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

Answer:

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

Step-by-step explanation:

The null hypothesis is:

$H_{0} = 4$

The alternate hypothesis is:

$H_{1} < 4$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{3.75 - 4}{\frac{1.2}{\sqrt{41}}}$

$z = -1.33$

$z = -1.33$ 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