Question

Hays Medical Center wants to estimate the percentage of adults who exercise at least 1 hour per week. How large a sample should be selected in order to estimate this proportion with a 90% confidence level and a margin of error of at most 5%? (Assume no preliminary estimate of p-hat is available.)

Answer 1

When there is no preliminary estimate, consider $\small \hat{p}$ = 0.5

Margin of error, E = Z*

V1-/n

For 90% confidence interval, Z* = 1.645

E = 5% = 0.05

Therefore, 0.05 = 1.645 x

V0.5(1 0.5)/n

$\small \sqrt{n}$ = 16.45

n = 271    (rounded up)

Sample size to be selected = 271