Question

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A simple random sample of size n = 400 individuals who are currently employed is asked if they work at home at least once per week. Of the 400 employed individuals surveyed, 42 responded that they did work at home at least once per week. Construct a 99% confidence interval for the population proportion of employed individuals who work at home at least once per week. The lower bound is (Round to three decimal places as needed.) The upper bound is (Round to three decimal places as needed.)

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 42 / 400 = 0.105

1 - $\hat p$ = 1 - 0.105 = 0.895

Z$\alpha$/2 = Z_0.005 = 2.576

Margin of error = E = Z_{$\alpha$ / 2} * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 2.576 ($\sqrt$((0.105 * 0.895) / 400)

= 0.039

A 99% confidence interval for population proportion p is ,

$\hat p$ ± E   

= 0.105  ± 0.039

= ( 0.066, 0.144 )

lower limit = 0.066

upper limit = 0.144