dependent Parameter SHP1 AREA SHP2 AREA SHP3 AREA

Variable: Y Estimate 2-1 0.61785714 2-1 0.24285714 2-1 0.23214286

Error 0.16044891 0.16044891 0.16044891

t Value 3.85 1.51 1.45

Pr > |t| 0.0003 0.1350 0.1527

Each confidence interval should be of the form estimate ± t65,0.01/6 ∗ Standard error The error degrees of freedom are n − b − v + 1 = 6 ∗ 14 − 14 − 6 + 1 = 65. The critical coefficient is t65,0.01/6 = 3.0477, so the three confidence intervals at overall level 99% are SHP1 AREA 2-1 0.61785714 ± 3.047 ∗ 0.16044891 = (0.129, 1.107) SHP2 AREA 2-1 0.24285714 ± 3.047 ∗ 0.16044891 = (−0.246, 0.732) SHP3 AREA 2-1 0.23214286 ± 3.047 ∗ 0.16044891 = (−0.257, 0.721) At overall level 99%, we conclude that there is no difference in the effects of the area for circle or triangle, but that for the square the small square causes the line to be drawn between 0.129 cm and 1.107 cm larger than the large square. e) Under what conditions would the contrasts in c(ii) and c(iii) be of interest? Do these conditions hold for this experiment? The contrasts in c(ii) and c(iii) would be of interest if the shape by area interaction is not significantly different from zero. From the following analysis of variance table, we see that we would fail to reject the null hypothesis of no interaction at any significance level smaller than .16 and so these contrasts would be of interest. The GLM Procedure Dependent Variable: Y Source Model Error Corrected Total

DF 18 65 83

Sum of Squares 32.81690476 11.71345238 44.53035714

Mean Square 1.82316138 0.18020696

F Value 10.12

Pr > F <.0001

Source SUBJECT SHAPE AREA SHAPE*AREA

DF 13 2 1 2

Type III SS 29.26869048 0.08589286 2.78678571 0.67553571

Mean Square 2.25143773 0.04294643 2.78678571 0.33776786

F Value 12.49 0.24 15.46 1.87

Pr > F <.0001 0.7886 0.0002 0.1617

16

Question 10.12 — Biscuit experiment a) State a suitable model for this experiment and check that the assumptions on your model hold for these data. The block-treatment model is

Yhijt = µ + bh + τij + hijt ; h = 1, .., 4; i = 1, 2, 3; j = 1, 2, 3; t = 1, 2 hijt ∼ N (0, σ 2 ) where bh is the effect due to block h and τij is the effect due to treatment combination ij (height, kneading time). The three residual plots below do not show any serious signs of heteroscedacity.

2 Z

0

-2

Plot of Z*PREDY. Symbol is value of BLOCK. | | 4 + 2 3 | 2 3 | 1 2 3 1 34 | 1 2 1 3 | 1142 1142 114 +---------1434-2-1-4------------------4--1--2-----| 332 243 2 3 | 4 21 3 2 2 | 433 1 2 4 | 3 + 1 | 4 1 | --+--------+--------+--------+--------+--------+--150 200 250 300 350 400 PREDY

Plot of Z*BLOCK. Legend: A = 1 obs, B = 2 obs, etc. | | A 2 + A A | A A Z | B A B B | B A A | G B A E 0 +-C------------C------------C------------D--------| C D A | A E B B | A A B B | A -2 + A | A A --+------------+------------+------------+--------1 2 3 4 BLOCK

17

Plot of Z*TC. Legend: A = 1 obs, B = 2 obs, etc. | | A 2 + A A | A A Z | A B A A A A | A A A A | A B D C B B A 0 +-B----A----B---------A----A----D---------B-------| A C A B A | A C A A A B A | A B B A | A -2 + A | A A | --+----+----+----+----+----+----+----+----+-------11 12 13 21 22 23 31 32 33 TC The Normality assumption appears satisfied. Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. | | | | A 2 + | AA | | AA Z | | BBBA | | CA | |BEDD 0 +----------------------AEDC-----------------------| DD | | CCD | | BBB | | A | -2 + A | | A A | | | -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 Rank for Variable Z b) Use an appropriate multiple comparisons procedure to evaluate which treatment combination is best. The Tukey pairwise comparisons at level 95% show that treatment combinations 11, 12, and 13 are all significantly different from the other treatment combinations (averaged over blocks). Since the differences are positive, we can conclude that a smaller initial height leads to a higher percentage increase in height and hence fluffier biscuits. For height 1 we see no significant differences in the effect of different kneading times. However, each contrast of treatment combination 13 versus each other treatment combination is positive and large. Thus, it looks as though a longer kneading time may be beneficial and this should perhaps be studied further in a follow-up experiment.

18

General Linear Models Procedure Tukey’s Studentized Range (HSD) Test for variable: Y NOTE: This test controls the type I experimentwise error rate. Alpha= 0.05 Confidence= 0.95 df= 60 MSE= 567.8419 Critical Value of Studentized Range= 4.550 Minimum Significant Difference= 38.337 Comparisons significant at the 0.05 level are indicated by ’***’.

TC Comparison 13 - 12 13 - 11 13 - 21 13 - 23 13 - 22 13 - 33 13 - 32 13 - 31 12 - 11 12 - 21 12 - 23 12 - 22 12 - 33 12 - 32 12 - 31 11 - 21 11 - 23 11 - 22 11 - 33 11 - 32 11 - 31 21 - 23 21 - 22 21 - 33 21 - 32 21 - 31 23 - 22 23 - 33 23 - 32 23 - 31 22 - 33 22 - 32 22 - 31 33 - 32 33 - 31 32 - 31

Simultaneous Simultaneous Lower Difference Upper Confidence Between Confidence Limit Means Limit -23.65 14.69 53.02 -17.44 20.90 59.24 71.64 109.98 148.31 *** 90.39 128.73 167.06 *** 95.08 133.41 171.75 *** 125.31 163.65 201.99 *** 126.86 165.20 203.54 *** 134.65 172.99 211.32 *** -32.12 6.21 44.55 56.95 95.29 133.62 *** 75.70 114.04 152.37 *** 80.39 118.73 157.06 *** 110.63 148.96 187.30 *** 112.18 150.51 188.85 *** 119.96 158.30 196.64 *** 50.74 89.08 127.41 *** 69.49 107.83 146.16 *** 74.18 112.51 150.85 *** 104.41 142.75 181.09 *** 105.96 144.30 182.64 *** 113.75 152.09 190.42 *** -19.59 18.75 57.09 -14.90 23.44 61.77 15.34 53.67 92.01 *** 16.89 55.22 93.56 *** 24.68 63.01 101.35 *** -33.65 4.69 43.02 -3.41 34.93 73.26 -1.86 36.48 74.81 5.93 44.26 82.60 *** -8.10 30.24 68.57 -6.55 31.79 70.12 1.24 39.57 77.91 *** -36.79 1.55 39.89 -29.00 9.34 47.67 -30.55 7.79 46.12

19

c) Evaluate whether blocking was worthwhile in this experiment. We see that relative to MSE, MSblock is larger. That is MSE/MSblock = 2.5. The plot below shows on average, biscuits in Block 1 were not as fluffy. Blocking was probably worthwhile. General Linear Models Procedure Dependent Variable: Y Source DF Sum of Squares F Value Model 11 317818.006250 50.88 Error 60 34070.516944 Corrected Total 71 351888.523194

Pr > F 0.0001

Source BLOCK TC

Pr > F 0.0682 0.0001

DF 3 8

Type I SS 4254.129306 313563.876944

F Value 2.50 69.03

Plot of AV_Y*BLOCK. Legend: A = 1 obs, B = 2 obs, etc. AV_Y | 290 + | | A | 280 + | A | | A 270 + | | A | 260 + ---+---------------+---------------+---------------+-1 2 3 4 BLOCK

20

Question 10.16 – Exam paper experiment a) Plot the data for each treatment combination in each block. Can you conclude anything from looking at the data plots? In block=1 Plot of score*tc. Legend: A = 1 obs, B = 2 obs, etc. score | 100 + | | A B A | C A A C 80 + A C A | A | A B A | A 60 + A A B | A C | A A | B A 40 + | A | A A | 20 + --+--------+--------+--------+11 12 21 22 tc In block=2 Plot of score*tc. Legend: A = 1 obs, B = 2 obs, etc. score | 100 + A | B E A | B A | A | B | A 75 + A A | E B B | A A A | A A B | C A B B | A 50 + A A | | | A A | | A 25 + A --+--------+--------+--------+11 12 21 22 tc

21

As seen in the above plots, the data in blocks 2 and 3 are mostly between 50 and 100 with a few low scores. In block 1, the data a re perhaps more evenly spread. it is not clear that the equal variance assumption is approximately satisfied. b) Fit a block-treatment model without block*treatment interaction. Using a computer package, calculate the analysis of variance table and state your conclusions. The model is

Yhit = µ + θh + τ1i + τ2j + hit hit N (0, (σ 2 ), hit ’s mutually independent h = 1, 2, 3; i = 1, 2; j = 1, 2; t = 1, ..., 49. The analysis of variance table from SAS is The GLM Procedure Dependent Variable: score Source Model Error Corrected Total

Source block color version

DF 4 125 129

Sum of Squares 1159.54816 42402.02876 43561.57692

Mean Square 289.88704 339.21623

F Value 0.85

Pr > F 0.4934

DF 2 1 1

Type III SS 259.3146437 27.5663936 818.7570356

Mean Square 129.6573219 27.5663936 818.7570356

F Value 0.38 0.08 2.41

Pr > F 0.6831 0.7761 0.1228

Individual tests for the effects of color and version are each insignificant since there p-values ore very large. Thus, we conclude that both color and version have negligible effect on the test scores. Blocking by teaching assistants did not increase the power of the tests for this experiment since msθ < msE..

22

c)

Check the assumptions on your model by plotting the standardized residuals.

Plot of z*nscore.


2 + | A A | AEBCBA | FFEB | FHA | GB 0 +-----------------------DHB-----------------------| CHE | FGE z | ADE | C | B -2 + ABA | AAA | | | A | -4 + -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 Rank for Variable z The normal probability plot shows a fairly straight line, although there is a short right tail. Plot of z*ypred.


2 + | A A | B B B B E A | C A C D B A A BA A | C A A D BB B | A A B A A C 0 + C A B A EB | B A D A A C A B A | C A A B A B CB C z | C A A A A C | B A | A A -2 + A A A A | A A A | | | A | -4 + -+-----------+-----------+-----------+-----------+65.0 67.5 70.0 72.5 75.0 ypred 23

The plot of the standardized residuals versus the predicted values shows no particular pattern. There is one value that could be an outlier since it has a z-score that is less than -3. Plot of z*block.


2 + | A A |B I C |G C I |H C D |B C D 0 +C I B |D G E |D G G z |D B D |B A | A A -2 +B B |A A A | | | A | -4 + -+-----------------------+-----------------------+1 2 3 block Plot of z*color.


2 + | B | E I | I J | K D | E D 0 + G G | H H | I I z | C G | A B | A A -2 + B B | A B | | | A | -4 + --+--------------------------------+--------------1 2 color

24

Plot of z*version.


2 + | A A | F H | H K | K D | E D 0 + J D | D L | K G z | E E | C | A A -2 + B B | A B | | | A | -4 + --+--------------------------------+--------------1 2 version We can now see that, although there was concern about the constant variance assumption in part (a), for tests of color and version, the equal variance assumption does appear reasonable. There appears to be an outlier for version 2, color 1 in block 3. Plot of z*order.


2 + | A A | AA B A AAAAAA BA | D AB AC AAA A CA | ACC BA A A A A A | A AA AA A A A A 0 + AAB B BA A A AB | A A A BA AA A A B BA A | AB A B BA AB BAAA A z | A A A BB B A | A AA | AA -2 + A A A A | A A A | | | A | -4 + --+---------+---------+---------+-----------------0 20 40 60 order The plot of the standardized residuals vs. order verifies that the independence assumption has not been violated. 25

d) If the same teaching assistant had been assigned to all three classes, should the experiment still have been designed as a block design? Discuss. Yes, the experiment should still be a block design. Even though the same teaching assistant would have proctored all three exams, there may still be an order effect. Suppose one exam was given at 9:00AM and another was given later in the afternoon, we must for the possibility that students from the earlier sections spoke with the students in the later sections, or that students may be tired later in the day.. Furthermore, we must for environmental differences between the three classrooms; for example, one classroom may be excessively hot or cold.

26



Question 12.6— Video Game Experiment Question 12.10 — Quantity Perception Experiment

1

Solution to Question 12.6 - Video Game Experiment (a) Model: Yhqi = µ + θh + φq + τi + hqi hqi ∼ N (0, σ 2 ) hqi ’s mutually independent h = 1, 2, 3, 4, 5; q = 1, 2, 3, 4, 5; i = 1, 2, 3, 4, 5; (h, q, i)inthedesign Below is a plot of the standardized residuals against the treatment levels, a plot of the standardized residuals again the days (column block), and a plot of the standardized residuals again the order (row block). Each plot is randomly distributed about zero, thus indicating that there does not exist a problem of lack of fit. This model appears to be appropriate for this data. The SAS System

6 16:06 Sunday, August 5, 2001 Legend: A = 1 obs, B = 2 obs, etc.

Plot of Z*trt. 4 + | | | | | 2 +A | A | A A Z | A | A B | A 0 +A-----------A-----------A-----------A-----------A|B A A | | B A A |A A | A -2 + -+-----------+-----------+-----------+-----------+1 2 3 4 5 trt

2

Plot of Z*day. Legend: A = 1 obs, B = 2 obs, etc. 4 + | | | | | 2 + A | A | A A Z |A | B A | A 0 +B-----------B-----------------------A------------|A A B | |A A A A | A A | A -2 + -+-----------+-----------+-----------+-----------+1 2 3 4 5 day

Plot of Z*order. Legend: A = 1 obs, B = 2 obs, etc. 4 + | | | | | 2 + A | A |A A Z | A | A A A | A 0 +B-----------------------------------------------C|A A A A | |A A B | A A | A -2 + -+-----------+-----------+-----------+-----------+1 2 3 4 5 order

3

Below is a plot of the standardized residuals vs. the predicted values. Again, the residuals appear to be randomly distributed about zero. There do not appear to be any strong increasing or decreasing trends to the data. Therefore, we conclude that the assumption of equal variances is valid. Plot of Z*pred. Legend: A = 1 obs, B = 2 obs, etc. 4 + | | | | | 2 + A | A | A A Z | A | A A A | A 0 +---------------------A--B--A--A------------------| B A A | | A A AA | A A | A -2 + --+------------+------------+------------+--------60 80 100 120 pred A normal probability plot of the residuals is shown below. The plot idicates a relatively straight line, indicating that the normality assumption is reasonable. Plot of Z*nscore. Legend: A = 1 obs, B = 2 obs, etc. 4 + | | | | | | | | | | | 2 + | A | | A | | A A Z | | A | | A AA | | A 0 +-----------------------AAAA-A--------------------| AAA A | | | | A A AA | | A A | |A | -2 + | -+-----------+-----------+-----------+-----------+-2 -1 0 1 2 Rank for Variable Z 4

(b) A plot of the adjusted data vs the treatment levels is presented below. Based on this plot, it appears that Professor Wardrop’s scores are lower for treatment 1 than for the other four treatmend; Professor Wardrop’s scores are approximately equal for treatments 2,3,4, and 5, after adjusting for the two blocking factors. Plot of yadj*trt. Legend: A = 1 obs, B = 2 obs, etc. 120 + | | A | | A B | B 100 + B A B |A | B A A yadj | B | A | 80 + A A |B |A | |A | 60 + -+-----------+-----------+-----------+-----------+1 2 3 4 5 trt (c) The analysis of variance table produced by SAS is presented below. The GLM Procedure Dependent Variable: Y Source Model Error Corrected Total

DF 12 12 24

R-Square 0.700738 Source order day trt

Sum of Squares 4094.720000 1748.720000 5843.440000 Coeff Var 12.93584

DF 4 4 4

Type III SS 514.240000 1711.440000 1869.040000

Mean Square 341.226667 145.726667

Root MSE 12.07173 Mean Square 128.560000 427.860000 467.260000

F Value 2.34

Pr>F 0.0774

Y Mean 93.32000 F Value 0.88 2.94 3.21

Pr>F 0.5033 0.0661 0.0523

(d) To evaluate whether or not blocking was effective, we must compare the mean square for each blocking factor to the msE. For the blocking factor of day, msφ/msE = 427.86/145.726667 = 2.9360 which is relatively large. So we would conclude that the day block was effective. Similarly, for the blocking factor of order, msθ/msE = 128.56/145.72667 < 1. Since the ratio is less than 1, we would conclude that the use of order for blocking was not effective. 5

(e) Using Scheffe’s Method, simultaneous 95% confidence intervals for all pairwise comparisons as well as the ”music vs. no music” and ”game sounds vs. no game sounds” contrasts are s X X X d i τi ∈ di τî ± w Vd ar( di τî ) i

i

i

p √ p where w = (v − 1)Fv−1,bc−b−c−v+2,α and, for these data, w = 4 ∗ F4,12,.05 = 4 ∗ 3.2592 = 3.6106 . From SAS, the 95% simultaneous confidence intervals for all pairwise differences are: τ1 − τ2 τ1 − τ3 τ1 − τ4 τ 1 − τ5 τ2 − τ3 τ2 − τ4 τ2 − τ5 τ3 − τ4 τ3 − τ5 τ4 − τ5

∈ (−49.366574, 5.766574) ∈ (−45.366574, 9.766574) ∈ (−43.566574, 11.566574) ∈ (−52.566574, 2.566574) ∈ (−23.566574, 31.566574) ∈ (−21.766574, 33.366574) ∈ (−30.766574, 24.366574) ∈ (−25.766574, 29.366574) ∈ (−34.766574, 20.366574) ∈ (−36.566574, 18.566574)

Since each of the confidence intervals contains zero, we conclude that there does not exist a statistically significant difference between any of the treatments, at an overall level of 95% The confidence interval for the ”music vs. no music” contrast 31 (τ1 + τ2 + τ3 ) − 12 (τ4 + τ5 ) is: s X X X ar( di τÎ )} d i τi ∈ { di τî ± w Vd i

=

i

i

− 7.30 ± 3.6106 ∗ 4.92826316 = (−25.0942, 10.4942) .

Since this confidence interval contains zero, we conclude that there does not exist a statistically significant difference between the music and no music treatment conditions. And the confidence interval for the ”games sound vs. no game sound” contrast 14 (τ1 + τ2 + τ3 + τ4 ) − τ5 is: s X X X d i τi ∈ { di τî ± w Vd ar( di τÎ )} i

=

i

i

− 11.1 ± 3.6106 ∗ 6.03586503 = (−32.8934, 10.6934) .

Since this confidence interval contains zero, we conclude that there does not exist a statistically significant difference between the games sounds and no game sounds treatment conditions.

The GLM Procedure Least Squares Means Adjustment for Multiple Comparisons: Scheffe

trt 1 2 3 4 5

LSMEAN Number 1 2 3 4 5

Y LSMEAN 77.200000 99.000000 95.000000 93.200000 102.200000

6

Least Squares Means for effect trt Pr > |t| for H0: LSMean(i)=LSMean(j)

i/j 1 2 3 4 5

Dependent Variable: Y 2 3 0.1528 0.3051 0.9903 0.9903 0.9621 0.9996 0.9958 0.9208

1 0.1528 0.3051 0.4014 0.0831

trt 1 2 3 4 5

i 1 1 1 1 2 2 2 3 3 4

Y LSMEAN 77.200000 99.000000 95.000000 93.200000 102.200000

j 2 3 4 5 3 4 5 4 5 5

4 0.4014 0.9621 0.9996

5 0.0831 0.9958 0.9208 0.8409

0.8409

95% Confidence Limits 65.437370 88.962630 87.237370 110.762630 83.237370 106.762630 81.437370 104.962630 90.437370 113.962630

Least Squares Means for Effect trt Difference Simultaneous 95% Between Confidence Limits for Means LSMean(i)-LSMean(j) -21.800000 -49.366574 5.766574 -17.800000 -45.366574 9.766574 -16.000000 -43.566574 11.566574 -25.000000 -52.566574 2.566574 4.000000 -23.566574 31.566574 5.800000 -21.766574 33.366574 -3.200000 -30.766574 24.366574 1.800000 -25.766574 29.366574 -7.200000 -34.766574 20.366574 -9.000000 -36.566574 18.566574

The GLM Procedure Dependent Variable: Y

Parameter music game sound

Estimate -7.3000000 -11.1000000

Standard Error 4.92826316 6.03586503

t Value -1.48 -1.84

Pr > |t| 0.1643 0.0908

(f) Based on the multiple comparisons of part (e), I would conclude that there does not exist a statistically significant difference between any of the treatment conditions, and thus no treatment condition produces better scores than any other treatment condition. Thus, it does not matter which sound mode Professor Wardrop uses.

7

*Chapter 12; *Exercise 6 - Video Game Experiment; options linesize=72; data video; input order day trt Y; cards; 1 1 1 94 1 2 3 100 1 3 4 98 1 4 2 101 1 5 5 112 2 1 3 103 2 2 2 111 2 3 1 51 2 4 5 110 2 5 4 90 3 1 4 114 3 2 1 75 3 3 5 94 3 4 3 85 3 5 2 107 4 1 5 100 4 2 4 74 4 3 2 70 4 4 1 93 4 5 3 106 5 1 2 106 5 2 5 95 5 3 3 81 5 4 4 90 5 5 1 73 ; proc glm; classes order day trt; model Y = order day trt / solution; output out=resids predicted =pred residual=Z; estimate ’music’ trt 2 2 2 -3 -3 /divisor = 6; estimate ’game sound’ trt 1 1 1 1 -4 /divisor=4; lsmeans trt / pdiff=all cl adjust=scheffe; proc standard std=1.0; var Z; proc rank normal=blom; var Z; ranks nscore; proc plot; plot Z*pred Z*trt Z*order Z*day / vref=0 vpos=19 hpos=50; plot Z*nscore / vref=0 href=0 vpos=19 hpos=50; *second run;

8

data video2; input day dhat @@; lines; 1 5.8 2 -6.6 3 -18.8 4 -1.8 5 0 ; proc means mean; var dhat; data video3; input order ohat @@; lines; 1 12 2 4 3 6 4 -.4 5 0 ; proc means mean; var ohat *third run; data video4; set video; if day=1 then yadj=Y-(5.8+4.28); else if day=2 then yadj=Y-(-6.6+4.28); else if day=3 then yadj=Y-(-18.8+4.28); else if day=4 then yadj=Y-(-1.8+4.28); else if day=5 then yadj=Y-(0+4.28); if order=1 then yadj=yadj-(12-4.32); else if order=2 then yadj=yadj-(4-4.32); else if order=3 then yadj=yadj-(6-4.32); else if order=4 then yadj=yadj-(-.4-4.32); else if order=5 then yadj=yadj-(0-4.32); proc plot; plot yadj*trt / vpos=19 hpos=50;

9

Solution to Question 12.10, Quantity Perception Experiment a) The subjects in this study may possibly be representative of the students from The Ohio State University. However, we should be cautious about this conclusion since only those students who frequent the Ohio Union had the opportunity to take part in the study. There may be slection bias on the part of the experimenters in recruiting the students (e.g. recruiting studernts who look friendly and unhurried). Presumably, no student was allowed to return for a second attempt and, since students were not allowed to view the experiment in progress with a previous subject, they did not have the chance to the true number of candies in advance. The conclusions of the study may not be relevant to people in general since the students who are from the Ohio Union hallway are likely to be different from the general population in the country because of the difference in average age, education, etc. b) The following plots are residual plots for the model shown in part (c) with Y being (true number - guessed number)/(true number). The residuals are approximately normally distributed and have approximately the same variance for each treatment apart from two two outliers from treatment 2 (subjects 3 and 14) whose standardized residuals are around 4. The SAS System 3 Plot of Z*TRTMT. Legend: A = 1 obs, B = 2 obs, etc. 5 + | A | | Z | |A B |A B A B A | C A C A C A C D |D C E E B D E A D 0 +F-----D-----B-----------H-----------F-----J------|D C E E B D E A D | C A C A C A C D |A B A B A |A B | | | | A -5 + -+-----+-----+-----+-----+-----+-----+-----+-----+1 2 3 4 5 6 7 8 9 TRTMT

10

Plot of Z*SUBJ. Legend: A = 1 obs, B = 2 obs, etc. 5 + | A | | Z | | A A A | A A A A A A A | A A C B B A A B B B A A | A B B A B E A A B C C A A A B C B 0 +---D-A---B-B-A-D-A-C-A-A-A-C-C-B-B-C-B-----------| B C C A A B A C C B A C A A B B B | A B B B B A A B A B B A | A A A A A A A | A B | | | | A -5 + --+---------+---------+---------+---------+-------0 5 10 15 20 SUBJ If we remove the two outliers, the residual plots look much better: Plot of Z*TRTMT. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | A 2 +A A A A B |A B A A A A D |A B C D B C A B C |E C C C C C C B B 0 +B-----B-----B-----B-----D-----------H-----H------|E C C C C C C B B |A B C D B C A B C |A B A A A A D -2 +A A A A B | A | | -4 + | -+-----+-----+-----+-----+-----+-----+-----+-----+1 2 3 4 5 6 7 8 9 TRTMT

11

Plot of Z*SUBJ. Legend: A = 1 obs, B = 2 obs, etc. Z | 4 + | | | A 2 + A A A A A A | A A A B A B A A A | A A B B B B B B B A A B A | B B B B A A B B A B B B B A C 0 +---A-A-B-C-A-B-B---B-A-C-A-B---A-B-B-B-----------| B C A C A C A C B A A A B B A | A B A A B A B C C A B B | A A A B A A A A A A -2 + A A A A A A | A | | -4 + | --+---------+---------+---------+---------+-------0 5 10 15 20 SUBJ Plot of Z*NSCORE. Legend: A = 1 obs, B = 2 obs, etc. Z | | 4 + | | | | | | | A 2 + | ABAAA | | AECB | | CGGD | |BJIF 0 +-----------------------IJI-----------------------| FIJB| | DGGC | | BCEA | -2 + AAABA | | A | | | | | -4 + | | | -+-----------+-----------+-----------+-----------+-4 -2 0 2 4 Rank for Variable Z

12

c) The model for the experiment is the row-column-treatment model for the 2-replicate Latin square design: Yhqi = µ + θh + φq + τi + (φτ )qi + hqi , hqi ∼ N (0, σ 2 ), hqi ’s are mutually independent, h = 1 . . . 18; q = 1 . . . 9; i = 1 . . . 9, where µ is a constant, θh is the effect of the hth row block (sublect), φq is the effect of the qth column block (time order), τi is the effect of the ith treatment, (φτ )qi is the effect of the interaction of the qth time order and ith treatment, Yhqi is the random variable representing the (true number - guessed number)/(true number) for treatment i observed for subject h and time order q, and hqi is the associated random error. The analysis of variance table obtained from SAS (after omitting the two outliers) is The GLM Procedure Dependent Variable: Y Source SUBJ ORDER TRTMT ORDER*TRTMT Error Corrected Total

DF 17 8 8 63 63 159

Type III SS 2.02359502 0.31489067 1.09980189 1.38403721 1.08946693 9.09581904

Mean Square 0.11903500 0.03936133 0.13747524 0.02196884 0.01729313

F Value 6.88 2.28 7.95 1.27

Pr > F <.0001 0.0330 <.0001 0.1724

We choose overall significant level α = 0.04, and split it into divide it between four hypotheses tests. i) H0intera : no interaction between time order and treatment. We reject H0intera if msOT /msE > F64,64,.01 = 1.8004. Since msOT/msE = 0.0220/0.0173 = 1.27 < 1.8004, we fail to reject H0 at significant level α = 0.01 and there is not sufficient eveidence to conclude an interaction between column block factor (order) and treatment. ii) H0T : all 9 treatments have the same efffect on the response. Reject H0T if msT/msE > F8,64,.01 = 2.8027. From the SAS output above, msT/msE = 0.1375/0.0173 =7.95 > 2.8027, we reject H0 at significance level α = 0.01 and conclude that the combination of true number of candies and color do not all have the same effect on the abilities of subjects to count them. iii) We would now like to test H0 : Quadratic trend of number equals zero and we would Reject H0 if ssC/msE > F1,64,.01 = 7.0483. Similarly, we would like to test H0 : Linear trend of color equals zero and we would Reject H0 if ssC/msE > F1,64,.01 = 7.0483. S When the two outliers are removed, the usual coeffcients in the Appendix are no longer the correct cooefficents due to the unequal number of observations per treatment. We would need to calculate the coefficients specially for this situation. If we make all tests including the two outliers, we find the same conclusions as above and the linear trend due to number is significantly different from zero at level 0.01.

13

Source SUBJ ORDER TRTMT ORDER*TRTMT Error Corrected Total

DF 17 8 8 64 64 161

Type III SS 1.97988929 0.32674684 1.09932801 1.54244613 1.57607923 9.61972136

Mean Square 0.11646408 0.04084336 0.13741600 0.02410072 0.02462624

F Value 4.73 1.66 5.58 0.98

Pr > F <.0001 0.1261 <.0001 0.5342

Contrast NUMBER LINEAR NUMBER QUADRATIC

DF 1 1

Contrast SS 0.90715809 0.06966527

Mean Square 0.90715809 0.06966527

F Value 36.84 2.83

Pr > F <.0001 0.0975

Although we are not testing whether the block effects equal zero or not, by comparing the mean square of ORDER and SUBJECT with mean square error, we conclude that it was worthwhile to create the two block factors in the model to reduce the experimental error. d) The analysis of variance in of number and color is given below (excluding the two outliers). The conclusions about the blocking factors remain the same. Using an overall significance level of 0.06 (0.01 for each test), we conclude that there is no significant interaction effect of ORDER by NUMBER by COLOR. There is no NUMBER by COLOR interaction effect. The interaction effects of ORDER by NUMBER and ORDER by COLOR are not significant at individual levels 0.01. The different numbers of candies do affect the subjects abilities to count them (consistent with the linear trend found above), whereas the different colors have no effect. The GLM Procedure Dependent Variable: Y Source DF SUBJ 17 ORDER 8 NUMBER 2 COLOR 2 NUMBER*COLOR 4 ORDER*NUMBER 16 ORDER*COLOR 16 ORDER*NUMBER*COLOR 31 Error 63 Corrected Total 159

Type III SS 2.02359502 0.31712431 0.94356973 0.03389871 0.08376415 0.45154695 0.47166600 0.46256507 1.08946693 9.09581904

Mean Square 0.11903500 0.03964054 0.47178487 0.01694935 0.02094104 0.02822168 0.02947912 0.01492145 0.01729313

F Value 6.88 2.29 27.28 0.98 1.21 1.63 1.70 0.86

Pr > F <.0001 0.0319 <.0001 0.3809 0.3151 0.0863 0.0689 0.6676

e) There are many possible answers to this question. One supplied by a student is as follows: Although the hypothesis test says the color itself does not have different effects on the response, from the standardized residual plot in part b), it seems that response from treatment 2, 5 and 8 seems have larger variance than the others in general. Thus one possible followup experiment we might want to study is that does ’Orange’ color especially influence the guessed number of candies. Checklist i) Define the objective of the experiment. The experiment is to study the effects of orange and other silimar shades with green on the accuracy of guessed number of candies for a certain true number of candies. ii) Identify all sources of variation. Timing of the experiment. We assume people do not concentrate well when they are hungry or in hurry, thus we select mid-afternoon for the study. Subjects should not have previously participated in the 14

study, so we will select a location at the other end of the campus, but still frequented by many different types of student. So we select the mall outside the bookstore and the central classroom building. Subjects. As before, subjects should not be allowed to view the experiment in progress with previous subjects in order to prevent them build up the knowledge of the true number of candies. Treatment factors and levels. Since we want to focus the study on the effect of different colors, we still have the two treatment factors as NUMBER and COLOR. But we will have only 2 levels of NUMBER 13 and 37, and 4 levels of COLOR as: yellow, brown, orange and green. Thus there are total 8 treatment combinations. Block factors. As before, we still have two block factors: subject and order. Since we have 8 treatment combinations now. We will set 8 levels of order and 16 subjects. iii) Still as before we will use 2-replicate Latin square design and assign the experimental unit to the treatment combination as before.

15

Solutions 2r1l2c

Overview 4q3b3c

More details 26j3b

Related Documents 171j1w

Solutions 2r1l2c

Solutions 2r1l2c

Solutions 2r1l2c

Solutions Hw5 6o6l4f

Forecasting Solutions 6vo4v

Rf Solutions 2r13n

More Documents from "Adnan Ali Raza" 4h6750

Solutions 2r1l2c

Molarity Calculator And Normality Calculator For Acids And Bases _ Sigma-aldrich.pdf 4w56s

003.030-evans Vanodine Handsan Alcohol Gel Hand Rub Pump Msds 5k4s6r

Crop Calendar Pakistan 5y6u5n

Survey Field Notes 6f3l2c

Pte Essays With Solution 5k5q4c