Study Guide
11.1a 2
11.1b
10
12
4
9
16
Chapter 11 Study Guide Solutions
Ideal Response
(a) H o
: p red
 18
38
, p black
 18
38
, p green
 2
38
versus H a
(b) There was a random sample of 200 spins. The expected count for black is
200
 
38

94.74
for red is 200
 
38

94.74
for green is 200
 
38

10.53
.
 2 
94.74
2

 2
94.74

 2
10.53

1.001 0.192
 
A chi-square goodness-of-fit test would not be appropriate here because the data for each day were not counts but rather means, and because the same 50 students were used each day (so the observations were not independent of each other).
A chi-square goodness-of-fit test would not be appropriate here because many of these students likely had same teachers (so the observations were not independent of each other).
State: We want to perform a test at the
H o
: p
Hispanic

0.28, p
Black

 
0.24, p
0.05
White

significance level of
0.35, p
Asian

0.12, p
Others

0.01
Plan: versus H a
.
We should use a chi-square goodness-of-fit test if the conditions are satisfied.
Random : A random sample was used. Large sample size : The total sample size is
800. The expected counts in each category are Hispanic:
  
224 ,
Black:
  
192 , White:
  
280 , Asian:
  
96 , and Others:
  
8 , all of which are at least 5. Independent : It seems
Do: reasonable that there would be more than 8000 residents of a large housing complex. The conditions are met.
The test statistic is:

2

 2

 2

 2

 2

224 and the distiribution has
192
5 1 4
280 96
 
df. The P-value is 0.00003.
 2
8

26.0625
Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the residents of this housing complex do not follow the distribution by ethnic background of New York City as a whole.
Follow-up analysis : The breakdown of the chi-square statistics is

2
   
24.5
. From this we see that the largest contributor to the statistic, by far, is the last amount. There were many more people in the “Other” category than we would have predicted.
State: We want to perform a test at the
 
0.05
significance level of H o
: p i

0.167
Plan:
Do: for all flavors versus H a p i
.
We should use a chi-square goodness-of-fit test if the conditions are satisfied.
Random : A random sample was used. Large sample size : The total sample size is
120. The expected counts in each category is 120(0.167) = 20 which is at least 5.
Independent : It seems reasonable that there would be more than 1200 Froot
Loops. The conditions are met.
The test statistic is:

2

 2
20


20 and the distiribution has
2

 2

 2


6 1 5
20 20 20
 
df. The P-value is 0.1618.
2

20
2

7.9
Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that the six flavors are not equally likely.
Follow-up analysis : Not necessary.

11.c
11.1 MC
11.2a
18
28
Chapter 11 Study Guide Solutions
(a) State: Our parameters of interest are p
1
 true proportion of older black men who fear crime and p
 true proportion of older black women who fear crime. We want to
2 estimate the difference p
1
 p
2
at a 90% confidence level.
Plan: We should use a two-sample z interval for p
1
 p
2
if the conditions are satisfied.
Random : Both samples were selected randomly and independent from one another. Normal : The number of successes and failures in both groups are at least
10 (Older black men: 46 successes, 17 failures. Older black women: 27 successes,
29 failures). Independent : Both samples are less than 10% of their respective populations (there are more than 630 older black men and 560 older black women
Do: in Atlantic City, NJ). The conditions are met.
From the data we find that n

,
1
63 p
1
 46 
63
0.73
So our 90% confidence interval is
, n

,
2
56
ˆ
2
 27 
56
0.482
0.73(0.27)  0.482(0.519)
63 56


(0.105,0.391)
Conclude: We are 90% confident that the interval between 0.105 and 0.391 captures the true difference in the proportions of older black men and older black women who fear crime. This interval suggests that between 10.5% and 39.1% more older black men than older black women fear crime.
(b) Since the interval does not contain 0, there is convincing evidence that the two proportions are not the same.
19. D 20. A 21. C 22. D
(a) There were 202 in each of the black parents and Hispanic parents groups and 201 in the white parents group so the conditional distributions are:
Survey result Black parents Hispanic parents White parents
Excellent
Good
Fair
Poor
Don’t know
0.059
0.342
0.371
0.119
0.109
0.168
0.272
0.302
0.119
0.139
0.109
0.403
0.299
0.119
0.070
(b)
(c) In general, Hispanic parents are more likely to think that the high schools are doing an excellent job. Whites are more likely than either of the other two groups to think that the high schools are doing either an excellent or a good job. Black parents are more likely to think that schools are only doing a fair job.

11.2b
11.2c
29
30
34
Chapter 11 Study Guide Solutions
(a) H o
: There is no difference in the distribution of sports goals for females and male undergraduates.
H a
: There is a difference in the distribution of sports goals for females and male undergraduates.
(b) First note that there are 67 + 67 = 134 students in the sample. Also, there are 14 + 31 = 45 total students who are classified as HSC-HM. So the expected count for the number of
(67)

22.5
. The other expected counts are computed in a similar fashion given in the table below:
Goal Females Males
HSC-HM
HSC-LM
22.5
12.5
22.5
12.5
LSC-HM
LSC-LM
13
19
13
19
(c) The test statistic is:

2

 2
22.5
 
 2
19

24.898
(a) H o
: There is no difference in the distribution of opinions about high school among black,
Hispanic, and white parents.
H a
: There is a difference in the distribution of opinions about high school among black,
Hispanic, and white parents.
(b) First note that there are 202 + 202 + 201 = 605 parents in the sample. Also, there are 12 +
34 + 22 = 68 total parents who thought the high schools are excellent. So the expected count for the number of black parents who think the high schools are excellent is computed
(202)

22.7
. The other expected counts are computed in a similar fashion given in the table below:
Survey result
Excellent
Good
Fair
Black parents
22.7
68.45
65.44
Hispanic parents
22.7
68.45
65.44
White parents
22.59
68.11
65.12
Poor
Don’t know
24.04
21.37
(c) The test statistic is:

2

2
 
22.7
(a) The two-way table of the data is given below:
Supplement Male
24.04
21.37
 2
21.26

22.426
21.26
21.26
PBM
NLCP
PL-LCP
TG-LCP
9
8
8
10
Female
11
11
11
9
Total
20
19
19
19
Total 35
The proportions (conditional distributions) are given in the table below. Note that the conditional distributions go across rows in this case.
Supplement Male Female
PBM
NLCP
9
20
8
19

0.45

0.421
0.55
0.579
PL-LCP
TG-LCP
8
19
10
19


0.421
0.526
It does appear that the groups are roughly balanced.
42 77
0.579
0.474

36
38
35
Chapter 11 Study Guide Solutions
(b) State: We want to perform a test at the
 
0.05
significance level of
H : There is no significant difference in the proportion of females in different o supplement groups.
H a
: There is a significant difference in the proportion of females in different supplement groups.
Plan: We should use a chi-square test for homogeneity if the conditions are satisfied.
Random : The data came from a randomized experiment. Large sample size : We used technology to get the following expected counts:
Supplement
PBM
Male
9.09
Female
10.91
NLCP
PL-LCP
8.64
8.64
10.36
10.36
TG-LCP 8.64 10.36
All of these counts are at least 5. Independent : Due to the random assignment,
Do: these three groups of infants can be viewed as independent. Individual observations in each group should also be independent—knowing whether one infant is female gives no information about any other infant’s gender. The conditions are met.
The test statistic is:
 2 square distribution with

2
 

2

0.568
. We use a chi-
9.09
(4 1)(2 1) 3
10.36
  
df and find a P-value of 0.9036.
Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to say that the groups differ significantly based on gender.
We cannot use a chi-square test with this data because we do not have the actual counts of the travelers in each category. We also do not know if the sample was taken randomly, nor if the same people could have been counted more than once (traveling both for leisure one time and business another).
The observations in this table are not independent of each other. Data on each woman occurs in each row.
(a) The data are in the table below: c
Placebo
Stroke
250
No stroke
1399
Total
1649
Aspirin
Dipyridamole
Both
Total
206
211
157
824
1443
1443
1493
5778
1649
1654
1650
6602
(b)
It appears that all 4 treatments have about the same success rate.
(c) The null hypothesis given says that each of the four treatments leads to the same probability

40
42
44
Chapter 11 Study Guide Solutions of success.
(d) There are a total of 6602 subjects in the sample. Also, there are 824 subjects who had strokes. So, the expected count for the number of who had strokes among those who got the placebo is computed as
824
6602
(1649)

205.81
. The other expected counts are computed in a similar fashion and are given in the table below.
Treatment Stroke
Placebo
Aspirin
205.81
205.81
Dipyridamole
Both
206.44
205.94
State: We want to perform a test at the
 
0.05
significance level of
No stroke
1443.19
1443.19
1447.56
1444.06
H : There is no significant difference in the stroke rates of the different treatments. o
Plan:
H a
: There is a significant difference in the stroke rates of the different treatments..
We should use a chi-square test for homogeneity if the conditions are satisfied.
Random : The data came from a randomized experiment. Large sample size : We computed the expected counts in Exercise 11.38. All of these counts are at least 5.
Independent : Due to the random assignment, these four groups of subjects can be viewed as independent. Individual observations in each group should also be independent—knowing whether one infant is female gives no information about any other infant’s gender. The conditions are met.
Do: The test statistic is:

2
 chi-square distribution with
 2
 
 2

24.243
. We use a
205.81
(4 1)(2 1) 3
1444.06
  
df and find a P-value of approximately 0.
Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that the stroke rate is different for at least one of the four treatments.
The statistic breaks down into its individual components as follows:

2        
24.243
The largest component of this equation comes from those who had strokes on both aspirin and
Dipyridamole. Far fewer were in this group than would have been expected. The next largest component comes from those who had strokes while on the placebo. Far more people were in this group than would have been expected. This means that the two drugs together far outperform the placebo and should probably be used if the side effects are not too great.
(a) The hypotheses are as follows:
H : There is no association between support for the death penalty and educational level. o
H a
: There is an association between support for the death penalty and educational level.
Assuming that there is no association between support for the death penalty and education level, the probability of observing a difference in support of the death penalty among those with a high school degree and those with a Bachelor’s degree as large or larger than the one found in this study is approximately 0. This suggests that there is an association between support for the death penalty and educational level in the population.
(b) The P-value for this test is identical to the P-value for the test in part (a). And since the hypotheses for the two-sample z test are the same (though usually stated in symbols rather than words), the conclusions to this test are the same as for the test in part (a).

11.2d 46 (a)
Chapter 11 Study Guide Solutions
47
The higher the degree earned, the less likely people are to think that astrology is scientific.
(b) State: We want to perform a test at the
 
0.05
significance level of
H : Education level and belief about astrology are independent. o
H a
: Education level and belief about astrology are not independent.
Plan: We should use a chi-square test for association/independence if the conditions are satisfied. Random : The data came from a randomized experiment. Large sample size : The computer printout show expected counts, and all of these counts are at least
5. Independent : The sample had 687 people in it. There are more than 6870 adults in the US who could have been part of the sample. The conditions are met.
Do: The test statistic is given by the output as being 10.582. We use a chi-square distribution with (4 1)(2 1) 3 df and find a P-value of 0.005.
Conclude: Since the P-value is less than 0.05, we reject the null hypothesis and conclude that education level and belief about the scientific nature of astrology are not independent.
(a) Of those who spend less than 2 hours on extracurricular activities per week, the proportion who earn a C or better is
11
20

0.55
and the proportion who earn a D or worse is 0.45. Of those who spend 2 to 12 hours on extracurricular activities per week, the proportion who earn a C or better is 0.747 and the proportion who earn a D or worse is 0.253. Of those who spend more than 12 hours on extracurricular activities per week, the proportion who earn a C or better is 0.375 and the proportion who earn a D or worse is 0.625. A bar graph is given below. It appears that those who are moderately involved in extracurricular activities (2 to
12 hours per week) earn higher grades than those who are very active and those who are not active.

48
50
Chapter 11 Study Guide Solutions
(b) A chi-square test should not be performed in this setting because there were only 8 students observed in the over 12 hours category. The expected count for at least one of these two cells will, necessarily, be less than 5. We also do not know how the students were chosen for this survey other than the fact that they were in a required chemical engineering course. We do not know if they were randomly selected or not, for instance.
(a) The second row in the table shown is just a continuation of the first row – that is, the same people are still being measured. Also, the table does not show the people who did not get rid of the warts, so this table only shows the successes, not the failures. The table we should use is as follows:
Result Treatment Control Total
Warts gone
Warts remain
25
7
3
20
28
27
Total 32 23 55
(b) It is still not appropriate to use a chi-square test because we do not know how the experiment was conducted. There is no mention of randomization so we have to wonder about that and independence.
(a)
In soft water, the new product is slightly preferred over the standard product and there is not much difference between the warm water wash and the hot water wash. In hard water, the new product has a larger majority of the preference and is slightly more preferred among those using a warm water wash.
(c) State: We want to perform a test at the
 
0.05
significance level of
H : There is no association between type of wash and support for the new product o among people who don’t’ use the established brand.
H a
: There is an association between type of wash and support for the new product among people who don’t’ use the established brand.
Plan: We should use a chi-square test for association/independence if the conditions are satisfied. Random : The data came from a random sample. Large sample size : We used technology to get the following expected counts.
Product preference Soft, warm Soft, hot Hard, warm
Standard 49.81 24.05 47.23
New 66.19 31.95 62.77
Hard, hot
30.92
41.08
All of these counts are at least 5. Independent : Our sample includes 354 people.
This is less than 10% of the population of the US who don’t currently use the established brand. The conditions are met.
Do: The test statistic is

2
 square distribution with
 2
 
 2

2.058
. We use a chi-
49.81
(4 1)(2 1) 3
41.08
  
df and find a P-value of 0.5605.
Conclude: Since the P-value is greater than 0.05, we fail to reject the null hypothesis We do not have enough evidence to say that there is an association between type of wash and support for the new product among people who don’t use the established brand..

11.2 MC
51
52
Chapter 11 Study Guide Solutions
(a) Since this represents one sample and the subjects were then classified by their answer and their gender, this would be a chi-square test of association/independence.
(b) The hypotheses are
H o
: Gender and where subjects live are independent.
H a
: Gender and where subjects live are not independent
(c) Random : This was a random sample. Large sample size : The expected counts are given in the output and they are all at least 5. Independent : This sample of 4854 young adults is less than 10% of all young adults who could have been sampled.
(d) If gender and place of living are independent, then we have a 1.2% chance of finding a sample with as much association or more than we found. This is fairly unlikely (less than a
5% chance) so we reject the null hypothesis and conclude that gender and where young adults live are not independent.
(a) Since there were two separate samples, one of American students and one of Asian students, we should perform a chi-square test for homogeneity.
(b) The hypotheses are
H : There is no difference in the reasons for shopping from catalogs between the two o groups of students.
H a
: There is a difference in the reasons for shopping from catalogs between the two groups of students.
(c) Random : This data came from random samples. Large sample size : The expected counts are given in the output and they are all at least 5. Independent : There are at least 950
American students and 620 Asian students.
(d) If the distribution of reasons for shopping from catalogs is the same for both Asian and
American students, we have a 0.01% chance of finding as big a difference or bigger than we found between the two samples. This is quite unlikely, so we reject the null hypothesis and conclude that there are differences between American and Asian students in terms of why they shop from catalogs.
53. E 54. C 55. A 56. D 57. B 58. B