Question

The manager of a fast-food restaurant wants to be sure that, on average, customers are served within 4 minutes from the time the order is placed. He is particularly concerned about the staff working during the early morning shift. From a random sample of 41 orders, the mean time was 3.75 minutes with a standard deviation of 1.2 minutes.

(a) Is there convincing evidence at the a = 0.05 significance level that the true mean service time during the early morning shift is less than 4 minutes?

Answer 1

There is not convincing evidence at the a = 0.05 significance level that the true mean service time during the early morning shift is less than 4 minutes

Explanation:

The null hypothesis is:

`H_(0) = 4`

The alternate hypothesis is:

`H_(1) < 4`

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.

From a random sample of 41 orders, the mean time was 3.75 minutes with a standard deviation of 1.2 minutes.

This means that
`n = 41, \mu = 3.75, \sigma = 1.2`

The test-statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (3.75 - 4)/((1.2)/(√(41)))`

`z = -1.33`

`z = -1.33` has a pvalue of 0.0918, looking at the z-table.

0.0918 > 0.05, which means that the null hypothesis is accepted, and that there is not convincing evidence at the a = 0.05 significance level that the true mean service time during the early morning shift is less than 4 minutes