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.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 would be _____.
A)[.059,.122]
B)[.090,.091]
C)[.000,.196]
D)[.240,.339]
E)[.118,.244]

Question

Quizplus · Accepted Answer

We are given a sample size of 319 patrons contacted and out of those, 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. We can use this information to calculate the proportion of patrons who still refuse to visit the restaurant.

Sample proportion (p̂) = 29/319 = 0.091

We can use the formula for a confidence interval for a proportion:

p̂ ± z*(√(p̂(1-p̂))/n)

where z is the z-score corresponding to the desired confidence level (95% in this case), n is the sample size, and p̂ is the sample proportion.

Using a z-score table or calculator, we find that the z-score for a 95% confidence level is 1.96.

Substituting the values, we get:

0.091 ± 1.96*(√((0.091)(0.909)/319))

Simplifying gives us the confidence interval:

[0.059, 0.122]

Therefore, the best choice is A) [.059, .122].

Recently,a Case of Food Poisoning Was Traced to a Particular