Question

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained physical activity. What sample size should be obtained if she wishes the estimate to be within four percentage points with 90​% ​confidence, assuming that

​(a) she uses the estimates of 22.8​% male and 19.4​% female from a previous​ year?

​(b) she does not use any prior​ estimates?

Answer 1

Ans:

a)Margin pf error,E=0.04

z=1.645 for 90% confidence level

Margin of error=0.04=1.645*sqrt((0.228*(1-0.228)/n₁)+(0.194*(1-0.194)/n₂))

Assume n₁=n₂=n

0.04=1.645*sqrt((0.228*(1-0.228)+0.194*(1-0.194))/n)

0.04=1.645*sqrt(0.33238/n)

n=0.33238*(1.645/0.04)^2

n=562.14

n₁=n₂=562

b)When no prior​ estimates are available,use 0.5

Margin of error=0.04=1.645*sqrt((0.5*(1-0.5)/n₁)+(0.5*(1-0.5)/n₂))

Assume n₁=n₂=n

0.04=1.645*sqrt((0.5*(1-0.5)+0.5*(1-0.5))/n)

0.04=1.645*sqrt(0.5/n)

n=0.5*(1.645/0.04)^2

n=845.63

n₁=n₂=846