Question

the board of a major credit card company requires that the mean wait time for customers when they call customer service is at most 3.50 minutes. to make sure that the mean wait time is not exceeding the requirement, an assistant manager tracks the wait times of 57 randomly selected calls. the mean wait time was calculated to be 3.76 minutes. assuming the population standard deviation is 0.80 minutes, is there sufficient evidence to say that the mean wait time for customers is longer than 3.50 minutes with a 98% level of confidence? step 2 of 3: compute the value of the test statistic. round your answer to two decimal places.

Explanation:

Step1: Identify the formula for z - test statistic

The formula for the z - test statistic in a one - sample z - test for the population mean when the population standard deviation $\sigma$ is known is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Step2: Substitute the given values

We are given that $\bar{x} = 3.76$, $\mu=3.50$, $\sigma = 0.80$, and $n = 57$.
First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{0.80}{\sqrt{57}}\approx\frac{0.80}{7.55}\approx0.106$.
Then, $z=\frac{3.76 - 3.50}{0.106}=\frac{0.26}{0.106}\approx2.45$.

Answer:

$2.45$