LibraryPirate

EXERCISES 157
old are drawn from these populations, find the probability that the difference in percent
of total natural teeth loss is less than 5 percent between the two populations.
Solution: We assume that the sampling distribution pN 1 - pN 2 is approximately normal.
The mean difference in proportions of those losing all their teeth is
and the variance is
m Np1 - Np 2
= .34 - .26 = .08
s 2 Np 1
- Np 2 = p 111 - p 1 2
n 1
+ p 211 - p 2 2
n 2
= 1.3421.662
250
+ 1.2621.742
200
= .00186
The area of interest under the curve of pN 1 - pN 2 is that to the left of .05. The
corresponding z value is
z =
.05 - 1.082
2.00186 = -.70
Consulting Table D, we find that the area to the left of z = -.70 is
.2420. ■
EXERCISES
5.6.1 According to the 2000 U.S. Census Bureau (A-8), in 2000, 9.5 percent of children in the state of
Ohio were not covered by private or government health insurance. In the neighboring state of
Pennsylvania, 4.9 percent of children were not covered by health insurance. Assume that these
proportions are parameters for the child populations of the respective states. If a random sample
of size 100 children is drawn from the Ohio population, and an independent random sample of
size 120 is drawn from the Pennsylvania population, what is the probability that the samples would
yield a difference, pN 1 - pN 2 of .09 or more
5.6.2 In the report cited in Exercise 5.6.1 (A-8), the Census Bureau stated that for Americans in the age
group 18–24 years, 64.8 percent had private health insurance. In the age group 25–34 years, the percentage
was 72.1. Assume that these percentages are the population parameters in those age groups
for the United States. Suppose we select a random sample of 250 Americans from the 18–24 age
group and an independent random sample of 200 Americans from the age group 25–34; find the probability
that pN 2 - pN 1 is less than 6 percent.
5.6.3 From the results of a survey conducted by the U.S. Bureau of Labor Statistics (A-9), it was estimated
that 21 percent of workers employed in the Northeast participated in health care benefits
programs that included vision care. The percentage in the South was 13 percent. Assume these
percentages are population parameters for the respective U.S. regions. Suppose we select a simple
random sample of size 120 northeastern workers and an independent simple random sample
of 130 southern workers. What is the probability that the difference between sample proportions,
pN 1 - pN 2 , will be between .04 and .20