# Exam 4 Corrective Questions Spring, 2009 E

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```							                 AP Statistics Exam 11 Corrective Assignment Spring, 2012 E.C. Hedstrom

Objective 22: Interpretation and Meaning of Confidence Intervals.

(1 – 3): For each situation below, do the following: (Assume that all required assumptions and conditions are
met so you do not need to state and/or check that they are met)

a) Give the interval being sure to show all work.
b) Interpret the interval in context.
c) Interpret the confidence level in context.

1. 39 out of a random sample of 117 registered voters approved of a bond proposal for a local school district.
Construct a 98% confidence interval to estimate the true proportion of the district’s registered voters who
support the proposal.

2. 289 out of a random sample of 567 members of a large trade union approved of the proposed contract from
management. Construct a 90% confidence interval to estimate the true proportion of the union’s members who
approve of the proposed contract.

3. 72 out of a random sample of 111 students who attend a college said that they think the price of their tuition
is too high. Construct a 96% confidence interval to estimate the true proportion of the college’s students who
think their tuition is too high.

4. A company that manufactures automobile engine parts would like to estimate the proportion of parts they
produce that do not fall within specifications so they take a simple random sample of 300 parts and note that 16
are nonconforming.

a) Assuming that all required conditions and assumptions are met, construct a 95% confidence interval for the
true proportion of this company’s parts that do not fall within specs.

Regarding the confidence interval you calculated in part (a), fill in the blanks below:

b) The name of the interval is:____________ c) The value of the sample proportion is:___________.

d) The value of the critical value is:_________ e) The value of the standard error is:______________

f) The value of the margin of error is:__________ g) The value of the width of the confidence interval is:____.

h) Interpret the meaning of the interval you constructed in (b) in the context of this question.

i) Interpret what is meant by “95% Confidence” in the context of this question.
5. A study of “adverse symptoms” in users of over-the-counter pain relief medications assigned subjects at
random to use one of two common pain relievers: acetaminophen and ibuprofen. 650 subjects took
acetaminophen and 44 experienced some adverse symptom.

a) Assuming that all required conditions and assumptions are met, construct a 99% confidence interval for the
true proportion users of over-the-counter acetaminophen who would experience adverse symptoms.

Regarding the confidence interval you calculated in part (a), fill in the blanks below:

b) The name of the interval is:____________ c) The value of the sample proportion is:___________.

d) The value of the critical value is:_________ e) The value of the standard error is:______________

f) The value of the margin of error is:__________ g) The value of the width of the confidence interval is:____.

h) Interpret the meaning of the interval you constructed in (b) in the context of this question.

i) Interpret what is meant by “99% Confidence” in the context of this question.

Objective 23: Confidence Intervals: Multiple-Choice.

1.    A researcher found that a 98% confidence interval for the mean hours per week spent studying by
college students was (13, 17). Which of the following is(are) true?

I. There is a 98% chance that the mean hours per week spent studying by college students is between 13
and 17.
II. 98% of college students study between 13 and 17 hours a week.
III. Students average between 13 and 17 hours per week studying on 98% of the weeks.

a) None b) I only             c) II only    d) III only      e) I and III only

2. A professor was curious about her students’ grade point averages (GPAs) so she took a random sample
of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534. Which of the
following formulas gives a 99% confidence interval for the mean GPA of the professors’ students?

0.534
a ) 3.01  2.947
15
0.534
b) 3.01  2.977
15
0.534
c) 3.01  2.576
15
0.534
d )3.01  2.947
14
0.534
e)3.01  2.977
14
3. Which of the following is(are) true about Student’s t-models?

I. They are unimodal, symmetric and bell-shaped.
II. They have fatter tails than the Normal model.
III. As the degrees of freedom increase, the t-models look more and more like the Normal Model.

a) I only           b) I and II only        c) I and III only      d) II and III only     e) I, II and III

4. At one vehicle inspection station, 13 of 52 trucks and 11 of 88 cars failed the emissions test. Assuming
these vehicles were representative of the cars and trucks in that area, what is the standard error of the
difference in the percentages of all cars and trucks that are not in compliance with air quality
regulations?

a) 0.025            b) 0.032       c) 0.049         d) 0.070       e) 0.095

5.    At one SAT test site, students taking the test for a second time volunteered to inhale supplemental
oxygen for 10 minutes before the test. In fact, some received oxygen but others (randomly assigned)
were given just normal air. Test results showed that 42 of 66 students who breathed oxygen improved
their SAT scores compared to only 35 out of 63 students who did not get the oxygen (but thought they
did). Which interval should we use to see if there is evidence that breathing extra oxygen can help test-
takers think more clearly?

a) One-proportion Z interval                b) Two-Proportion Z interval           c) 1-sample t-interval
d) 2-sample t-interval                      e) Matched-pairs t-interval

6. A survey asked people, “On what percent of days do you get more than 30 minutes of vigorous
exercise?” Using their responses we want to estimate the difference exercise frequency between men
and women. We should use a

a) 1-prop Z interval       b) 2-prop Z interval     c) 1-sample t-interval d) 2-sample t-interval
e) matched-pairs t-interval

7.    The manager of an orchard expects about 70% of his apples to exceed the weight requirement for
“Grade A” designation. At least how many apples must he sample to be 90% confident of estimating
the true proportion within plus/minus 4%?

a) 19               b) 23          c) 89            d) 356         e) 505

8. We have calculated a confidence interval based on a sample of size n = 100. Now we want to get a
better estimate with a margin of error that is only one-fourth as large. How large does our new sample
need to be?

a) 25               b) 50          c) 200           d) 400         e) 1600
9. A certain population is bimodal. We want to estimate its mean so we will collect a sample. Which
should be true if we use a large sample rather than a small one?

I. The distribution of our sample data will be more clearly bimodal.
II. The sampling distribution of the sample means will be approximately bimodal.
III. The variability of the sample means will be smaller.

a) I only            b) II only                c) III only       d) II and III only   e) I, II and III

10. A certain population is strongly skewed to the right. We want to estimate its mean so we will collect a
sample. Which should be true if we use a large enough sample rather than a small one?

I. The distribution of our sample data will be closer to normal.
II. The sampling model of the sample means will be closer to normal.
III. The variability of the sample means will be greater.

a) I only            b) II only                c) III only       d) I and III only    e) II and III only

11. A relief fund is set up to collect donations for the families affected by recent storms. A random sample
of 400 people shows that 28% of those 200 who were contacted by telephone actually made
contributions compared to only 18% of the 200 who received first class mail requests. Which formula
calculates the 95% confidence interval for the difference in the proportions of people who make
donations if contacted by telephone or first class mail?

 0.23 0.77 
a)    0.28  0.18  1.96
200
 0.23 0.77    0.23 0.77 
b)    0.28  0.18  1.96
200               200
 0.23 0.77 
c)    0.28  0.18  1.96
400
 0.28 0.72   0.18 0.82
d)    0.28  0.18  1.96
200               200
 0.28 0.72   0.18 0.82
e)    0.28  0.18  1.96
400               400

12. Which is true about a 95% confidence interval based on a given sample?

I. The interval contains 95% of the population.
II. Results from 95% of all samples will lie in the interval.
III. The interval is narrower than a 98% confidence interval would be.

a) None              b) I only                 c) II only        d) III only          e) II and III only
13. Trainers need to estimate the level of fat in athletes to ensure good health. Initial tests were based on a
small sample but now the trainers double the sample size for a follow-up test. The main purpose of the
larger sample is to

a) Reduce response bias        b) decrease the variability in the population c) reduce non-response bias
d) reduce confounding due to other variables e) decrease the standard deviation of the sampling model

14. We want to know the mean winning score at the US Open golf championship. An Internet search gives
us all the scores for the history of that tournament, and we create a 95% confidence interval based on a t-
distribution. This procedure was not appropriate because

a)   Since these are the best players in the world, the scores are probably skewed.
b)   The entire population of scores was gathered so there is no reason to do inference.
c)   Tiger Woods’ recent poor performance is probably an outlier.
d)   The population standard deviation is known se we should have used a z-model.
e)   In big golf tournaments the players are not randomly selected.

Objective 24. Confidence Intervals: Free Response.

1. People who run are generally quite thin. You might wonder – what is the typical weight of a runner? To
estimate this, you take a simple random sample of 24 male runners and record their weight, in kg, which is
recorded in the table below:
67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9
60.9 69.2 63.7 68.3 64.7 65.6 56.0 57.8
66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7
Estimate the true mean weight of male runners with a 95% confidence interval. Be sure to state and check all
required conditions.

2. Crop researchers plant 16 plots with a new variety of corn. The yields in bushels per acre are

138.0 139.1 113.0 132.5 140.7 109.7 118.9 134.8
109.6 127.3 115.6 130.4 130.2 111.7 105.5 106

Estimate the true mean yield in bushels per acre for this variety of corn using a 90% confidence interval.

3. Does cocaine use by pregnant women cause their babies to have low birth weight? To study this question,
birth weights of babies of women who tested positive for cocaine/crack during a drug-screening test were
compared with a group we call “other”. Here are the summary statistics. The birth weights are measured in
grams. (Data from a study conducted at the Medical University of South Carolina in 1989).

Group        n   x    s
Positive Test 134 2733 599
Other      5974 3118 672

Construct a 99% confidence interval to answer the question – does cocaine use by pregnant women cause their
babies to have low birth weight?
4. 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. Do these results mean that there is a difference in the
proportion of men and women who exercise regularly? Construct a 96% confidence interval to answer the

5. Each year people who have income file income tax reports with the government. In some instances people
to lower the percentage of taxes paid to the government each year. A random sample of people who filed tax
reports resulted in the data in the table below. Do these data indicate that people should seek tax advice from an
accountant or financial advisor? Construct a 95% confidence interval for the true difference in the proportion of
people who paid lower taxes between those who asked for advice and those who did not.

Paid Lower % of taxes  No      Yes
No             48       19
Yes            24       86

6. Some of the cigarettes sold in the US claim to be “low tar”. How much less tar would a smoker get by
smoking low tar brands instead of the regular cigarettes of the same brand? Samples of 15 brands of each type
were randomly chosen from the 1206 varieties that are marketed. Their tar contents (mg/cig) are listed in the
table below. Find a 95% confidence interval for the difference between regular and low tar brands.

Type                  Milligrams of tar per cigarette
Regular    18 10 14 15 15 12 17 11 14 17 12 14 15 15 12
Low Tar    9 5 10 4 8 9 9 3 7 12 6 10 8 11 8
Difference 9   5 4 11 7 3 8 8 7 5 6 4 7 4 4

7. 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? Answer
the question by constructing a 90% confidence interval for the true mean difference between grapes from vines
that were fertilized and those that were not.
8. The table shows the average number of points scored in home and away games by 8 randomly selected NFL
teams during the 2002 season. Assuming that the offensive performance of these teams is representative of
other teams during this season and others, do these data provide evidence of a home field advantage when it
comes to scoring? Construct a 99% confidence interval to answer the question.

Average Points Scored
Team      Home         Away
Jets      24.3         21.2
Ravens      19.8         19.8
Texans      13.5         13.1
49ers      25.2         20.6
Giants      18.9         23.0
Bucs      26.5         20.8
Eagles     24.1         25.5
Packers     24.6         23.0

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