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 occured 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. What sample size would be needed in order to be 99% confident that the sample proportion is within .02 of the true proportion who refused to go back to the restaurant?
To determine the sample size needed to be 99% confident that the sample proportion is within 0.02 of the true proportion, we can use the formula for sample size determination in proportion estimation:
n = (Z^2 * p * (1 - p)) / E^2
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence
p = estimated proportion of population that will exhibit the characteristic of interest (refusing to go back to the restaurant)
E = margin of error (difference between the estimated proportion and the true proportion)
In this case, we want a 99% level of confidence, which corresponds to a Z-score of approximately 2.58. The estimated proportion (p) is 29/319, as 29 out of the 319 patrons contacted indicated they would not go back to the restaurant. The margin of error (E) is 0.02.
Let's substitute these values into the formula:
n = (2.58^2 * (29/319) * (1 - 29/319)) / 0.02^2
Simplifying the equation gives us:
n = (6.6564 * (29/319) * (1 - 29/319)) / 0.0004
n = (0.1628 * 0.8372) / 0.0004
n ≈ 0.1361 / 0.0004
n ≈ 340.25
Therefore, a sample size of at least 340 would be needed to be 99% confident that the sample proportion is within 0.02 of the true proportion of patrons who refused to go back to the restaurant.