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 \(\boxed{0.072}\).