Stat 240 Homework 3 Solutions
6.5
a.
A on midterm
Not A on midterm
Total
6.9
6.25
6.29
Prefer inclass
15
20
35
Total
25
50
75
b. The response variable is preference for type of final exam. The explanatory variable is grade on
the midterm.
c. Among students who got an A on the midterm, (10/25)  100% = 40% prefer a take-home final
exam (and 60% prefer an in-class exam).
Among students who did not get an A on the midterm, (30/50)  100% = 60% prefer a take-home
final exam (and 40% prefer an in-class exam)
There is relationship between the two variables. Students who did not get an A on the midterm are
more likely to prefer a take-home final exam than are the students who got an A on the final.
a. Percent ever bullied = (42/92)  100% = 45.65% among short students
b. Percent ever bullied = (30/117)  100% = 25.64% among students not short
c. (omitted, but here’s the answer): Yes, there is a relationship. Short students are much more
likely to have been bullied than are students who are not short.
a. Frequency of religious activities (i.e. frequency of attending church, praying, and so on).
b. Blood pressure.
c. Answers may vary. Some possibilities mentioned in the Case Study are smoking habits, diet,
and extent of social network.
a. The new treatment was the more successful treatment in Hospital A. The percent surviving
with the new treatment was 100/1000 or 10%, compared to 5/100 or 5% with the standard
treatment.
b. The new treatment was the more successful treatment in Hospital B. The percent surviving
with the new treatment was 95/100 or 95%, compared to 500/1000 or 50% with the standard
treatment.
c. The combined contingency table is:
Standard
New
Total
6.32
Prefer
take-home
10
30
40
Survive
505 (5+500)
195 (95+100)
600
Die
595 (95+500)
905 (900+5 )
1600
Total
1100
1100
2200
The standard treatment is more successful than the new treatment in the combined data set. The
percent surviving with the standard treatment is (505/1100)  100% = 45.9%, compared to
(195/1100)  100% = 17.7%. This is an example of Simpson’s Paradox.
a. null:
H0: There is no relationship between gender and opinion on the death penalty.
alternative: Ha: There is a relationship between gender and opinion on the death penalty.
6.32 continued:
b.
Gender
Male
Femal
e
Total
Oppose
Favor
648  337
 147
1488
840  337
 190
1488
648  1151
 501
1488
840  1151
 650
1488
337
1151
Total
648
840
1488
6.38
a. Null hypothesis: age and having seen a ghost (or not) are not related
Alternative hypothesis: age and having seen a ghost (or not) are related
b. There is a statistically significant relationship because p-value = 0.001 is less than .05. This
means that we believe the relationship observed in the sample also holds in the population.
Regarding the hypotheses stated in part (a), the alternative hypothesis is selected.
c. The probability is .001 that the relationship observed in the sample would been have been as
strong as it is if there is no relationship between the variables in the population. More specifically,
the probability is .001 that the chi-square statistic would be as large as it is, or larger, if there is no
relationship in the population.
6.39
a. Expected count =
6.59
Row total  Column tot al 1525  677

 174 .93 for the “age 18-29 yes” cell.
Total n
5902
b. Expected counts must have the same row and column totals as the observed counts.
So, expected "age 30+ yes” = Total “yes”  expected "age 18-29 yes” = 677  174.93 = 502.07
c. In the age 18-29 group, percent expected “yes” = (174.93/1525)  100% = 11.47%.
In the age 30+ group, percent expected “yes” = (502.07/4377)  100% = 11.47%.
The percents are the same, and this is a general property of expected counts.
a. The given information is summarized in this table:
Positive SalEst
Negative SalEst
All
59%
41%
n = 615
33%
67%
n = 27
Delivered before
42nd week
Delivered 42nd
week or later
Applying the given percents to the given sample sizes, and rounding to the nearest integer, gives
the two-way table of counts for the relationship between time of delivery and SalEst test result.
The table of counts is
Delivered before
42nd week
Delivered 42nd
week or later
All
Positive SalEst
Negative SalEst
All
363 (59% of 615)
252 (41% of 615)
615
9 (33% of 27)
18 (67% of 27)
27
372
270
642
b. The biologically correct view would be that an impending delivery will cause a positive SalEst
test, so the explanatory variable would be the time until delivery and the response variable would
be the result of the test. The view of a test user, however, might be that the test result is used to
predict the time of delivery and with this view the explanatory variable would be the test result
and the time of delivery would be the response variable.
c. For all women in the study: Percent delivering before 42 weeks = (615/642)  100% = 95.79%
6.59 continued:
d. For women with a positive test: Percent delivering before 42 weeks = (363/372)  100% =
97.58%, which rounds up to 98%.
e. For women with a negative test: Percent delivering before 42 weeks (252/270)  100% =
93.33%
f. Opinions may differ, but generally the test doesn’t predict delivery much better than what might
be predicted without doing the test.. Notice that overall almost 96% of the women delivered
before the 42nd week. So, if we predicted that every woman would deliver before the 42 nd week,
we would be right almost 96% of the time. Knowledge of a positive SalEst test only improves this
predictive accuracy to 98%. And, about 93% of the women with a negative test delivered before
the 42nd week so a negative test doesn’t provide the information that a woman will have a later
delivery.
6.64
a. Null hypothesis: There is not a relationship between gun ownership (or not) and opinion about
required police permits for guns.
Alternative hypothesis: There is a relationship between gun ownership (or not) and opinion about
required police permits for guns.
b. The two-way table of counts, along with the percent favoring and opposing permits for each
gun ownership group, is:
Favor permits Oppose permits
Total
No gun in home
544 (90.2%)
59 (9.8%)
603
Yes, have gun in home
323 (72.4%)
123 (27.6%)
446
Total
867
182
1049
In the whole sample, percent owning a gun is (603/1049)  100% = 57.5%.
Notice that the sample size for this table is much less than the overall sample size of the data set.
These questions were not asked of all respondents.
c. Gun in home: (323/446)  100% =72.4% favor permits
No gun in home: (544/603)  100% = 90.2% favor permits.
The difference in percent favoring permits is almost 18%, and the sample size is relatively large,
so there probably is a real relationship between gun ownership and opinion about permits.
d. p-value  0. Yes, the relationship is statistically significant. Minitab output is shown below
(with expected counts shown below observed counts).
Rows: owngun
Minitab output for Exercise 6.64d
Columns: gunlaw
Favor
544
498.38
Oppose
59
104.62
All
603
603.00
Yes
323
368.62
123
77.38
446
446.00
All
867
603.00
182
182.00
1049
1049.00
No
Chi-Square = 56.609, DF = 1, P-Value = 0.000
e. An observed relationship is statistically significant if it is unlikely that a relationship as strong,
or stronger, would be observed in a sample if there were no relationship in the population. In the
context of this exercise, it is nearly impossible (p-value  0) that the difference in the sample
percents calculated in part (b) would be so large if there were no difference in opinion between
those who have a gun and do not have a gun in the population.