To calculate the test statistic for this sample, you can use the following formula:

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

Where:

• $M$ is the sample mean (76)
• $\mu$ is the population mean under the null hypothesis (73.7)
• $\sigma$ is the population standard deviation (6.6)
• $n$ is the sample size (55)

Substitute the values into the formula:

$z = \frac{76 - 73.7}{\frac{6.6}{\sqrt{55}}}$

Now let's calculate it:

$z = \frac{2.3}{\frac{6.6}{\sqrt{55}}}$

$z = \frac{2.3}{\frac{6.6}{7.416198487}}$

$z = \frac{2.3}{0.8903} \approx 2.583$

So, the test statistic $z$ is approximately 2.583 (to three decimal places).

If you have any questions or would like further details, feel free to ask.

Here are 8 related questions you might be interested in:

1. What does a test statistic of 2.583 indicate about the sample mean in this context?
2. How does the sample size affect the test statistic in hypothesis testing?
3. What is the p-value corresponding to this test statistic?
4. How would the test statistic change if the population standard deviation were larger?
5. What is the importance of the significance level ($\alpha$) in hypothesis testing?
6. How do you interpret the critical value in this test?
7. What would happen if the sample mean were exactly equal to the population mean?
8. How do you determine if the null hypothesis should be rejected based on the test statistic?

Tip: Always check if the assumptions of the test (e.g., normality of the population distribution) are satisfied before performing hypothesis testing.