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 former customers 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.

Answer 1

Answer:

The 95% confidence interval for the true proportion is (0.057, 0.1192).

Step-by-step explanation:

The formula to estimate a population's true proportion using a confidence interval is given by:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}$

Here:

$n=319$

$\hat{p}=\frac{29}{329}=0.0881$

$1-\alpha=0.95 \Rightarrow z_{\alpha/2}=1.96$

Therefore replacing in the formula with the respective values we have:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.0881\pm 1.96\times\sqrt{\frac{0.0881\times(1-0.0881)}{319}}=(0.057, 0.1192)$

And (0.057, 0.1192) is the 95% 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.