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.What sample size would be needed in order to be 95% confident that the sample proportion is within 0.02 of p,the true proportion of customers who refuse to go back to the restaurant?
A)14
B)38
C)129
D)2401
E)1,777

Question

Quizplus · Accepted Answer

To determine the sample size needed for a given confidence level and margin of error, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score for the confidence level (1.96 for 95%), $p$ is the estimated proportion (if unknown, 0.5 is used for maximum sample size), and $E$ is the margin of error (0.02 in this case). Substituting the values, $n = \frac{(1.96)^2 \cdot 0.5 \cdot (1-0.5)}{(0.02)^2} = 2401$.

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