Microwave Cooking A survey of 90 households found that 67% use the microwave to prepare their meals. Find the 95% confidence interval for the proportion. Use Excel and round the answers to three decimal places.

Answer:
Given:
The sample proportion $(\hat{p})=0.67$
The sample size $(n)=90$
The confidence interval level $=95\mathrm{\%}$

Solution:
The significance level $(\alpha ):$
$\alpha =1–0.95$

$=0.05$

The critical value $(z_c):$
$z_c=z_{\alpha /2}$

$=z_{0.05/2}$

$=1.96$

The confidence interval:

$CI=\hat{p}\pm z_c\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=0.67\pm 1.96\times \sqrt{\frac{0.67(1-0.67)}{90}}$

$=(0.573<p<0.767)$

Final Answer:
The 95% confidence interval $=(0.573<p<0.767)$