Week 9 Assignment
1. Problem 12, page 523
Effectiveness of Smoking Bans
The Joint Commission on Accreditation of Healthcare Organizations mandated that
hospitals ban smoking by 1994. In a study of the effects of this ban, subjects who smoke
were randomly selected from two different populations. Among 843 smoking employees
of hospitals with the smoking ban, 56 quit smoking one year after the ban. Among 703
smoking employees from workplaces without a smoking ban, 27 quit smoking a year
after the ban (based on data from “Hospital Smoking Bans and Employee Smoking
Behavior” by Longo, Brownson, et al., Journal of the American Medical Association,
Vol. 275, No. 16). Is there a significant difference between the two proportions at a 0.05
significance level? Is there a significant difference between the two proportions at a 0.01
significance level? Does it appear that the ban had an effect on the smoking quit rate?
H 0 : p1 = p2
H 0 : p1 ¹ p2
a = 0.05 or 0.01
Critical region (a = 0.05) : |Z| > 1.96
Critical region (a = 0.01) : |Z| > 2.576
Under H0 we have that the best estimate for p is pˆ =
Test statistic Z =
pˆ1 - pˆ 2
n+ m
pˆ (1- pˆ )
nm
x+ y
56 + 27
.
=
n + m 843 + 703
= 2.43
Conclusions:
Since 2.43 > 1.96 we reject H0 at a = 0.05 and conclude that the difference is
significant.
Since 2.43 < 2.576 we fail to reject H0 at a = 0.01 and conclude that the difference is not
significant.
The P-value is 0.015 so it appears that the ban had an effect on the smoking quit rate.
2. Problem 6, page 537
Confidence Interval for Effects of Marijuana Use on College Students
Refer to the sample data used in Exercise 5 and construct a 98% confidence interval for
the difference between the two population means. Does the confidence interval include
zero? What does the confidence interval suggest about the equality of the two population
means?
98% CI:
m1 - m2 = x1 - x2 ± t ×s p
1 1
1
1
+ = 53.3 - 51.3 ± 2.356 ×4.0714
+
= (0.311, 3.689)
n m
65 64
The interval does not contain zero. Since it does not contain zero it suggests that the
difference in means is significant (at a = 0.02 ).
3. Problem 6, page 548
Self-Reported and Measured Male Heights
As part of the National Health and Nutrition Examination Survey conducted by the
Department of Health and Nutrition Services, self-reported heights and measured heights
were obtained for males aged 12-16. Listed below are sample results.
a. Is there sufficient evidence to support the claim that there is a difference between
self-reported heights and measured heights of males aged 12-16? Use a 0.05
significance level.
b.
n = 12
D= - 1
sD = 3.52
H0 : m= 0
H1 : m ¹ 0
a = 0.05
df = n - 1 = 12 - 1 = 11
Critical region: |T| > 2.201
Test statistic T =
D
sD
n
- 1
= - 0.98
3.52 12
=
Conclusion: Since |-0.98| = 0.95 < 2.201 we fail to reject H0 and conclude that the
difference is not significant.
c. Construct a 95% confidence interval estimate of the mean difference between
reported heights and measured heights. Interpret the resulting confidence interval,
and comment on the implications of whether the confidence interval limits contain 0.
95% CI:
m= D ± t
sD
n
= - 1 ± 2.201
3.52
= (- 3.24, 1.24)
12
Since it contains zero we cannot claim that the difference is significant.
Reported Height
Measured Height
Difference D
68
71
67.9 69.9
0.1
1.1
63
64.9
-1.9
70
68.3
1.7
71
70.3
0.7
60 65
60.6 64.5
-0.6 0.5
64
67
-3
54
63
55.6 74.2
-1.6 -11.2
66
65
1
72
70.8
1.2
4. Example 4, page 559
Hypothesis Test for Effect of Marijuana Use on College Students
In a study of the effect of marijuana use, light and heavy users of marijuana in college
were tested for memory recall, with the results given below (based on data from “The
Residual Cognitive Effects of Heavy Marijuana Use in College Students” by Pope and
Yurgelun-Todd, Journal of the American Medical Association, Vol. 275, No. 7). Use a
0.05 significance level to test the claim that the population of heavy marijuana users has a
standard deviation different from that of light users.
Items sorted correctly by light marijuana users:
Items sorted correctly by heavy marijuana users:
n = 64, x = 53.3, s = 3.6
n = 65, x = 51.3, s = 4.5
H 0 : s 12 = s 22
H 0 : s 12 ¹ s 22
a = 0.05
Critical value: F < F(0.025, 63, 64) = 0.61 or F > F(0.975, 63, 64) = 1.64
Test statistic F =
s12 3.62
=
= 0.64
s22 4.52
Conclusion: Since 0.61 < 0.64 < 1.64 we fail to reject H0 and conclude that the observed
difference in standard deviations is not significant.
BONUS: (Maximum of 5 points)
Problem 9, page 559