Question

Please Show Work. Thanks.

3. In a random sample of 193 engineers, 15.2% of them were women. a) Construct an 85% confidence interval for the population proportion, p. Round to tenths of a percent and write a sentence answer. b) Find the margin of error, round to tenths of a percent

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = $\hat p$ = 0.152

1 - $\hat p$ = 1 - 0.152 = 0.848

Z$\alpha$/2 = Z_0.075 = 1.44

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

= 1.44 ($\sqrt$((0.152 * 0.848) / 193)

= 0.037

A 85% confidence interval for population proportion p is ,

$\hat p$ ± E   

= 0.152  ± 0.037

= ( 0.115, 0.189 )

= ( 11.5%, 18.9% )

We are 85% confident that true proportion of them were women between 11.5% and 18.9%

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

= 1.44 ($\sqrt$((0.152 * 0.848) / 193)

= 0.037

= 3.7%