Answer to Question #347058 in Statistics and Probability for anx

Question

You are the manager of a fast-food restaurant. The business problem is to determine

whether the population mean waiting time to place an order has changed in the past month

from its previous population mean value of 4.5 minutes. From past experience, you can

assume that the population is normally distributed, with a population standard deviation of

1.2 minutes. You select a sample of 36 orders during a one-hour period. The sample mean is

5.1 minutes. To test this statistical hypothesis, with a = 0.05

you have then testing hypothesis (null and alternate hvpothesises) is

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=4.5$

$H_1:\mu\not=4.5$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{5.1-4.5}{1.2/\sqrt{36}}=3$$

Since it is observed that $|z|=3>1.96=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=2P(z>3)= 0.0027,$ and since $p= 0.0027<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is different than 4.5, at the $\alpha = 0.05$ significance level.