Recently,a case of food poisoning was traced to a particular restaurant chain.The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur.Despite the response from the restaurant chain,many consumers refused to visit the restaurant for some time after the event.A survey was conducted three months after the food poisoning occurred,with a sample of 319 patrons contacted.Of the 319 contacted,29 indicated that they would not go back to the restaurant because of the potential for food poisoning.Construct a 95 percent confidence interval for the true proportion of the market who still refuse to visit any of the restaurants in the chain three months after the event.
A) $\left[ \begin{array} { l l }
.059 & .122
\end{array} \right]$
B) $[ .090 \quad .091 ]$
C) $[ .000 \quad .196 ]$
D) $[ .240 \quad .339 ]$
E) $\left[ \begin{array} { l l }
.118 & .244
\end{array} \right]$

Question

Recently,a case of food poisoning was traced to a particular restaurant chain.The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur.Despite the response from the restaurant chain,many consumers refused to visit the restaurant for some time after the event.A survey was conducted three months after the food poisoning occurred,with a sample of 319 patrons contacted.Of the 319 contacted,29 indicated that they would not go back to the restaurant because of the potential for food poisoning.Construct a 95 percent confidence interval for the true proportion of the market who still refuse to visit any of the restaurants in the chain three months after the event.
A) $\left[ \begin{array} { l l } 
.059 & .122
\end{array} \right]$
B) $[ .090 \quad .091 ]$
C) $[ .000 \quad .196 ]$
D) $[ .240 \quad .339 ]$
E) $\left[ \begin{array} { l l } 
.118 & .244
\end{array} \right]$

Quizplus · Accepted Answer

The sample proportion is 29/319. Using the formula for a confidence interval for a proportion, the interval is calculated as $\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = 29/319$, $Z$ is the Z-score for 95% confidence (approximately 1.96), and $n = 319$. This calculation results in the interval $[0.059, 0.122]$.

Recently,a Case of Food Poisoning Was Traced to a Particular [.059.122]\left[ \begin{array} { l l } .059 & .122\end{array} \right][.059​.122​]

Recently,a Case of Food Poisoning Was Traced to a Particular $\left[ \begin{array} { l l } .059 & .122\end{array} \right]$