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erastova [34]
2 years ago
14

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p

hysical activity. What sample size should be obtained the wishes the estimate to be within three percentage points
with 90% confidence, assuming that
(a) he uses the estimates of 21.3% male and 19.5% female from a previous year?
(b) he does not use any prior estimates?
(a)n (Round up to the nearest whole number)
(b) (Round up to the nearest whole number.)
Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
5 0

Using the z-distribution and the formula for the margin of error, it is found that:

a) A sample size of 54 is needed.

b) A sample size of 752 is needed.

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 of:

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

90% confidence level, hence\alpha = 0.9, z is the value of Z that has a p-value of \frac{1+0.9}{2} = 0.95, so z = 1.645.

Item a:

The estimate is \pi = 0.213 - 0.195 = 0.018.

The sample size is <u>n for which M = 0.03</u>, hence:

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

0.03 = 1.645\sqrt{\frac{0.018(0.982)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.018(0.982)}

\sqrt{n} = \frac{1.645\sqrt{0.018(0.982)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.018(0.982)}}{0.03}\right)^2

n = 53.1

Rounding up, a sample size of 54 is needed.

Item b:

No prior estimate, hence \pi = 0.05

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

0.03 = 1.645\sqrt{\frac{0.5(0.5)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.645\sqrt{0.5(0.5)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.645\sqrt{0.5(0.5)}}{0.03}\right)^2

n = 751.7

Rounding up, a sample of 752 should be taken.

A similar problem is given at brainly.com/question/25694087

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