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 three percentage points with 90% confidence, assuming that (a) she uses the estimates of 22,8% male and 18.6% female from a previous year? (b) she does not use any prior estimates? (Round up to the nearest whole number.) () = (Round up to the nearest whole number) Th Enter your answer in each of the answer box hilar 30 MacBook Pro S 2 4 & 5 6 7 8 Q W E R. Т. Y U A S F G H Z Х C

Answer 1

Solution :

Here we want determine the difference in proportion of men and women who participated in the regular sustained physical activity. We want to find the sample size such that she wishes the estimate to be within three percent point with 90% confidence interval.

(a). The estimate of 22.8% of men and 18.6% of women from previous year then the formula for the sample size is given by;

$n = \hat p*(1-\hat p) *( \frac{Z_{\alpha /2}}{me})^2$

Where $\hat p = \frac{\hat p_1 + \hat p_2}{2} = \frac{0.228 + 0.186}{2}=0.207$ ​​​​​​

$Z_{\alpha /2} = 1.64 ~~ for~ 90\% ~Confidence ~ interval$​​​​​

me (margin of error) = 0.03.

And hence we get

$n =0.207 *(1-0.207)* (\frac{1.64}{0.03})^2 = 490.5561 \approx 491$

(b). If there is no any prior estimate then we use $\hat p= 0.5$ ​​​​​​

And hence we get, $n =0.5 *(1-0.5)* (\frac{1.64}{0.03})^2 = 747.1111 \approx 748$