Q:

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 he wishes the estimate to be within two percentage points with 90​% ​confidence, assuming that ​(a) he uses the estimates of 21.6​% male and 19.2​% female from a previous​ year? ​(b) he does not use any prior​ estimates?

Accepted Solution

A:
Answer:Step-by-step explanation:The standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean(μ)  and standard deviation(σ)The general formula for the sample size is given below: [tex]n=p^{'}(1-p^{'})(\frac{Z_{\frac{a}{3} } }{E} )^{2}[/tex]The formular for finding sample size is given as:[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]a.) it is given that [tex]E=±0.02, p^{'}_{1}=0.216, p^{'}_{2}=0.192[/tex] The confidence level is 0.90For (1 - ∝) = 0.90∝=0.10; ∝/2 = 0.05frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.216(1-0.216)+0.192(1-0.192))\\=351.22≅352[/tex]The required sample size is 352 (nearest whole number)b.) it is given that [tex]E=±0.02, p^{'}_{1}=0.5, p^{'}_{2}=0.5[/tex] The confidence level is 0.90For (1 - ∝) = 0.90∝=0.10; ∝/2 = 0.05frm the standard normal table, the required [tex]Z_{0.05}[/tex] value for 90% confidence is 1.645. The sample size is as shown:[tex]n=(\frac{Z_{\frac{a}{3} } }{E} )^{2} * (p_{1}(1-p_{1})+p_{2}(1-p_{2}))[/tex]=[tex]n=(\frac{1.645}{0.05} )^{2} * (0.5(1-0.5)+0.5(1-0.5))\\=541.205≅542[/tex]The required sample size is 542 (nearest whole number)