The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 86.7 seconds. A manager devises a new drive-through system that he believes will decrease wait time. As a test, he initiates the new system at his restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts (a) and (b) below.
109.0
69.9
56.3
76.5
65.5
82.0
95.3
84.0
69.6
81.9

(A) Find the test statistic
(Round to two decimal places as needed.)

(B) Find the P-value

(Round to three decimal places as needed.)

Step 1 ：We are given a set of wait times: 109.0, 69.9, 56.3, 76.5, 65.5, 82.0, 95.3, 84.0, 69.6, 81.9. The population mean is 86.7 seconds.

Step 2 ：We are testing the null hypothesis that the mean waiting time is 86.7 seconds against the alternative hypothesis that the mean waiting time is less than 86.7 seconds.

Step 3 ：We calculate the sample mean to be 79.0 seconds and the sample standard deviation to be approximately 15.21 seconds.

Step 4 ：We use these values to calculate the t-statistic using the formula: \(t = \frac{{\text{{sample mean}} - \text{{population mean}}}}{{\text{{sample standard deviation}} / \sqrt{{\text{{sample size}}}}}}\). This gives us a t-statistic of approximately -1.60.

Step 5 ：We then find the P-value by using the cumulative distribution function (CDF) for the t-distribution with degrees of freedom equal to the sample size minus 1. The P-value is the probability that a t-statistic under the null hypothesis is as extreme as, or more extreme than, the observed t-statistic. This gives us a P-value of approximately 0.072.

Step 6 ：Final Answer: The test statistic is \(\boxed{-1.60}\) and the P-value is \(\boxed{0.072}\).