Exhibit 8-2
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 checkout time is one minute.
-Refer to Exhibit 8-2. The 95% confidence interval for the average checkout time of all customers is
A)3 to 5
B)1.36 to 4.64
C)2.804 to 3.196
D)1.04 to 4.96

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for a population mean with known standard deviation:
CI = x̄ ± z*(σ/√n)
where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z is the critical value for the desired confidence level. With a 95% confidence level, z = 1.96.
Plugging in the given values, we get:
CI = 3.0 ± 1.96*(1/√100)
CI = 3.0 ± 0.196
CI = (2.804, 3.196)
Therefore, the 95% confidence interval for the average checkout time of all customers is (2.804, 3.196).

Exhibit 8-2 The Manager of a Grocery Store Has Taken a Random