In the past, the average waiting time per customer at a fast food restaurant has been 7.5 minutes. The customer waiting time has a normal distribution. The manager claims that the use of a new cashier system will decrease the average customer waiting time in the store. A random sample of 16 customer transactions produces a mean waiting time of 6.3 minutes and a standard deviation of 2 minutes.
-Give a range for the p-value for this test.
A)0.01

Question

In the past, the average waiting time per customer at a fast food restaurant has been 7.5 minutes. The customer waiting time has a normal distribution. The manager claims that the use of a new cashier system will decrease the average customer waiting time in the store. A random sample of 16 customer transactions produces a mean waiting time of 6.3 minutes and a standard deviation of 2 minutes.
-Give a range for the p-value for this test.
A)0.01 < p-value < 0.025
B)0.001 < p-value < 0.005
C)0.005 < p-value < 0.01
D)0.05 < p-value < 0.10
E)0.025 < p-value < 0.05

Quizplus · Accepted Answer

The test statistic for this hypothesis test is calculated using the formula for a one-sample t-test, since the population standard deviation is not known. The formula is $t = \frac{\bar{x} - \mu}{(s/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $s$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values, we get $t = \frac{6.3 - 7.5}{(2/\sqrt{16})} = \frac{-1.2}{0.5} = -2.4$. Looking up a t-value of -2.4 on the t-distribution table (or using software) for a one-tailed test with 15 degrees of freedom (since n-1 = 16-1 = 15), we find that the p-value falls between 0.01 and 0.025, indicating that there is statistical evidence to support the manager's claim at a significance level of less than 0.025 but more than 0.01.

In the Past, the Average Waiting Time Per Customer at a Fast