Question

In a recent study, your town's Chamber of Commerce found that 73% of town residents believed that local businesses overcharged for their products. The source of this information was a telephone survey of 1,000 town residents. Find a 95% confidence interval for the average percentage of everyone in your town who believes that local businesses overcharge.

Answer 1

Solution: The 95% confidence interval is given below:

$\dpi{100} \large \hat{p} \pm z_{\frac{0.05}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

$\dpi{100} \large \hat{p}=0.73$

$\dpi{100} \large z_{\frac{0.05}{2}}=1.96$ is the critical value at 0.05 significance level

$\dpi{100} \large n=1000$ is the sample size

Therefore, we have:

$\dpi{100} \large 0.73 \pm 1.96\sqrt{\frac{0.73(1-0.73)}{1000}}$

$\dpi{100} \large 0.73 \pm 0.0275$

$\dpi{100} \large \left ( 0.73 -0.0275,0.73 +0.0275 \right )$

$\dpi{100} \large \left ( 0.7025,0.7575 \right )$

$\dpi{100} \large \left ( 70.25\%,75.75\% \right )$

Therefore, the 95% confidence interval is $\dpi{100} \small \boldsymbol{\left ( 70.25\%,75.75\% \right )}$