We will compute the value of the test statistic using a z-test since the population standard deviation is known.

Formula for the z-test statistic:

$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

Where:

• $\bar{x}$ is the sample mean (i.e. 3.87 minutes)
• $\mu$ is the population mean (i.e. 3.50 minutes)
• $\sigma$ is the population standard deviation (i.e. 1.54 minutes)
• $n$ is the sample size (i.e. 37 calls)

Let's compute the test statistic now.

Given values

x_bar = 3.87 # sample mean in minutes mu = 3.50 # population mean in minutes sigma = 1.54 # population standard deviation in minutes n = 37 # sample size

Calculate the z-test statistic

z = (x_bar - mu) / (sigma / (n ** 0.5)) round(z, 2)The value of the test statistic is $1.46$, rounded to two decimal places.

Would you like to proceed with the next steps or have any questions?

Here are 5 related questions for further exploration:

1. What does the positive test statistic imply in this scenario?
2. How do we find the critical value at a 95% confidence level for this test?
3. How do we determine the p-value corresponding to the test statistic of $1.46$?
4. What conclusion can be made if the test statistic exceeds the critical value?
5. How would the test statistic change if the sample size were larger or smaller?

Tip: Confidence levels and significance levels are complementary; a 95% confidence level corresponds to a 5% significance level (α = 0.05).