Question

Marketers want to know about the differences between​ men's and​ women's use of the Internet. A research poll in April 2009 from a random sample of adults found that

2375 of 3036 men use the​ Internet, at least​ occasionally, while 2352 of 3182 women did.

​a) Find the proportions of men and women who said they use the Internet at least occasionally.

​b) What is the difference in​ proportions?

​c) What is the standard error of the​ difference?

​d) Find a​ 95% confidence interval for the difference between percentages of usage by men and women nationwide.

Answer 1

a)
p1cap = 0.7823 , p2cap = 0.7392
n1 = 3036 , n2 = 3182

b)

diff in proportion = 0.7823 - 0.7392 = 0.0431

c)

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.7823 * (1-0.7823)/3036 + 0.7392*(1-0.7392)/3182)
SE = 0.0108

d)

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.7823 - 0.7392 - 1.96*0.0108, 0.7823 - 0.7392 + 1.96*0.0108)
CI = (0.0219 , 0.0643)