A major airline is concerned that the waiting time for customers at its ticket counter may be exceeding its target average of 190 seconds. To test this, the company has selected a random sample of 100 customers and times them from when the customer first arrives at the checkout line until he or she is at the counter being served by the ticket agent. The mean time for this sample was 202 seconds with a standard deviation of 28 seconds. Given this information and the desire to conduct the test using an alpha level of 0.02, which of the following statements is true?
A) The chance of a Type II error is 1 - 0.02 = 0.98.
B) The test to be conducted will be structured as a two-tailed test.
C) The test statistic will be approximately t = 4.286, so the null hypothesis should be rejected.
D) The sample data indicate that the difference between the sample mean and the hypothesized population mean should be attributed only to sampling error.

Question

Quizplus · Accepted Answer

The test statistic is calculated using the formula $t = \frac{\bar{x} - \mu}{(\sigma/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $\sigma$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values ($\bar{x} = 202$, $\mu = 190$, $\sigma = 28$, $n = 100$), we get $t = \frac{202 - 190}{(28/\sqrt{100})} = \frac{12}{2.8} = 4.286$. Given the alpha level of 0.02, this high t-value suggests that the null hypothesis (that the mean waiting time is 190 seconds) should be rejected, indicating that the actual mean waiting time is significantly different from 190 seconds.

A Major Airline Is Concerned That the Waiting Time for Customers