Download Review Chapter 12/13

MULTIPLE CHOICE
1. An SRS of 384 people in one city revealed that 112 people worked more than one job. A different SRS
of 432 people in another city showed that 91 people worked more than one job. Find a 99% confidence
interval for the difference between the proportions of workers in the two cities who work more than one
job.
A. (.003, .159)
B. (.021, .141)
C. (-.159, .004)
D. (.031, .131)
E. Sample sizes are not large enough to justify using a z procedure.
2. Which of the following is a characteristic of the t distribution?
A. It contains a finite number of terms
B. Its shape depends on the number of degrees of freedom.
C. It is identical to a normal curve.
D. Its median is positive.
E. It can be used with a small sample only if there are outliers.
3. You’d like to give an estimate for the mean difference in the length between the right foot and the left
foot on people within a certain population. Which is the best way to do this?
A. Find the differences in the lengths of the left and right feet for each person you measured. Use the
mean and standard deviation of your sample to conduct a hypothesis test of whether the right foot tends to
be bigger than the left foot.
B. Find the differences in the lengths of the left and right feet for each person you measured. Report the
average difference.
C. Find the differences in the lengths of the left and right feet for each person you measured. Use the
mean and standard deviation of your sample to construct a confidence interval for the difference.
D. Find the average length of all the right feet. Find the average length of all the left feet. Construct a
confidence interval for each average and report both of them.
E. Find the average length of all the right feet. Find the average length of all the left feet. Conduct a
hypothesis test on the average length of the left feet with a null hypothesis that the man of the left foot is
equal to the mean of the right foot.
4. An SRS of grades reveals the following:
First Semester: 89, 98, 78, 86, 95, 83, 90, 87, 85, 80, 96, 93, 90, 91, 81, 87, 93, 90, 87 and 88
Second Semester: 87, 85, 89, 78, 79, 77, 82, 90, 91, 93, 88, 87, 86, 83, 85, 81, 90, 86, 86 and 84
Compute the statistics necessary to construct a confidence interval for the difference between these two
populations:
A. x = 88.35
s = 5.314
n = 20
B. x = 85.35
s = 4.368
n = 20
C. x = 85.35
s = 4.368
n = 20 (group one)
s = 4.841
n = 20 (group two)
x = 86.85
D. x = 88.35
s = 5.314
n = 20 (group one)
s = 4.368
n = 20 (group two)
x = 85.35
E. There is no way to tell but the grades are too high.
5. After receiving several complaints from his customers about the store being closed on Sundays, a
storekeeper decided to conduct a survey. He randomly selected 100 female customers and 120 male
customers and asked them, “Are you interested in shopping at this store on Sundays?” He counted the
number of customers answering “yes” and constructed a 95 percent confidence interval for the idfference
in the proportions of male and female customers in favor of shopping on Sundays. The resulting interrval
was (-.023, -0.18). Which of the following is a correct interpretation of the interval?
a) We are 95 percent confident that the proportion of women interested in shopping on Sundays
exceeds the proportion of men interested in shopping on Sundays.
b) We are 95 percent confident that the proportion of men interested in shopping on Sundays exceeds
the proportion of women interested in shopping on Sundays.
c) We are 95 percent confident that the proportion of women interested in shopping on Sundays is
equal to the proportion of men interested in shopping on Sundays.
d) Because the interval contains negative values, it is invalid and should not be interpreted.
e) Because the interval does not contain zero, the interval is invalid, and should not be interpreted.
6. A company is interested in comparing the mean sales revenue per salesperson at two different locations.
The manager takes a random sample of 10 salespersons from each location independently and records the
sales revenue generated by each person during the last 4 weeks. He decides to use a t-test to compare the
mean sales revenue at two locations. Which of the following assumptions is necessary for the validity of
the test?
a) The population variances at both locations are equal.
b) The population variances at both locations are not equal.
c) The population variances at both locations are known.
d) The population of the sales records at each location is normally distributed.
e) The population of the difference in sales records computed by pairing one salesperson from each
location is normally distributed.
7. A manufacturer of motor oil is interested in testing the effects of a newly developed additive on the
lifespan of an engine. Twenty-five different engine types are selected at random and each one is tested
using oil with the additive and oil without the additive. What type of alalysis will yield the most useful
information?
a) Matched pairs comparison of population proportions.
b) Matched pairs comparison of population means.
c) Independent samples comparison of population proportions.
d) Independent samples comparison of population means.
e) Chi-square test of homogeneity.
8. A skeptic decides to conduct an experiment in ESP in which a blindfolded subject calls out the color of
a card dealt from a regular deck of cards (half the cards are red; the other half, black). One hundred cards
are dealt from a well-shuffled pack, with each card being replaced after a deal. Using a 5 percent level of
significance, what is the lowest number of cards that the subject needs to call out correctly in order ot
show that he is doing better than he would if he were simply guessing?
a) 51
b) 59
c) 75
d) 96
e) 98
9. Which of the following are true?
I. In order to use t procedures, you must know the population standard deviation.
II. t procedures are considered to be robust.
III. The standard error of the mean is an estimate of the standard deviation of the sampling distribution
of a sample mean.
A. II and III only
B. I only
C. II only
D. III only
E. All of these
10. Which of the following is NOT a matched pair situation?
A. Teachers attending a summer language course are tested on their language skills at the beginning and the end of
the course.
B. A sample of 50 people agree to participate in a study on Vitamin C. Half the group take Vitamin C supplements,
while the other half take a placebo. During a six-month period, researchers record the number of colds each subject
has.
C. People at a store are asked to try wiping up a spill using two different brands of paper towels. The time it takes
to wipe up the spill with each paper towel is recorded.
D. An inhaler that may inhibit yawning is being tested. People are asked to record the number of times they yawn
on the day before the inhaler is used, and then to record the number of times they yawn on the day they use the
inhaler.
E. In a test that looks at two types of lubricating oil, two identical machines are placed on two identical assembly
lines. One machine is lubricated with oil A and the other is lubricated with oil B. The number of breakdowns per
year is recorded for each machine.
11. You're designing a new mouse for computers. You want to see if it's faster for left-handed people to use it with
their right or their left hand. Which of the following best describes a matched pairs scenario for this study?
A. You divide the left-handed people in your company into two groups. The employees complete a series of timed
tasks using the mouse, with one group using their left hand and the other their right.
B. You give each left-handed employee one of two new mouse designs. You ask them to try it for a week and
report their findings.
C. You design two versions of the mouse, one left-handed and one right-handed. Each left- handed employee is
randomly assigned to use one of the two mouse designs. You have each employee complete a series of timed tasks
using the mouse they were given.
D. You give a mouse to left-handed employees and ask them to use it. After three weeks, you distribute a survey
asking each employee which hand they prefer.
E. Left-handed employees are asked to use the mouse to complete a series of timed tasks on each hand. For each
employee, the starting hand is selected randomly.
12. The following is a list of differences between first and second semester final grades in chemistry (the first
semester grade was subtracted from the second):
-6, 10, -3, 5, 4, 8, -4, 9, -2, 0, -5, -5, 5, 12, 10.
Interpret the hypotheses in the one-sided test Ho:  (2-1) = 0 and Ha:  (2-1) > 0.
A. Ho:  (2-1) = 0 (The mean first semester grade is 0.) Ha:  (2-1) > 0 (The mean first semester grade is greater than
0.)
B. Ho:  (2-1) = 0 (The mean second semester grade is 0.) Ha:  (2-1) > 0 (The mean second semester grade is greater
than 0.)
C Ho:  (2-1) = 0 (The mean difference between the first and second semester grade is 0.) Ha:  (2-1) > 0 (The mean
first semester grade is higher than the mean second semester grade.)
D Ho:  (2-1) = 0 (The mean difference between the first and second semester grade is 0.) Ha:  (2-1) > 0 (The mean
second semester grade is higher than the mean first semester grade.)
E. Ho:  (2-1) = 0 (The mean difference between the first and second semester grade is 0.) Ha:  (2-1) > 0 (The mean
first semester grade isn't equal to the mean second semester grade.)
13. The following list shows first and second semester final student grades for a chemistry class:
Students: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1st smstr 96 80 83 80 90 70 74 80 82 90 95 85 70 80 70
2nd smstr 90 90 80 85 94 78 70 89 80 90 90 80 75 92 80
Calculate the test statistic and p-value for the one-sided test Ho:  (2-1) = 0 and Ha:  (2-1) > 0.
A. t = 1.533, p = .074
B. t = 2.68, p = .009
C. t = 1.19, p = .254
D. t = 2.145, p = .025
E. t = 1.57, p = .069
14. You're going to construct a confidence interval for the difference between two population means. The
sample sizes are 15 and 17. How many degrees of freedom do you base your analysis on if you don't pool
your variances? How many if you do pool your variances?
A. 30, 14
B. 15, 32
C. 14, 30
D. 16, 30
E. 14, 31
15. Suppose we compare the data on two samples, A and B, and come up with the following data:
na = nb = 10, x a = 25, sa = 3.21, x b = 22.2, sb = 3.09. What's the standard error of the difference
between the means (that is, what is s x1  x 2 )? (For this question, don't pool. )
A. 1.99
B. 3.15
C. 1.49
D. 1.41
E. You need to use a confidence interval on the calculator to find this value.
Questions 16-19 use the following information: You draw two random samples form two distinct
populations and calculate the following:
x1 = 23.4, s1 = 4.2, n1 = 25 and x 2 = 25.3, s2 = 3.9, n2 = 27
16. Using the conservative method (no calculator, use table), what are the degrees of freedom for this
problem?
A. 24
B. 26
C. 51
D. 50
E. 24
17. Using the formula for a confidence interval for the difference between two means, (estimate) 
(margin of error), what's the correct value for (estimate)?
A. 23.4
B. 25.3
C. -1.9
D. 3
E. -2
18. The estimate of the degrees of freedom, k, equals n1 - 1, or 24, and t* equals 2.064. For the formula
for a confidence interval, (estimate)  (margin of error), what's the correct value for (margin of error)?
A. 1.126
B. 1.15
C. 2.325
D. 1.734
E. 2.62
19. Construct a 95% confidence interval for the difference between these two populations and draw a
conclusion based on this confidence interval.
A. The confidence interval is (-6.699, 2.899); there's a difference between the two population means. .
B. The confidence interval is (-6.699, 2.899); there's no difference between the two population means.
C. The confidence interval is (-3.275, 1.375); there's a difference between the two population means.
D. The confidence interval is (-4.225, .425); there's a difference between the two population means.
E. The confidence interval is (-4.225, .425); there's no difference between the two population means.
20. Suppose biologists have been studying the question of whether squirrels produce more offspring in colder
climates. The theory goes like this: In colder climates, fewer offspring will live to maturity, so the squirrels need to
produce more offspring to make up for the additional loss. In a study on one species, scientists collected data on the
number of offspring for squirrels in both states. Here are the summary statistics. (The study and statistics are
hypothetical. )
smpl size
mean # per year
std dev
CA Squirrels
73
11.49
4.28
OR Squirrels
49
13.57
3.71
What's the p- value of this test?
A. -2.85
B. .0052
C. .0026
D. 112.31
E. .0032
21. Suppose you generate a two-sample t statistic of 2.59 by hand in a one-sided hypothesis test. If you were testing
the hypothesis at  = .01, and if the samples are of equal size, how small could each sample be and still result in a
null hypothesis?
A. 3
B. 16
C. 17
D. Can't determine without knowing the sample standard deviations
E. Can't determine without knowing the sample means
22. You take two random samples from two distinct populations and calculate the following statistics:
x1 = 52.6, s1 = 3.2, n1 = 28 and x2 = 49.3, s2 = 2.9, n2 = 29 with k = degrees of freedom = 27
You get t = 4.07 and p = 0.0002
State a conclusion, based on the p-value and a = .05, for the hypothesis test Ho: 1  2 , Ha: 1  2
A. On average, the difference between the two populations is 4.07.
B. We don't have enough evidence to reject the null hypothesis that there's no difference between the means of the
two populations.
C. Reject the null hypothesis in favor of the alternative that there's a difference between the two population means.
D. There's a difference between the two population means;
E. Reject the null hypothesis in favor of the alternative, which states that the mean of population one is greater than
the mean of population two.
23. Researchers randomly assign subjects to two groups, with each group receiving a different asthma medication.
For one month, researchers record the number of asthma attacks suffered by each subject and compute the summary
statistics as follows: x1 = 3.8, s1 = 1.7, n1 = 23 and x2 = 4.2, s2 = 1.5, n2 = 24
Compute a t statistic and p-value for the hypothesis test Ho: ( 2  1 ) = 0,Ha: ( 2  1 ) < 0. Use the conservative
method to calculate your estimate for degrees of freedom, and don't pool or use your calculator. What did you get
for your t statistic and your p-value, and what's your conclusion based on a  = .05?
A. t = -.859; p = .8002; do not reject the null hypothesis that there's no difference between the two means.
B. t = -.854; p = .2012; do not reject the null hypothesis that there's no difference between the two means.
C. t = -.189; p = .1974; do not reject the null hypothesis that there's no difference between the two means.
D. t = -1.823; p = ,959; do not reject the null hypothesis that there's no difference between the two means.
E. t = -1.823; p = .0375; reject the null hypothesis that there's no difference between the two means in favor of the
alternative hypothesis that the first mean is less than the second.
FREE RESPONSE
A study of iron deficiency in infants compared samples of infants whose mothers chose different ways to
feed them. One group contained breast fed infants, the other contained infants who were fed a standard
baby formula without any iron supplements. Here is a summary of the results on the blood hemoglobin
levels at 12 months of age: Groups
n
s
x
Breast-fed
23
13.3 1.7
Formula
19
12.4 1.8
24. State the assumptions that your procedure requires in order to be valid. State your hypotheses. Carry
out a complete test of significance. (Include a conclusion).
25. Give a 95% C.I. for the mean difference in hemoglobin levels. Show work!
26. Researchers randomly assign subjects to two groups, with each one receiving a different asthma medication.
For one month, researchers record the number of asthma attacks suffered by each subject and compute the summary
statistics as follows:
Group number one
Group number two
Compute a test statistic and a p value for H0:
x = 3.8
x = 4.2
1  2 ,
Ha:
s = 1.7
n = 23
s = 1.5
n = 24
1  2 and let  = .05
27. Construct a 99% confidence interval for the difference between the two populations described above.
(Hint: Don't pool) Show this interval TWO TIMES. Once by doing the calculations “by hand” (show
work) and the other using the TI 83 to do the work.
28. Do a test of significance for these data sets (Show all work). Draw a conclusion about the difference
between these populations.
29. Explain (in 20 words or less) the difference between a matched pair t test and a two sample t test.
30. You take two random samples from two distinct populations and calculate the following statistics:
s = 3.2
n = 28 (group one)
x = 52.6
s = 2.9
n = 29 (group two)
x = 49.3
State a full conclusion using  = .05 for the hypothesis test H0: 1  2 , Ha: 1  2
31. We are given the following information about simple random samples from two populations:
From population 1: n1 = 3, x-bar1=20, s1=3
From population 2: n2 = 4, x-bar2=21, s2=2
a) We want to test for the equality of means. Begin by stating the null and alternative hypothesis.
b) Calculate the appropriate test statistic.
c) What is the p-value?
d) Do you reject the claim of equality at the 5% level of significance?
e) State your conclusion and comment on the most obvious reason you got the results you did.
Chapter 12/13 Review Solutions
1. A
2. B
3. C
4. D
5. A: The entire interval for
that it is likely that
pm  p f
falls below 0. This indicates that p
pm  p f
m
 p f < 0, which means
6. D: Equality of variances is not necessary for a t-test to be valid. One of the conditions of a t-test is
that the underlying populations be normally distributed.
7. B: Both engine oils, with the additive and without the additive, are tested on each car. Therefore,
each car is acting as a block, or in other words, each car is matched with itself.
8. B: The hypothesis used in this experiment are
Ho: The subject is guessing: p = .050
Ha: The subject is not guessing: p> 0.50
Where p = the proprotion of correct answers. Using a 0.05 significance level, the null hypothesis
will be rejected if
pˆ  p0
> 1.645, i.e. if
p0 (1 p0 )
n
p̂ > 0.50 +1.645
100
 0.5823
0.50(1  0.50)
The subject needs to identify at least 58.23 percent correctly. Since 100 cards are dealt, the subject
needs to identify at least 59 cards correctly.
9. Answer A. II and III only. Generally, a t procedure should be used for normal distributions. But since t
procedures are robust, you can also use them in situations where the distribution isn't quite normal, as long
as you don't have outliers or significant skewness. The standard error of the mean is an estimate of the
standard deviation of the sampling distribution of the mean.
10. Answer B. In all the other examples, individual measurements were paired. In this case, there's no
pairing.
11. Answer E. left-handed employees are asked to use the mouse to complete a series of timed tasks on
each hand. For each employee, the starting hand is selected randomly. In this case, the experimental units
(hands) are paired. You analyze the difference in performance between hands for each pair of hands. This
scenario also uses randomization of the starting hand to reduce bias or confounding.
12. Answer D. Ho:  (2-1) = 0 (the mean difference between the first and second semester grade is 0),
Ha:  (2-1) > 0 (the mean second semester grade is higher than the mean first semester grade). Since the
first semester grade was subtracted from the second semester grade, the difference will be greater than a if
the second grade is higher.
13. Answer A use t = (2.533-0)/(6.4/root15) = 1.533. To find the p-value on a calculator, use
tcdf(1.533,100,14) = .074.
14. Answer C. 14, 30. If you don't pool, the number of degrees of freedom is one less than the n for the
smaller of the two samples, which is 15 - 1 = 14. If you do pool, the number of degrees of freedom is two
less than the sum of the sample sizes, which is 15 + 17 - 2 = 30.
15. D. 1.41. The standard error of the difference between the means using the conservative method is
3.212 3.092
= 1.41

10
10
16. E. conservative: 24, pooled: 50. To compute the degrees freedom with the conservative method, use
the first sample size minus one or the second sample size minus one, whichever is smaller. In this case, nl
-1 = 25 - 1 = 24, and n2 - 1 = 27 - 1 = 26, so you'd use 24. If you were to pool the variances, you'd take the
sum of the two sample sizes and subtract two. In this case, nl + n2 - 2 = 25 + 27 - 2 = 50.
17. Answer C. -1.9. You're looking for a confidence interval for the difference between two means, so
you'd calculate the estimate by subtracting the second sample mean from the first: x 1 – x2 = 23.4 - 25.3 = 1.9
18. Answer C. 2.325. The value for (margin of error) is equal to half the width of the confidence interval,
and is computed as follows: t* = 2.064
4.22 3.92

25
27
19. E. the confidence interval is (-4.225, .425); there's no difference between the two population means.
Calculate the confidence interval as follows: (23.4 - 25.3) ::f:: 2.325 = -1.9::f:: 2.325 = (-4.225, .425)
Since zero is included in this interval, we can't conclude that there's a difference between the means.
20. Answer C. .0026. Using the TI-83 two-sample t test, enter the numbers as given. The alternative
hypothesis is that mean 1 < mean 2, and you shouldn't pool the variances. The calculator gives the p-value
for this test to be approximately .0026.
21. B. 16. Looking at the t table, you need a p-value of less than .01. This first occurs at 15 degrees of
freedom, which means each sample needs to be at least 16. Look at table.
22. Answer E. reject the null hypothesis in favor of the alternative, which states that the mean of
population one is greater than the mean of population two. We reject a null hypothesis when p < a. In this
case, we're rejecting the null hypothesis in favor of the alternative hypothesis, which states that the mean
of the first population is greater than the second.
23. Answer B. t = -.854; p = .2012; do not reject the null hypothesis that there's no difference between
3.8  4.2
the two means. The t statistic is computed as follows: t =
=-.854
1.7 2 1.52

23
24
You can find the p-value on your TI-83 by using tcdf( -100,-.854,22) = .2012. Since p isn't less than a, we
can't reject the null hypothesis.
24. .Assume: Both groups are SRS and are normally distributed. Ho: means are equal, Ha: means are not
equal.
Hand: df = 18, t = 1.6537, p = .1154
Calc: df = 37.5976, t = 1.6537, p = .1065
Reasonable evidence to fail to reject Ho. Not significant at .05 level. We conclude that breastfed babies
and formula babies have same amount of iron in blood.
25. Calc: (-.2021, 2.0021) using t*=2.0251
Hand: (-.2434, 2.0434) using t* = 2.101
26. Calc: t = -.8539, p=.8011, df = 43.7695
Fail to reject Ho. Not significant at .05 level so both asthma medications are the same.
27. 11Calc: (-1.179, 7.1791)
Hand: (-1.4007, 7.4007)
Df = 36.6281
28.Ho: Means are equal, Ha: means are not equal
Assumptions: Show normal distribution
Calc: t = 1.9503, p = .0588, df = 36.6281
Hand: t = 1.9503, .05<pvalue<.10, df = 19
Pvalue is not less than .005, so Fail to Reject Ho.
Alternatively, you could say that 0 is in the Confidence Interval (-1.179, 7.1791),
So you can’t reject Ho.
29. Matched pairs t-test: -Comparing 2 responses to 2 treatments, Not independent
2-sample t-test – Comparing responses to 2 treatments or comparing characteristics of two
populations; independent of each other; separate sample from each treatment or each population.
30. 14 df = 27, t = 4.07, p = .0002. Strong evident to reject Ho. It is significant at the .05 level, so we agree
with the alternative hypothesis.
31. a) Ho:
1  2 , Ha: 1  2
b) t = -.5
c) Table: pvalue > .25*2 = .50 (pvalue off left side of table)
Calc: p = .6484
d) Fail to reject since the p-value is not less than the set signif. Level of .05
e) There is no evidence to conclude the means are not the same.
f) The sampel sizes are extremely small. When small sample, area under the cuve is large nad the
result is not far out from the mean. Small sample sizes don’t generally produce results that are
significant.