The population mean wait time to check out of a supermarket has been 10.73 minutes. Recently in an effort to improve the waiting time, the supermarket has experimented with a system in which there is a single waiting line with multiple checkout servers. A sample of 25 customers was selected, and their mean wait time to check out was 12.1 minutes. Assume a population standard deviation of 5.8 minutes. At the 0.05 level of significance, is there evidence that the population mean wait time to check out is more than 10.73 minutes?

Since the p value is higher than the significance level we FAIL to reject the null hypothesis so then we can conclude that the true mean is not more than 10.73 minutes for this case.

Step-by-step explanation:

Information provided

[tex]\bar X=12.1[/tex] represent the sample mean

[tex]\sigma=5.8[/tex] represent the population standard deviation

[tex]n=25[/tex] sample size

[tex]\mu_o =10.73[/tex] represent the value to verify

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic

[tex]p_v[/tex] represent the p value for the test

System of hypothesis

We want to test if the population mean wait time to check out is more than 10.73 minutes, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 10.73[/tex]

Alternative hypothesis:[tex]\mu > 10.73[/tex]

Since we know the population deviation the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

Replacing the info given we got:

[tex]z=\frac{12.1-10.73}{\frac{5.8}{\sqrt{25}}}=1.181[/tex]

Now we can calculate the p value taking in count that we are conducting a right tailed test with this probability:

[tex]p_v =P(z>1.181)=0.119[/tex]

Since the p value is higher than the significance level we FAIL to reject the null hypothesis so then we can conclude that the true mean is not more than 10.73 minutes for this case.