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. The 95% confidence interval for the true average checkout time (in minutes) is
A)3:00 to 5:00
B)1.36 to 4.64
C)1.00 to 5.00
D)2.804 to 3.196

Question

Quizplus · Accepted Answer

We can use the formula for the confidence interval of a population mean:
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 z-score for the desired confidence level. 
For a 95% confidence interval, the z-score is 1.96.
Substituting the values given in the problem, we have:
CI = 3.0 ± 1.96*(1/√100)
CI = 3.0 ± 0.196
CI = (2.804, 3.196)
Therefore, the 95% confidence interval for the true average checkout time is 2.804 to 3.196 minutes. Choice D is the correct answer.

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