Question

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

Refer to Exhibit 9-4. At 95% confidence, it can be concluded that the mean of the population is
Select one:
a.
significantly greater than 3
b.
not significantly greater than 3
c.
significantly less than 3
d.
significantly greater then 3.18

Answer 1

Explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 3

For the alternative hypothesis,

H1: µ > 3

This is a right tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 3.1

µ = population mean = 3

s = samples standard deviation = 0.5

n = number of samples = 100

t = (3.1 - 3)/(0.5/√100) = 2

We would determine the p value using the t test calculator. It becomes

p = 0.024

Alpha = 1 - confidence level = 1 - 0.95 = 0.05

Since alpha, 0.05 > than the p value, 0.024, then we would reject the null hypothesis. Therefore, at 95% confidence level, it can be concluded that the mean of the population is significantly greater than 3.