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 64 orders. The sample mean waiting time is 4.03 ​minutes, with a sample standard deviation of 0.8 minutes. Complete parts​ (a) and​ (b) below.
a. Determine the test statistic.(two decimal places)
b. Find the​ p-value.(three decimal places)
c. What is the conclusion?

Answer 1

Answer:

(a) The test statistics is 1.30.

(b) The P-value is 0.198.

(c) We conclude that the mean waiting time was equal to 3.9 minutes.

Step-by-step explanation:

We are given that 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 64 orders. The sample mean waiting time is 4.03 ​minutes, with a sample standard deviation of 0.8 minutes.

Let $\mu$ = mean waiting time at the​ drive-through window for branches in your geographical​ region.

So, Null Hypothesis, $H_0$ : $\mu$ = 3.9 minutes     {means that the mean waiting time was equal to 3.9 minutes}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 3.9 minutes     {means that the mean waiting time was different from 3.9 minutes}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                        T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean waiting time = 4.03 minutes

            s = sample standard deviation = 0.8 minutes

            n = sample of orders = 64

So, test statistics  =  $\frac{4.03-3.9}{\frac{0.8}{\sqrt{64} } }$  ~ $t_6_3$

                              =  1.30

(a) The value of t test statistics is 1.30.

(b) The P-value of the test statistics is given by;

            P-value = P( $t_6_3$ > 1.30) = 0.099

For two tailed test, p-value is calculated as = 0.099 $\times$ 2 = 0.198

(c) Since, in the question we are not given with the level of significance so we assume it to be 5%.

Now, P-value of the test statistics is more than the level of significance as 0.198 > 0.05, so we have insufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean waiting time was equal to 3.9 minutes.