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 2399 of 3038 men use the​ Internet, at least​ occasionally, while 2354 of 3185 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)

Here, , n1 = 3038 , n2 = 3185
p1cap = 0.7897 , p2cap = 0.7391

b)

difference in​ proportions = 0.7897 - 0.7391 = 0.0506

c)

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.7897 * (1-0.7897)/3038 + 0.7391*(1-0.7391)/3185)
SE = 0.0107

d)

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.7897 - 0.7391 - 1.96*0.0107, 0.7897 - 0.7391 + 1.96*0.0107)
CI = (0.0296 , 0.0716)