Question

In a sample of 58 men, 37 said that they had less leisure time today than they had 10 years ago. In a sample of 58 women, 45 women said that they had less leisure time today than they had 10 years ago. At =α0.05, is there a difference in the proportions?

Use p1 for the proportion of men with less leisure time.

Find the 95% confidence interval for the difference of the two proportions. Round the answers to three decimal places.

Answer 1

Sol:

p1^=sample proportion of men with less leisure time=x1/n1=37/58= 0.637931

p2^=sample proportion of women with less leisure time=x2/n2=45/58= 0.7758621

95% confidence interval for the difference of the two proportions

=(p1^-p2^)+-z*sqrt(p1^*(1-p1^)/n1+p2^(1-p2^)/n2)

=(0.637931-0.7758621)-1.96*sqrt((0.637931*(1-0.637931)/58)+(0.7758621*(1-0.7758621)/58))

-0.3016891,0.02582694

ANSWER"

95% confidence interval for the difference of the two proportions.

(-0.302,0.026)