Question

A recent survey showed that 65% of U. S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 500 companies. Find the 95% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage.

Answer 1

Solution :

Given that,

$\hat p$ = 65% = 0.65

1 - $\hat p$ = 1 - 0.65 = 0.35

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96

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

= 1.96 * ($\sqrt$((0.65 * 0.35) / 500)

= 0.042

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < P < $\hat p$ + E

0.65 - 0.042 < p < 0.65 + 0.042

0.608 < p < 0.692

(0.608 , 0.692)