Errata for AP Statistics: Preparing for the Advanced Placement Examination
 

by Jim Bohan with Beth Chance
Amsco School Publications, Inc
 

Please email your corrections to mathmanjfb@aol.com or bchance@calpoly.edu


(page numbers refer to location of error)

p. 19, second paragraph

The formula for the weighted mean should read 


p. 82, problem 2

Answer C should be the HS principals from large urban school districts.
p. 108, problem 2
(B) P(A|B) = P(A)
(C) P(A|B)=P(B)
p. 117, example 5

                 Last line should read (30/36)(30/36)(6/36) = 25/126

p. 137, 3rd line from end of example 5

Should read P(X=4 and Y=0) =? P(X=4).P(Y=0)
p. 139, Section 5.8 Review problem 3
The AMC-12 has a total of 150 points but to qualify for the AIME test, you must score 100 or more.
p. 145, problem 1
Probability of finding a special prize in a box of popcorn should be .20.
p. 146, Section 5.10, problem 4
Change failure rate to 20% to match answer in back.
p. 151, multiple-choice question
 In item II, replace "may" by "must" to read "The standard deviations of two different samples from the same population must be the same."
p. 167, free response problem 2
Change the problem to read "Find the minimum sample size so that the probability that a sample mean is less than 55 (for this distribution with mean 57 and standard devation 5.6) equals .01."  See corrections to solution as well.
p. 175-6, formula for exact df
Should read:
p. 177, Multiple-Choice, problem 4
"The probability that samples of these sizes have a difference in means of 2.8 or more..."
p. 181, problem 3
The mean overhead cost for the pizza stores should be $764.50, not $763.50.
p. 198, problem 3
margin of error is 2.1, not 2.1%; use n=30 weeks as the sample of the population to compute the margin of error.
p. 208, two-sample t-test (unpooled)
formula approximating degrees of freedom should be df=min(n1,n2) -1
p. 220, problem 8
change (C) to read "The 99% confidence interval does not contain the mean specified by the null hypothesis"
p. 229
In the formula for two-tailed test, the term (za/2+zb) should be squared.
p. 233, example 3
 In the confidence interval formula, t8 for 90% confidence should equal 1.860 instead of 1.397.
p. 234, example 4
Standard error of slope should be .0488

p. 245, problem 4 (D) Increase sample size and decrease confidence level p. 245, problem 9
Answer E should be Ho: mb = mg; Ha: mbımg.  This leaves C as the correct answer.
Also there should be no hats on the subscripts for the hypotheses.
p. 265, problem 12
the units on smpg should be mi/gal
p. 266, problem 21
Answer A should read "For a fixed p, the variance increases as n increases".
p. 268, problem 32
Change (D) to read "We expect the 95% confidence interval of the sample result would contain the expected population value."
p. 271, problem 3, question a
Change the question to read "In a test of significance to determine the validity of the company's claim, why is it inappropriate to use the normal approximation to the binomial? Explain your answer."
p. 271, problem 4, question d
Change the question to read "Estimate the approximate life of the car in the sample whose purchase price was $25,000."
p. 273, problem 6
Part (e) was not included in the scoring guide.
P. 274, problem 2
Answer C should read "Consecutive values of a discrete variable may or may not differ by a fixed(constant) amount."
p. 277, problem 18
Change answer A to read "Honesty of the respondent".
p. 298, problem 30 Question should read "Which of the following is a valid conclusion based on this information?" p. 299, problem 32
Start question as "In a test of the equality of the proportions..."
p. 304, Model Exam 5, problem 3
Answer C should read "bimodal".
p. 305, problem 11
The equation should read  = 10.2 + 4.15x
p. 305, problem 14
(A) Long distance service is predicted to cost 15 cents per minute.
(B) The monthly fixed cost of long distance service is, on average, $4.70
p. 306, problem 22
Add "The restaurant further claims that you have a chance to win on every visit."
p. 327, Multiple Choice, problem 4
Note, that if the z-score is not rounded in the middle of the calculation, then the student must increase his score by 9.02 points.  To ensure improvement (not just staying the same) then this could be rounded up, resulting in a different answer.
p. 328, Multiple Choice, problem 5
Change explanation to read that the percentile of a z-score of 1 is approximately 84%
p. 329, Multiple Choice, problem 5
The minimum is at 32.
p. 329, Multiple Choice, problem 6 standard deviation is 19.49 and variance is 381.94 p. 329, Free Response, problem 2 numerator should be a subtraction p. 329, Investigative Task, problem 1 p. 332, Free Response Question 4
Using height to predict weight, the regression equation is: weight = -91.9 + 3.25 height
p. 332, Free Response Question 5
The y-intercept should refer to the value of -91.9 not -58.
p. 332, Free Response Question 6
The correct prediction is -91.9+3.24(65)=119.2
p. 332, Multiple Choice, Question 1
Answer should be D, as the only option that is not a correct conclusion.
p. 332, Multiple Choice, Question 4
Answer should be E, none of these lines of fit would produce a residual plot similar to the given residual plot.
p. 341, Multiple-Choice Question 5
Choice should read D, explanation is correct.
p. 342, Multiple Choice Question 2
P(A|B) = P(A) if A and B are independent
P(A|B) = P(B) if P(A and B) = [P(B)]2
p. 342, Multiple-Choice Question 3
Answer should read D.  One explanation: P(A|B) = P(A and B)/P(B) = .5/.3 > 1, not a legitimate value.  In fact, P(A and B) must always be less than (or equal to) P(B) for this fraction to be < 1.  Similarly, need P(A and B) to be less than or equal to P(A).
p. 342, Section 5.2, Free Response Question 3
P(sum=8 and both even) = 3/36, should answer is 5/36 + 9/36 - 3/36 = 11/36
p. 342, Section 5.2, Free Response Question 7
4/52(4/51) = .006 not .06.
p. 344, Section 5.6, Multiple Choice Question 2
z = -1.28 = (38-50)/or z = 1.28 = (62-50)/s
p. 345, Section 5.8, Free-Response Question 4b
Note: to obtain the probabilities in (a), we use the fact that the random variables are independent as stated in the problem.

E(2L+M) = 5(.01)+6(.01)+7(.04)+8(.08)+9(.07)+10(.15)+11(.10)+12(.22)+13(.08)+14(.24) = 11.3 = 2E(L)+E(M)
V(2L+M) = (5-11.3)2(.01) + (6-11.3)2(.01) + (7-11.3)2(.04) + (8-11.3)2(.08) + (9-11.3)2(.07) + (10-11.3)2(.15) + (11-11.3)2(.10) + (12-11.3)2(.22) + (13-11.3)2(.08) + (14-11.3)2(.24)  = 5.01
= 4V(L)+V(M) since the two random variables are independent
SD(2L+M)=sqrt(5.01) = 2.238

p. 346, Chapter 5 Assessment, problem 4 Complement probability: P(at least one works) = 1-P(none work) p. 347, problem 9
B is the correct answer, but the explanation is off. Delete the "not"s in the independence explanation.
p. 347, Investigative task, problem 2 Graph is wrong. Should be a rectangle bounded on the x-axis by (8.5, 17.25) and 1/(17.25-8.5)=1/8.75 on the y-axis. The shade rectangle should have x-bounds of 12 and 15. p. 350, Section 6.5, free-response quesetion 2
The z value given is incorrect, should be -.253. For z=-.253, any value of n will work.  If the problem statement is changed as suggested above (value of n so probability that sample mean is less than 55 is .01) the answer is n>42.56 or 43.
p. 352, Section 6.7, multiple-choice question 4
Explanation should being P(1-2> 2.8)...
p. 353, problem 9
Explanation is correct, but answer should be D.
p. 357, Section 7.4, Free-response question 2
final statement should read P(>2.8| m=2.4)
p. 357, Section 7.4, Free-response, question 4
P(boys’ sample times would be at least 4.9 seconds faster than girls’ times if the...)
p. 358, multiple choice, question 8
The correct answer should be "The 99% confidence interval does not contain the mean specified by the null hypothesis"
p. 359, step 10
The demoninator of the z statistic should be square root of 1/6(5/6)/100
p. 361, step 5
Ha: (words) There is a difference by gender in science placement exam performance in 1988
p. 362, Section 7.5, Free Response, problem 13
z-statistic for a two-sided test would have to exceed 1.96. So n> (1.96)2(.25)/(.03)2 = 1068
p. 362, Section 7.5, Free Response, problem 14b
p-value=P(>65 or < 59 if m=62)

p. 362, Section 7.6, Multiple Choice, problem 1

Answer should be C

p. 363, Free Response, problem 1

The answer to 1b is missing: The slope indicates that the predicted temp 2 increases by an average of 1.05651 degrees for each additional degree of temp 1.
The printed answer of 1b belongs to question 1c.
The printed answer to 1c belongs to question 1d.
In answer 1d, there is a typo: 90-2=88
p. 363, Free response, problem 2 The answer for 2b is missing: The total score is about 4 points higher on average for each additional point of the problem-solving subscore.
The printed answer of 2b belongs to question 2c. Also, R2=62% (in this course, we will use R2 not R2(adj))
The printed answer of 2c belongs to question 2d.
The printed answers for questions 2d and 2d belong to question 2e.  Also, critical value when df=34 is 2.032.
The printed answer of 2e belongs to question 2f.
Note, without residual plots, we cannot truly "verify" the assumptions for this problem.
p. 368, item 9
2(.25) = .5 (end of p-value calculation)
p. 369
3. Check of Assumptions: For sample sizes, the smallest is 50(.40) = 20 > 5
7. Test statistic: =[.4(100)+.44(50)]/150 = .4133, so then z = -.469
8. Sketch should shade -.469 and below
9. p-value = .319
p. 374, Model Exam 1, problem 4
Explanation is correct, but answer should be A.
p. 374, problem 7 Answer is A p. 374, problem 8
Technically if you allow the two points to fall at the same x value, then r is undefined and if they have the same y value, then r=0.
p. 375, problem 24
Answers should be A. Explanation:  Since there is no replacement of the kings in the deck, the maximum number of rotten fruit is 4
for every shipment of 24 fruit.  Since the cards are dealt without replacement, the probability of a rotten fruit is not constant.
p. 375, problem 28
Answer should be D (explanation is correct)
p. 375, problem 32
Answer should be B.  Answer D is correct since the two-sided p-value would be .032. We would fail to reject the hypothesized value at the .05 level, indicating that the hypothesized value is contained in a 95% confidence interval. Note, this is exactly true when analyzing means, but you could see a discrepancy between the CI and test with proportions since a different formula is used for the standard deviation.
p. 375, problem 36
Answer should be E, with explanation that an expected count is less than 5 and therefore these data do not meet the requirements for the chi-square test.
p. 375, problem 38
Since 14.5 is the claimed value for the population standard deviation, we could recommended using the t test (with s=12.5 hours) instead of the z test. However, in order to use either the z-test or the t-test here, we would need to know/assume/believe that the population distribution is normal. There is no indication of this in the problem statement.  This leaves E as the correct answer.

p. 376, problem 3, part (c)

Can use Ho: p <= .02, Ha: p > .02 with p-value = .056

p. 377, problem 6

Part (e) was not included in the scoring guide.  However, answers will vary. Most students will obtain an interval that contains m which should not surprise them - 95% of all intervals will contain m. An interval that does not contain m would be surprising.
p. 378, problem 2
Add to the explantion "On the other hand, the numbers of houses in a set of housing developments may be any set of integers, consecutive values of which do not have to differ by a fixed amount."
p. 378, problem 35 Answer is C p. 379, problem 19

                    Answer is D, 3*3*3 = 27

p. 382, problem 13

                    Answer is E

p. 382, problem 15

Answer is D p. 382, problem 40 Answer is C, need to subtract one before multiplying the number of rows by the number of columns. p. 386, problem 32
"... gives no insight into the form of the alternative..."
p. 388, problem 5d
Add to the answer: Since these are discrete data, could instead interpret in terms of the "average deviation from the expected payment."  This would also help in comparing these results to another benefit plan payout scheme.
The data are somewhat symmetric, so applying the empirical rule is not too aggregious.
Page Last Updated: April 28, 2009