g The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. The p-value is

Step-by-step explanation:

Null hypothesis: u <= 3 minutes

Alternative hypothesis: u > 3minutes

Using a z test formula, z score is

z = (x - u) / (s/√n)

Where x = 3.1, u = 3, s = 0.5, n = 100

z = (3.1 - 3) / (0.5/√100)

z = (0.1) / (0.5/10)

z = 0.1/0.05

z = 2

To calculate for the p value assuming a 0.05 level of significance, the p value is 0.02275 which is less than 0.05. Thus we will reject the null and conclude that there is enough statistics evidence to show that the mean waiting time of all customers is significantly more than 3 minutes.

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step-by-step explanation: