# A survey of 600 women shoppers found that 22 % enjoy shopping for shoes. What is the 95 %...

## Question:

A survey of {eq}600 {/eq} women shoppers found that {eq}22 \ \% {/eq} enjoy shopping for shoes. What is the {eq}95 \ \% {/eq} confidence interval for the true proportion of women shoppers who enjoy shopping for shoes?

## Confidence interval

The confidence interval contains lower and upper limits which include an unknown population parameter. The confidence interval affected the sample size when sample size increased then the width of the confidence interval will be decreased.

Given information

• Sample size: 600
• Sample proportion: 0.22

The critical value is obtained from a standard normal table at the level of significance (0.05)

The critical value is 1.96

The 95% confidence interval for the true proportion of women shoppers who enjoy shopping for shoes is calculated as follow.

{eq}\begin{align*} P\left( {\hat P - {Z_{\alpha /2}}\sqrt {\dfrac{{\hat P\left( {1 - \hat P} \right)}}{n}} < p < \hat P + {Z_{\alpha /2}}\sqrt {\dfrac{{\hat P\left( {1 - \hat P} \right)}}{n}} } \right) &= 0.95\\ P\left( {0.22 - 1.96\sqrt {\dfrac{{0.22\left( {1 - 0.22} \right)}}{{600}}} < p < 0.22 + 1.96\sqrt {\dfrac{{0.22\left( {1 - 0.22} \right)}}{{600}}} } \right) &= 0.95\\ P\left( {0.186 < p < 0.253} \right) &= 0.95 \end{align*}{/eq}

Therefore, required confidence interval is (0.186 to 0.253) 