The manager of a grocery store has taken a random sample off 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 .5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

(A) calculate the test statistic

(B) calculate the p-value

(C) at a .05 level significance it can be concluded that the mean of the population is:

- significantly greater than 3

- Not significantly greater than 3

- Significantly less than 3

- Significantly greater than 3.18

c) Since the p value is lower than the significance level provided of [tex]\alpha=0.05[/tex] we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 3 minutes

- significantly greater than 3

Step-by-step explanation:

Part a

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes so then the system of hypothesis are:

Null hypothesis: [tex]\mu \leq 3[/tex]

Alternative hypothesis: [tex]\mu>3[/tex]

The info given by the problem:

[tex]\bar X=3.1, \sigma =0.5 , n =100[/tex]

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

[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And replacing we got:

[tex] z = \frac{3.1-3}{\frac{0.5}{\sqrt{100}}}= 2[/tex]

Part b

The p value taking in count that we are conducting a right tailed test can be calculated like this:

[tex] p_v = P(z>2)= 0.0228[/tex]

Part c

Since the p value is lower than the significance level provided of [tex]\alpha=0.05[/tex] we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 3 minutes

- significantly greater than 3