Here’s the breakdown of the hypothesis testing problem in the image:

(a) State the null and alternative hypotheses:

• $H_0: \mu \leq 10$ (The mean wait time is less than or equal to 10 minutes.)
• $H_1: \mu > 10$ (The mean wait time is greater than 10 minutes.)
  This is a one-tailed test.

(b) Perform the Z-test:

The test statistic formula is:

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

• Sample mean ($\bar{x}$) = 11.3 minutes
• Population mean ($\mu$) = 10 minutes
• Population standard deviation ($\sigma$) = 3.9 minutes
• Sample size ($n$) = 41

Now compute $Z$:

$Z = \frac{11.3 - 10}{\frac{3.9}{\sqrt{41}}} = \frac{1.3}{0.609} \approx 2.14$

Find the p-value:

Using standard normal distribution, the p-value is the area to the right of $Z = 2.14$.

• From Z-tables or a calculator, the p-value for $Z = 2.14$ is approximately 0.0163.

(c) Conclusion at the 0.05 level of significance:

• Since $p = 0.0163$ is less than 0.05, reject the null hypothesis.
• Conclusion: There is enough evidence to conclude that the mean wait time is now greater than 10 minutes.

Let me know if you'd like details on how to use Z-tables, or have further questions! Here are related questions for further exploration:

1. What happens if the p-value is very close to the significance level?
2. How do sample size changes affect the Z-test outcome?
3. Why is the test one-tailed and not two-tailed in this context?
4. How would a change in population standard deviation alter the result?
5. Can a similar problem use a t-test instead? When?

Tip: Always verify assumptions like normality or known standard deviation before choosing a Z-test!