Exhibit 8-3
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the population of checkout times is one minute.
-Refer to Exhibit 8-3. With a .95 probability, the sample mean will provide a margin of error of
A)1.96
B)0.10
C)0.196
D)1.64

Question

Quizplus · Accepted Answer

The margin of error is given by E = z* (sigma/sqrt(n)), where z* is the critical value of the standard normal distribution corresponding to the desired level of confidence (0.95 in this case), sigma is the population standard deviation, and n is the sample size. Substituting the given values, we have E = 1.96*(1/sqrt(100)) = 0.196. Therefore, the best choice is C.

Exhibit 8-3 The Manager of a Grocery Store Has Taken a Random