You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $12 ( = 42, = 12). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is $44.50 from tips. Test for a difference between this value and the population mean at the = .05 level of significance.

To test for a difference between your average nightly tips and the population mean, we'll use a hypothesis test for a single sample mean. Here are the steps:

### 1. State the Hypotheses

- **Null Hypothesis ($H_0$)**: There is no difference between your average tips and the population mean.
  \[
  H_0: \mu = 42
  \]
- **Alternative Hypothesis ($H_a$)**: There is a difference between your average tips and the population mean.
  \[
  H_a: \mu \neq 42
  \]

### 2. Determine the Significance Level

The significance level ($\alpha$) is 0.05.

### 3. Calculate the Test Statistic

Since the population standard deviation ($\sigma$) is known, we use the Z-test for the sample mean. The test statistic is given by:
\[
z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
\]
Where:
- $\bar{x}$ is the sample mean = 44.50
- $\mu$ is the population mean = 42
- $\sigma$ is the population standard deviation = 12
- $n$ is the sample size = 16

First, we calculate the standard error of the mean:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{16}} = \frac{12}{4} = 3
\]

Now, calculate the Z value:
\[
z = \frac{44.50 - 42}{3} = \frac{2.50}{3} \approx 0.833
\]

### 4. Determine the P-value

For a two-tailed test at the $\alpha = 0.05$ significance level, we need to compare the test statistic with the critical values from the standard normal distribution.

The critical Z-values for a two-tailed test at $\alpha = 0.05$ are approximately $\pm 1.96$.

Using the Z-table or a calculator, the p-value corresponding to $z = 0.833$ is:
\[
P(Z > 0.833) \approx 0.202
\]

Since this is a two-tailed test, we multiply the p-value by 2:
\[
p = 2 \times 0.202 = 0.404
\]

### 5. Compare the P-value with the Significance Level

Since the p-value (0.404) is greater than the significance level ($\alpha = 0.05$), we fail to reject the null hypothesis.

### Conclusion

There is not enough evidence to conclude that your average nightly tips are different from the population mean of $42 at the 0.05 significance level.

Would you like more details or have any questions?

Here are 5 more questions you might want to ask:
1. How do I interpret the p-value in hypothesis testing?
2. Can you explain the difference between one-tailed and two-tailed tests?
3. What are Type I and Type II errors in hypothesis testing?
4. How does sample size affect the results of a hypothesis test?
5. What is the difference between the Z-test and the T-test?

**Tip:** When conducting hypothesis tests, always ensure your sample size is adequate to achieve reliable results.

Learn how to test for a difference between average nightly tips and a population mean using hypothesis testing with a Z-test. Calculate the test statistic, determine the p-value, and interpret the results. Suitable for advanced college-level students and professionals.

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