Question

A random sample of 400 shoppers in a grocery store were asked whether they liked the store's new layout.140 of them (that is, 35% of the sample) said that they did like it.Find the 90% confidence interval of the true proportion of shoppers who like the store's new layout.Give your answer rounded to three places after the decimal point (the nearest .001, or nearest tenth of a percent).

Answer 1

`\begin{gathered} \hat{p}=(140)/(400)=0.35 \\ \hat{p}+z=\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}}=0.35\pm1.645\cdot\sqrt[]{(0.35(1-0.35))/(400)} \\ \text{ Lower Bound} \\ 0.35-\sqrt[]{(0.35(1-0.35))/(400)}=0.32615152\rightarrow0.326\text{ rounded off to three decimal point} \\ \text{ Upper Bound} \\ 0.35+\sqrt[]{(0.35(1-0.35))/(400)}=0.37384848\rightarrow0.374\text{ rounded off to three decimal point} \\ \\ \text{The 90\% confidence interval of the true proportion of shopeers is} \\ (0.326,0.374) \end{gathered}`