Question

Suppose that in a random sample of 500 employed Americans, there are 59 individuals who say that they would fire their boss if they could. Calculate a 90% confidence interval for the population proportion who would fire their boss if they could. (Round the answers to three decimal places.)

Answer 1

Solution :

Given that,

n = 500

x = 59

$\hat p$ = x / n = 59/500 = 0.118

1 - $\hat p$ = 1 - 0.118 = 0.882

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.1

$\alpha$ / 2 = 0.1/ 2 = 0.05

Z_$\alpha$/2 = Z_0.05 =1.645

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

= 1.645 * $\sqrt$[(0.118 * 0.882) / 500]

= 0.024

A 90% confidence interval for population proportion p is ,

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

0.118 - 0.024 < p < 0.118 + 0.024

0.094 < p < 0.142

The 90% confidence interval for the population proportion is 0.094 < p < 0.142