Question

You are the manager of a restaurant for a​ fast-food franchise. Last​ month, the mean waiting time at the​ drive-through window for branches in your geographical​ region, as measured from the time a customer places an order until the time the customer receives the​order, was 3.9 minutes. You select a random sample of 81 orders. The sample mean waiting time is 4.05 ​minutes, with a sample standard deviation of 0.9 minute.

At the 0.01 level of​ significance, is there evidence that the population mean waiting time is different from 3.9 ​minutes? State the null and alternative hypotheses

Determine the test statistic.(two decimal places)

Find the​ p-value.(three decimal places)

What is the conclusion

Do i need to be concerned about the shape of the population distribution?

Answer 1

Answer:

There is not evidence that the population mean waiting time is different from 3.9 ​minutes.

Step-by-step explanation:

From the question we know that the size (n), the mean (x) and the standard deviation (s) of the sample are 81, 4.05 minutes and 0.9 minutes. Additionally, we are going to decide if the waiting time is different or not form 3.9 minutes, so the null and alternative hypotheses are:

H0: m=3.9

H1:  m≠3.9

Where m is the mean of the population.

Then, we don't need to be concerned about the shape of the population distribution because the value of the n is bigger than 30 and we can use the statistic z as:

$z=\frac{x-m}{\frac{s}{\sqrt{n}}}$

So, replacing the values, the test statistic is:

$z=\frac{4.05-3.9}{\frac{0.9}{\sqrt{81}}}=1.50$

On the other hand, the p value for this test is calculated as:

p value = 2P(z>1.50) = 2(0.0668) = 0.134

Taking into account that the p value is bigger than the level of significance 0.01, the null hypothesis is not reject, and there is not evidence that the population mean waiting time is different from 3.9 ​minutes.