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 he wishes the estimate to be within three percentage points with 99 ​% ​confidence, assuming that ​(a) he uses the estimates of 22.3 ​% male and 18.9 ​% female from a previous​ year? ​(b) he does not use any prior​ estimates?

Answer 1

Using the z-distribution, it is found that the needed sample sizes are:

a) 242

b) 1842

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

99% confidence level, hence$\alpha = 0.99$, z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$, so $z = 2.575$.

Item a:

The estimate is:

$\pi = 0.223 - 0.189 = 0.034$

The sample size is n for which M = 0.03, then:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 2.575\sqrt{\frac{0.034(0.966)}{n}}$

$0.03\sqrt{n} = 2.575\sqrt{0.034(0.966)}$

$\sqrt{n} = \frac{2.575\sqrt{0.034(0.966)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.034(0.966)}}{0.03}\right)^2$

$n = 241.97$

Rounding up, a sample of 242 is needed.

Item b:

No prior estimates, hence $\pi = 0.5$ is used.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 2.575\sqrt{\frac{0.5(0.5)}{n}}$

$0.03\sqrt{n} = 2.575\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{2.575\sqrt{0.5(0.5)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.5(0.5)}}{0.03}\right)^2$

$n = 1841.8$

Rounding up, a sample of 1842 is needed.

For more on the z-distribution, you can check brainly.com/question/25404151