Question

Exhibit 9-2 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. The population standard deviation is known to be 0.5 minute. 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 9-2. At a .05 level of significance, it can be concluded that the mean of the population is _____.

Answer 1

At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.

Explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

At the null hypothesis, we test if the mean is of at most 3 minutes, that is:

`H_0: \mu \leq 3`

At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:

`H_1: \mu > 3`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

3 is tested at the null hypothesis:

This means that
`\mu = 3`

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. The population standard deviation is known to be 0.5 minute.

This means that
`n = 100, X = 3.1, \sigma = 0.5`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (3.1 - 3)/((0.5)/(√(100)))`

`z = 2`

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228

The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.