Question

A local eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. The owner of the store has determined that home delivery will be successful if the average time spent on the deliveries does not exceed 34 minutes. The owner has randomly selected 23 customers and has delivered pizzas to their homes in order to test if the mean delivery time actually exceeds 34 minutes. Suppose the P-value for the test was found to be 0.0290. State the correct conclusion

Answer 1

Answer:

We conclude that the mean delivery time actually exceeds 34 minutes.

Step-by-step explanation:

We are given that the owner of the store has determined that home delivery will be successful if the average time spent on the deliveries does not exceed 34 minutes.

The owner has randomly selected 23 customers and has delivered pizzas to their homes in order to test if the mean delivery time actually exceeds 34 minutes. Suppose the P-value for the test was found to be 0.0290.

Let $\mu$ = average time spent on the deliveries.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 34 minutes    {means that the mean delivery time is less than or equal to 34 minutes}

Alternate Hypothesis, $H_A$ : $\mu$ > 34 minutes     {means that the mean delivery time actually exceeds 34 minutes}

Now, the decision rule based on the P-value to whether null hypothesis should be accepted or rejected is given by;

• If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis.

• If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis.

Here, we assume that the level of significance is 5%.

Since, the P-value of test statistics is less than the level of significance as 0.029 < 0.05, so we have sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean delivery time actually exceeds 34 minutes.