Question

A physical therapist wants to determine if there is a difference in proportion of men and women who participate in regular sustained physical activity. The physical therapist took two independent random samples of men and women, each containing 800 observations. The number of men who participate in regular sustained physical activity was 380, and the number of women was 420. At a = 0.05 can you reject the claim that the proportion is the same for men and women?

A) state the null and alternative hypothesis

B) Check the assumptions to perform the significance

C) Calculate the test statistic and conlude

D) Conclude in context

E) Create a 95% confidence interval for the difference in proportions

F) Interpret the confidence interval in context

Answer 1

Solution-A:

Ho:p1=p2

proportion is the same for men and women

H1:p1 not =p2

proportions are different for men and women

B) Check the assumptions to perform the significance

Randomly selected samples,n1=n2=800

independent samples

n1p1=800*380/800=380>5

n1(1-p1)=800*(1-380/800)=420>5

n2p2=800*420/800=420>5

n2(1-p2)=800*(1-420/800)=380>5

Approximately normal

C) Calculate the test statistic and conlude

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

pbar=x1+x2/n1+n2=(380+420)/(800+800)=0.5

p1^=380/800=0.475

p2^=420/800=0.525

z=(0.475-0.525)/sqrt(0.5*(1-0.5)*(1/800+1/800))

z=-2

D) Conclude in context

p value (lef tail )in excel

=NORM.S.DIST(-2,TRUE)

=0.022750132

2*0.022750132=0.04550026

p value=0.04550026

p<0.05

Reject Ho

Accept Ha

Conclusion:
There is suffcient statistical evidence at 5% level of significance to

reject the claim that the proportion is the same for men and women