by Jim Bohan with Beth Chance
Amsco School Publications, Inc
Please email your corrections to email@example.com or firstname.lastname@example.org
(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. 117, example 5(B) P(A|B) = P(A)
Last line should read (30/36)(30/36)(6/36) = 25/126
p. 137, 3rd line from end of example 5
p. 139, Section 5.8 Review problem 3Should read P(X=4 and Y=0) =? P(X=4).P(Y=0)
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
p. 146, Section 5.10, problem 4Probability of finding a special prize in a box of popcorn should be .20.
p. 151, multiple-choice questionChange failure rate to 20% to match answer in back.
p. 167, free response problem 2In item II, replace "may" by "must" to read "The standard deviations of two different samples from the same population must be the same."
p. 175-6, formula for exact dfChange 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. 177, Multiple-Choice, problem 4Should read:
p. 181, problem 3"The probability that samples of these sizes have a difference in means of 2.8 or more..."
p. 198, problem 3The mean overhead cost for the pizza stores should be $764.50, not $763.50.
p. 208, two-sample t-test (unpooled)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.
formula approximating degrees of freedom should be df=min(n1,n2) -1p. 220, problem 8
p. 229change (C) to read "The 99% confidence interval does not contain the mean specified by the null hypothesis"
In the formula for two-tailed test, the term (za/2+zb) should be squared.
p. 234, example 4In the confidence interval formula, t8 for 90% confidence should equal 1.860 instead of 1.397.
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.
the units on smpg should be mi/galp. 266, problem 21
p. 268, problem 32Answer A should read "For a fixed p, the variance increases as n increases".
p. 271, problem 3, question aChange (D) to read "We expect the 95% confidence interval of the sample result would contain the expected population value."
p. 271, problem 4, question dChange 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. 273, problem 6Change the question to read "Estimate the approximate life of the car in the sample whose purchase price was $25,000."
P. 274, problem 2Part (e) was not included in the scoring guide.
p. 277, problem 18Answer C should read "Consecutive values of a discrete variable may or may not differ by a fixed(constant) amount."
p. 298, problem 30Change answer A to read "Honesty of the respondent".
p. 304, Model Exam 5, problem 3Start question as "In a test of the equality of the proportions..."
p. 305, problem 11Answer C should read "bimodal".
p. 305, problem 14The equation should read = 10.2 + 4.15x
p. 306, problem 22(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. 327, Multiple Choice, problem 4Add "The restaurant further claims that you have a chance to win on every visit."
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
p. 329, Multiple Choice, problem 5Change explanation to read that the percentile of a z-score of 1 is approximately 84%
p. 329, Multiple Choice, problem 6The minimum is at 32.
p. 332, Free Response Question 5Using height to predict weight, the regression equation is: weight = -91.9 + 3.25 height
p. 332, Free Response Question 6The y-intercept should refer to the value of -91.9 not -58.
p. 332, Multiple Choice, Question 1The correct prediction is -91.9+3.24(65)=119.2
p. 332, Multiple Choice, Question 4Answer should be D, as the only option that is not a correct conclusion.
p. 341, Multiple-Choice Question 5Answer should be E, none of these lines of fit would produce a residual plot similar to the given residual plot.
p. 342, Multiple Choice Question 2Choice should read D, explanation is correct.
p. 342, Multiple-Choice Question 3P(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, Section 5.2, Free Response Question 3Answer 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 7P(sum=8 and both even) = 3/36, should answer is 5/36 + 9/36 - 3/36 = 11/36
p. 344, Section 5.6, Multiple Choice Question 24/52(4/51) = .006 not .06.
p. 345, Section 5.8, Free-Response Question 4bz = -1.28 = (38-50)/s or z = 1.28 = (62-50)/s
p. 346, Chapter 5 Assessment, problem 4Note: 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
B is the correct answer, but the explanation is off. Delete the "not"s in the independence explanation.
p. 352, Section 6.7, multiple-choice question 4The 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. 353, problem 9Explanation should being P(1-2> 2.8)...
Explanation is correct, but answer should be D.
final statement should read P(>2.8| m=2.4)p. 357, Section 7.4, Free-response, question 4
p. 358, multiple choice, question 8P(boys’ sample times would be at least 4.9 seconds faster than girls’ times if the...)
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)/100p. 361, step 5
Ha: (words) There is a difference by gender in science placement exam performance in 1988p. 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 = 1068p. 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
p. 3692(.25) = .5 (end of p-value calculation)
p. 374, Model Exam 1, problem 43. 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
Explanation is correct, but answer should be A.
p. 375, problem 24Technically 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 28Answers 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 32Answer should be D (explanation is correct)
p. 375, problem 36Answer 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.
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.
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
p. 378, problem 2Part (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 35Add 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."
Answer is E
p. 382, problem 15
p. 388, problem 5d"... gives no insight into the form of the alternative..."
Page Last Updated: April 28, 2009Add 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.