Ap Statistics Practice Exam 531j67
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Statistics Independent Study Page 1 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Multiple-Choice Section
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C
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 2 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Free-Response Section Introduction These questions simulate the Free-Response section of the AP Exam. Questions 1-5 should take about 10-15 minutes each. Question 6 is an "Investigative Task," and should take about 20-30 minutes. You should do all that you can do in no more than 90 minutes. 1. A. Answer:
C U M R E L F R E Q
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
1234567 NUMBER OF ATTEMPTS
B. Answer: µ = (1)(40/75) + (2)(12/75) + (3)(8/75) + (4)(9/75) + 5(5/75) + 6(0/75) + 7(1/75) = 2.08 attempts. The median is found at the fiftieth percentile. More than half ed on the first trial, so the median is 1. 2. A. Answer: Blocks would be desirable here. You believe that men and women might have differing reactions to the medication, so it would be appropriate to block by gender here. You are also concerned with differences between those 60 and over and those under 60, so you would also need to divide those under 60 and those 60 and over into different blocks.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 3 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B. Answer: Divide your 180 volunteers into two groups based on their gender. Then break each of these two groups into two additional blocks of those 60 and over and those under 60. This results in four blocks: men under 60, men 60 and over, women under 60, and women 60 and over. For the sake of discussing the design, assume there are equal numbers in each block. This gives you four blocks of 45 people each. Within each block, randomly assign 1/3 of the subjects to a control group that will receive a placebo, 1/3 to a treatment group that will receive 350 mg tablets, and 1/3 to a treatment group that will receive 500 mg tablets. The study should be double-blind: Neither the subjects nor the person istering the tests should know which subjects are receiving which treatment. After a designated period of time, compare pain-relief scores for each of the three groups within each block. C. Answer: In any one block, there are approximately 15 people that receive the 350 mg pill and 15 that receive the 500 mg pill. Compute an average relief score for each group. If randomization has been done correctly, the assumption that the two groups are independent is reasonable. You must also check to see that the distribution of scores for each group does not indicate that the samples might have been drawn from non-normal populations. If these conditions are satisfied, you would want to conduct a two-sample t test for the difference between population means. The computation you would use for the standard error of the difference would depend on whether you were using the conservative method, whether you pool your estimates of the population variance, or whether you use computer software. 3. Answer: H0 : µ d = 0 (The mean difference between the before and after scores is 0.)
H a : µ d > 0 (The mean difference is greater than zeroscores were higher after training.) Since the scores are matched (that is, there are two different measurements for each individual), a one-sample t test is appropriate for the differences between the sample values. A boxplot of the data shows no outliers. There is some skewness to the left, but not enough to conclude that this sample is not reasonably normal. We can use a onesample t procedure. x = 1.333, s = 1.969, n = 12 t =
1.333 − 0 = 2.345 1.969 12
df = 11 è .01 < p < .02 (On the TI-83, you would find a P-value of .019.) _____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 4 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers This P-value is quite low and gives us strong evidence against the null hypothesis. It is reasonable to conclude that the program was effective in improving mathematics skills in the subjects. Note: If you used a significance level of .01, you would reach a different conclusion. 4. A. Answer:
æ 51ö (6/51)(5/50)(4/49)(3/48)(2/47)(1/46) = 1 /çç ÷÷ = 1/18009460 = .0000000555. è6 ø B. Answer: Now, the probability of a match remains constant at 6/51. P(at least one match in first five drawings) = 1 – P(no matches in the first five drawings) 5
æ 45 ö =1– ç ÷ = .465. è 51 ø C. Answer: (6/51)(1 – 6/51)4 = .071 5. A. Answer: It is a one-sided test, since the school believes that their students will do better. H0 : p = .6 (The proportion of ing grades is .6.)
Ha : p > .6 (The proportion of ing grades is larger than .6.) A one-proportion z-test is appropriate here. We are justified in using a z-test here because 35(.4) and 35(.6) are both larger than 10. ˆ = 26/35 = .743, z = p
.743 − .6 (.6)(.4) 35
= 1.73 è P-value = 1 – .9582 = .0418
This value of P is less than the significance level of .05, so we have good evidence to reject our null hypothesis. The claim that the school's students do better than the national average of .6 is ed by these data.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 5 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B. Answer: The worst thing that could happen in of the school's claim is that the missing .722 − .6 ˆ = 26/36 = .722; z = = 1.494 student would not have ed. In this case, p .6(.4) 36 è P-value = 1 – .9324 = .0676. This value is greater than the significance level of .05. Under these conditions, we would not have rejected the null hypothesis and would conclude that the evidence was unconvincing that the school is actually doing better than the national average. 6. Answer: A histogram of these data (below) shows a pattern that seems to be quite normal.
If we assume the data are from a normal distribution, we can use a χ 2 goodness-of-fit test to compare the observed frequencies in each interval with the expected frequencies. We need to find the area under the normal curve for each interval by using either a calculator or standard z-score techniques. Assuming that the distribution from which the sample was drawn has a mean of 275 and a standard deviation of 47, we would find, for example, that the area corresponding to the interval 189.5 – 219.5 could be calculated as follows: z 219.5 =
219.5 − 275 = –1.18 è Upper area = .119 47
z189.5 =
189.5 − 275 = –1.82 è Lower area = .034 47
Area in the interval (189.5, 219.5) = .119 – .034 = .085. (If you use the TI-83 you will get .084.) Expected frequency in the interval (189.5, 219.5) = .085(500) = 42.5. (If you use the TI-83 you will get 42.19.)
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 6 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Proceeding similarly for each interval and using the results from the TI-83, we can add a third column to the table: Cholesterol Level
Less than 189.5 189.5 – 219.5 219.5 – 249.5 249.5 – 279.5 279.5 – 309.5 309.5 – 339.5 339.5 – 369.5 369.5 – 399.5 Over 399.5
Observed Frequency 17 41 80 134 118 71 24 13 2
Expected Frequency 17.22 42.19 87.44 122.21 115.20 73.24 31.40 9.07 2.02
For these data, X 2 = 5.39, df = 9 – 1 = 8 è p > .25 (p = .715 on the TI-83). A P-value this high indicates that a set of observations such as this is quite likely if the sample were in fact drawn from a normal population. Thus we have no evidence to argue against the claim that the cholesterol levels of American men are normally distributed.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 7 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Scoring
To determine your score, use the following guidelines. Multiple-Choice Section
Determine your score on this section as follows: Score = number of correct multiple-choice questions – 1/4(number of incorrect multiplechoice questions). Free-Response Section
Questions 1–5 are graded on a four-point scale. Question 6 is graded on an eight-point scale. Determine your score on this section as follows: Score = question 1 + question 2 + question 3 + question 4 + question 5 + question 6. Overall Score
Overall score = (multiple-choice score • 7) + (free-response score • 10). You can roughly approximate your score on the AP Exam by using the table below. Overall Score 390 – 560 305 – 389 225 – 304 160 – 224 0 – 159
_____________
AP Approximation 5 4 3 2 1
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
_____________
1.
C
21.
B
2.
B
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D
3.
C
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A
4.
D
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E
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D
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C
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B
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9.
D
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B
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A
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11.
D
31.
D
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A
32.
E
13.
E
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A
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C
34.
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15.
A
35.
A
16.
B
36.
E
17.
E
37.
C
18.
A
38.
B
19.
E
39.
D
20.
E
40.
C
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 2 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Free-Response Section Introduction These questions simulate the Free-Response section of the AP Exam. Questions 1-5 should take about 10-15 minutes each. Question 6 is an "Investigative Task," and should take about 20-30 minutes. You should do all that you can do in no more than 90 minutes. 1. A. Answer:
C U M R E L F R E Q
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
1234567 NUMBER OF ATTEMPTS
B. Answer: µ = (1)(40/75) + (2)(12/75) + (3)(8/75) + (4)(9/75) + 5(5/75) + 6(0/75) + 7(1/75) = 2.08 attempts. The median is found at the fiftieth percentile. More than half ed on the first trial, so the median is 1. 2. A. Answer: Blocks would be desirable here. You believe that men and women might have differing reactions to the medication, so it would be appropriate to block by gender here. You are also concerned with differences between those 60 and over and those under 60, so you would also need to divide those under 60 and those 60 and over into different blocks.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 3 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B. Answer: Divide your 180 volunteers into two groups based on their gender. Then break each of these two groups into two additional blocks of those 60 and over and those under 60. This results in four blocks: men under 60, men 60 and over, women under 60, and women 60 and over. For the sake of discussing the design, assume there are equal numbers in each block. This gives you four blocks of 45 people each. Within each block, randomly assign 1/3 of the subjects to a control group that will receive a placebo, 1/3 to a treatment group that will receive 350 mg tablets, and 1/3 to a treatment group that will receive 500 mg tablets. The study should be double-blind: Neither the subjects nor the person istering the tests should know which subjects are receiving which treatment. After a designated period of time, compare pain-relief scores for each of the three groups within each block. C. Answer: In any one block, there are approximately 15 people that receive the 350 mg pill and 15 that receive the 500 mg pill. Compute an average relief score for each group. If randomization has been done correctly, the assumption that the two groups are independent is reasonable. You must also check to see that the distribution of scores for each group does not indicate that the samples might have been drawn from non-normal populations. If these conditions are satisfied, you would want to conduct a two-sample t test for the difference between population means. The computation you would use for the standard error of the difference would depend on whether you were using the conservative method, whether you pool your estimates of the population variance, or whether you use computer software. 3. Answer: H0 : µ d = 0 (The mean difference between the before and after scores is 0.)
H a : µ d > 0 (The mean difference is greater than zeroscores were higher after training.) Since the scores are matched (that is, there are two different measurements for each individual), a one-sample t test is appropriate for the differences between the sample values. A boxplot of the data shows no outliers. There is some skewness to the left, but not enough to conclude that this sample is not reasonably normal. We can use a onesample t procedure. x = 1.333, s = 1.969, n = 12 t =
1.333 − 0 = 2.345 1.969 12
df = 11 è .01 < p < .02 (On the TI-83, you would find a P-value of .019.) _____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 4 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers This P-value is quite low and gives us strong evidence against the null hypothesis. It is reasonable to conclude that the program was effective in improving mathematics skills in the subjects. Note: If you used a significance level of .01, you would reach a different conclusion. 4. A. Answer:
æ 51ö (6/51)(5/50)(4/49)(3/48)(2/47)(1/46) = 1 /çç ÷÷ = 1/18009460 = .0000000555. è6 ø B. Answer: Now, the probability of a match remains constant at 6/51. P(at least one match in first five drawings) = 1 – P(no matches in the first five drawings) 5
æ 45 ö =1– ç ÷ = .465. è 51 ø C. Answer: (6/51)(1 – 6/51)4 = .071 5. A. Answer: It is a one-sided test, since the school believes that their students will do better. H0 : p = .6 (The proportion of ing grades is .6.)
Ha : p > .6 (The proportion of ing grades is larger than .6.) A one-proportion z-test is appropriate here. We are justified in using a z-test here because 35(.4) and 35(.6) are both larger than 10. ˆ = 26/35 = .743, z = p
.743 − .6 (.6)(.4) 35
= 1.73 è P-value = 1 – .9582 = .0418
This value of P is less than the significance level of .05, so we have good evidence to reject our null hypothesis. The claim that the school's students do better than the national average of .6 is ed by these data.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 5 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers B. Answer: The worst thing that could happen in of the school's claim is that the missing .722 − .6 ˆ = 26/36 = .722; z = = 1.494 student would not have ed. In this case, p .6(.4) 36 è P-value = 1 – .9324 = .0676. This value is greater than the significance level of .05. Under these conditions, we would not have rejected the null hypothesis and would conclude that the evidence was unconvincing that the school is actually doing better than the national average. 6. Answer: A histogram of these data (below) shows a pattern that seems to be quite normal.
If we assume the data are from a normal distribution, we can use a χ 2 goodness-of-fit test to compare the observed frequencies in each interval with the expected frequencies. We need to find the area under the normal curve for each interval by using either a calculator or standard z-score techniques. Assuming that the distribution from which the sample was drawn has a mean of 275 and a standard deviation of 47, we would find, for example, that the area corresponding to the interval 189.5 – 219.5 could be calculated as follows: z 219.5 =
219.5 − 275 = –1.18 è Upper area = .119 47
z189.5 =
189.5 − 275 = –1.82 è Lower area = .034 47
Area in the interval (189.5, 219.5) = .119 – .034 = .085. (If you use the TI-83 you will get .084.) Expected frequency in the interval (189.5, 219.5) = .085(500) = 42.5. (If you use the TI-83 you will get 42.19.)
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 6 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Proceeding similarly for each interval and using the results from the TI-83, we can add a third column to the table: Cholesterol Level
Less than 189.5 189.5 – 219.5 219.5 – 249.5 249.5 – 279.5 279.5 – 309.5 309.5 – 339.5 339.5 – 369.5 369.5 – 399.5 Over 399.5
Observed Frequency 17 41 80 134 118 71 24 13 2
Expected Frequency 17.22 42.19 87.44 122.21 115.20 73.24 31.40 9.07 2.02
For these data, X 2 = 5.39, df = 9 – 1 = 8 è p > .25 (p = .715 on the TI-83). A P-value this high indicates that a set of observations such as this is quite likely if the sample were in fact drawn from a normal population. Thus we have no evidence to argue against the claim that the cholesterol levels of American men are normally distributed.
_____________
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
Statistics Independent Study Page 7 of 7 Mixed Practice with Multiple-Choice and Free-Response Questions Answers Scoring
To determine your score, use the following guidelines. Multiple-Choice Section
Determine your score on this section as follows: Score = number of correct multiple-choice questions – 1/4(number of incorrect multiplechoice questions). Free-Response Section
Questions 1–5 are graded on a four-point scale. Question 6 is graded on an eight-point scale. Determine your score on this section as follows: Score = question 1 + question 2 + question 3 + question 4 + question 5 + question 6. Overall Score
Overall score = (multiple-choice score • 7) + (free-response score • 10). You can roughly approximate your score on the AP Exam by using the table below. Overall Score 390 – 560 305 – 389 225 – 304 160 – 224 0 – 159
_____________
AP Approximation 5 4 3 2 1
© Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of ed s only. No portion of these materials may be reproduced or redistributed in any form without the express written permission of Apex Learning Inc.
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