Question

Question He Marketers want to know about the differences between men's and women's use of the met. A research polt in April 2009 from a random sample of adults found that 2378 of 30 Internet, at least occasionally, while 2382 of 3169 women did. a) Find the proportions of men and women who said they use the internet a host osionaly. b) What is the difference in proportions? c) What is the standard error of the diference? d) Find a 95% confidence interval for the difference between percentages of usage by men and women ratio de alep, be the sample proportion of men who use the more at least occasional and lot , be the sample proportion of women (Round to two decimal places as needed --Found to four decimal places as needed SE(-6) - Rond to tour decimal places as needed di The confidence interval is Enter your answer in each of the answer boxes

Answer 1

This is a question for binomial proportions.

  

=1 - 0.95 = 0.05

Therefore the critical value at

= Z_0.025

= 1.96 .............using normal percentage tables with p = 0.025

SE =

Margin of error = Critical value * SE

= *

(1- )% CI for population proportion

   

		Men	Women
	x	2378	2382
	n	3049	3169
1)	Sample	0.7799	0.7517
2)	Sample p diff	0.0283
	pooled	0.7655
3)	SE (diff)	0.0107
	C.V.	1.96
	MOE	0.02107
4)	Lower L	0.00721
Upper L	0.04934