Math 221 ________________________
Exam 2 Name
For each problem:
(1) State the parameter of interest.
(2) State the appropriate test.
(3) Check the assumptions (use complete sentences). If a problem fails one or more
assumptions, say so, but continue with the problem anyway.
(4) Compute the P-value and/or the confidence interval. Show your work.
(5) Clearly state your conclusions.
thThis exam is due by the beginning of class on Tuesday, August 12. If you have any
questions, feel free to email me at GregK@spokanefalls.edu. Good luck.
1. Waiters are expected to keep track of their income from tips and report it on their income tax forms. The Internal Revenue Service suspects that one waiter (we’ll call him “Fred”) has been under-reporting his income, so they are auditing his tax return. An IRS agent goes through the restaurant’s files and obtains a random sample of 80 credit card receipts from people Fred served. The average tip size shown on these receipts was $9.68 with a standard deviation of $2.72. Create a 90% confidence interval for the mean size of all of Fred’s credit card tips. On his tax return Fred had claimed that his tips averaged $8.73. Based on their confidence interval, does the IRS have a case against him? Explain.
(1) Mean size of all of Fred’s credit card tips
(2) 1-sample t-interval
(3) Randomization: The credit card receipts were randomly selected.
10% condition: The 80 credit card receipts must be no more than 10% of all John’s tips from credit
cards this year. That means John must get tips from at least 800 credit cards, which seems reasonable.
Nearly normal condition: No histogram is shown, but due to the large sample only extremely skewed
data would violate this condition. We should still check a histogram.
Under these conditions, the sampling distribution of the mean can be modeled by Student’s t with df =
80 – 1 = 79.
，~2.72(4) , or ( $9.17, $10.19 ) 9.68；1.665：?：?80；?
(5) The IRS may have a case against John because John claimed that his tips averaged $8.73, but the
interval shows that his average tip was between $9.17 and $10.19 with 90% confidence. However, there
may be a problem with the methodology used by the IRS. The interval estimated only tips from credit
cards. If the average claimed by John includes non-credit card tips, it may be reasonable for John’s
average to be lower than the interval estimate. Bills paid by credit card tend to be larger than those paid
in cash, so they also get larger tips. The IRS estimate may be biased because it didn’t include tips paid
2. An insurance company advertises that 90% of their accident claims are settled within 30 days. A consumer group randomly selects 104 of last year’s claims from the company’s files, and finds that only 89 of them were settled within 30 days. Is the company guilty of false advertising?
(1) Proportion of accident claims settled within 30 days.
(2) 1-propotion z-test
(3) Independence: We assume the time to settle one claim does not affect the time to settle another.
Random condition: The claims in the sample were selected randomly.
10% condition: If it’s a large company, this sample of 104 claims should be less than 10% of all claims
Success/Failure: 89 claims were settled within 30 days, 15 claims were not. Both are at least 10.
Under these conditions the sampling distribution of the proportion can be modeled by a Normal model.
We will find a one-proportion z-test.
890.856？0.9ˆ(4) , , , , H:p！0.9H:p，0.9p！！0.856z！！？1.520A104(0.9)(0.1)
(5) Since the P-value is more than 0.05, we fail to reject the null hypothesis. There is not enough evidence
to prove that the company settles fewer than 90% of its claims within 30 days.
3. The Prince County Bottling Company supplies bottles of lemonade labeled 12 oz. When the Prince County Department of Weights and Measures tests a random sample of bottles, the amounts listed below are obtained. Using a 0.05 significance level, is there sufficient evidence to file a charge that the bottling company is cheating consumers by giving amounts with a median less than 12 oz?
11.4 11.8 11.7 11.0 11.9 11.9 11.5 12.0 12.1 11.9 10.9 11.3
11.5 11.5 11.6 12.0 12.2 11.5 11.9 12.9 11.1 11.2 11.8 12.5
(1) Median amounts of lemonade in each bottle.
(2) Sign test
(3) Independence: We assume bottles were filled independently of each other.
Random condition: Data came from a random sample.
10% condition: It seems safe to assume the company has bottled more than 240 bottles.
Success/Failure: Under the null hypothesis, there would be 11 bottles above the median, and 11 bottles
below the median.
Under these conditions the sampling distribution of the proportion below the median can be modeled by
a Normal model. We will find a one-proportion z-test.
18ˆ(4) , , , H:median！12H:median，12p！！0.8180A22
(5) Since the P-value is so low, we reject the null hypothesis. There is strong evidence that the company
fills bottles to less than 12 oz.
4. Before you took this course, you probably heard many stories about statistics courses. Oftentimes parents of students have had bad experiences with statistics courses and pass on their anxieties to their children. To test whether actually taking statistics decreases students’ anxieties about the
subject, a statistics instructor gave a test to rate student anxiety at the beginning and end of his course. Anxiety levels were measured on a scale 0-10. Here are the data for 16 randomly chosen students from a class of 180 students.
Pre-course anxiety level 7 6 9 5 6 7 5 7 6 4 3 2 1 3 4 2
Post-course anxiety level 4 3 7 3 4 5 4 6 5 3 2 2 1 3 4 3
Difference (Post – Pre) -3 -3 -2 -2 -2 -2 -1 -1 -1 -1 -1 0 0 0 0 1
Do the data indicate that anxiety levels about statistics decrease after students take a statistics course? Test an appropriate hypothesis and state your conclusion.
(1) Differences in anxiety level between post-course and pre-course tests
(2) Matched pairs t-test
(3) Paired data: The data are paired because they are measurements on the same individuals both before and
after the intro stats course.
Independence: The anxiety level of any student is independent of the anxiety level of any other student,
so the differences are independent.
Randomization: We are told this is a random sample from the class.
10% condition: Our sample of 16 students is less than 10% of all students who took this statistics class.
Nearly normal condition: The histogram of the differences is unimodal and roughly symmetric.
Under these conditions the sampling distribution of the differences can be modeled by a Student’s t-
model with 15 degrees of freedom, and we will use a paired t-test.
(4) , , , , H:(！0H:(，0s！1.1475d！？1.125d0dAd
？1.125？0, P-value < 0.005 t！！？3.92151.1475
(5) With a P-value this small, we can reject the null hypothesis. We have strong evidence that taking this
statistics class reduced the anxiety level of students about statistics. (Since we sampled from only one
instructor’s class, we cannot generalize our result to all statistics classes.)
5. An agronomist hopes that a new fertilizer she has developed will enable grape growers to increase the yield of each grapevine by more than 5 pounds. To test this fertilizer she applied it to 44 vines and used the traditional growing strategies on 47 other vines. The fertilized vines produced a mean of 58.4 pounds of grapes with standard deviation 3.7 pounds, while the unfertilized vines yielded an average of 52.1 pounds with standard deviation 3.4 pounds of grapes. Do these experimental results confirm the agronomist’s expectations?
(1) Differences in the means of fertilized and unfertilized grapevines.
(2) Two-sample t-test
(3) Independent group assumption: Different vines were used in each group. As long as the two sets of
vines were not close enough to affect each other, they should be independent.
Randomization condition: If the vines were randomly assigned to receive fertilizer or not, this is
satisfied. (The description does not say how the vines were selected for treatment.)
10% condition: The population of all vines is very large. These vines are much less than 10% of all
Nearly normal condition: No histograms are shown, so this condition cannot be checked. However,
since both groups are large (more than 40 vines each), only extreme skewness will affect the results.
Still we should always check histograms.
*(4) For 43 degrees of freedom, , t！2.016743
223.73.4(58.452.1)2.0167 A 95% confidence interval is , or (4.79, 7.81). ？；？4447
(5) Since 5 is inside the 95% confidence interval, we fail to reject the null hypothesis. The new fertilizer
may not have produced an average of more than 5 pounds per vine higher than traditional methods.
6. A random sample of 150 men found that 88 of the men exercise regularly, while a random sample of 200 women found that 130 of the women exercise regularly. Based on the results, construct a 95% confidence interval for the difference in the proportions of women and men who exercise regularly. A friend says that she believes that a higher proportion of women than men exercise regularly. Does your confidence interval support this conclusion? Explain.
(1) Difference in proportions between men and women who exercise regularly.
(2) Two-sample z-test for difference in proportions
(3) Randomization condition: We are told that we have random samples.
10% condition: We have less than 10% of all men and less than 10% of all women.
Independent samples condition: The two groups are clearly independent of each other.
Success/Failure condition: Of the men, 88 exercise regularly and 62 do not; of the women,
130 exercise regularly and 70 do not. The observed number of both successes and failures in
both groups is at least 10.
With the conditions satisfied, the sampling distribution of the difference in proportions is
approximately normal with a mean of , the true difference between population p？pMW
proportions. We can find a two-proportion z-interval.
88130(4) , p！！0.587p！！0.650MW150200
A 95% confidence interval is
( -16.6%, 4.0%)
(5) Since zero is contained in my confidence interval, I cannot say that a higher proportion of
women than men exercise regularly. My confidence interval does not support my friend’s
7. Could eye color be a warning signal for hearing loss in patients suffering from meningitis? British researcher Helen Cullington recorded the eye color of 130 deaf patients, and noted whether the patient’s deafness had developed following treatment for meningitis. Her data are summarized in
the table below. Test an appropriate hypothesis and state your conclusion.
Deafness related to…
Eye color meningitis Other
Light 30 72
Dark 2 26
(1) Association between deafness and eye color
(2) Chi-square test
(3) These are counts of categorical data, assumed to be representative of deaf patients in Britain,
with expected counts (25.1, 76.9, 6.9, and 21.1) all at least 5. OK to do a chi-square test for
independence, with df = 1.
(4) Deafness and eye color are independent. H:0
There is an association between deafness and eye color. H:A
P-value = 0.01539
(5) Since the P-value is low, I reject the null hypothesis. There is strong evidence that hearing loss is
associated with eye color. It appears that people with dark-colored eyes are at less risk of suffering
deafness from meningitis.
8. When two competing teams are equally matched, the probability that each team wins any game is 0.5. The NBA championship goes to the team that wins four games in a best-of-seven series. If the teams were equally matched, the probability that the final series ends with one of the teams
4sweeping four straight games would be , or 12.5%. Further probability 2(0.5)！0.125
calculations indicate that 25% of these series should last five games, 31.25% should last six games, and the other 31.25% should last the full seven games. The table shows the number of games if took to decide each of the last 62 NBA champs. Do you think the teams are usually equally matched? Give statistical evidence to support your conclusion.
Length of series 4 games 5 games 6 games 7 games
NBA finals 8 14 24 16
(1) Teams in the NBA finals are equally matched.
(2) Chi-square goodness of fit test
(3) Counted data: These are counts of categorical data.
Randomization: We will assume that these 62 series are typical of past and future NBA championships.
Expected cell frequency: The expected values are all greater than five.
Under these conditions, we can perform a chi-square goodness-of-fit test with 3 degrees of freedom.
(4) Degrees of freedom = # categories – 1 = 4 – 1 = 3
Length of series Observed counts Expected counts
4 8 0.125(62) = 7.75
5 14 0.25(62) = 15.5
6 24 0.3125(62) = 19.375
7 16 0.3125(62) = 19.375
The distribution of length of series is consistent with a 50-50 chance for each team to win each H:0
The distribution of length of series isn’t consistent with a 50-50 chance for each team to win H:Aeach game.
2222(8？7.75)(14？15.5)(24？19.375)(16？19.375)2, ，！？？？！1.84527.7515.519.37519.375P-value = 0.605
(5) The high P-value means we do not reject the null hypothesis, there is no evidence that the
NBA championship series are inconsistent with the conjecture that the teams are evenly