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This is a hypothesis testing problem where we need to determine if the claim that the average income is $24,500 is correct based on a sample. The hypothesis test will be carried out using a **p-value approach** at a significance level $\alpha = 0.05$.

Here are the steps:

### Step 1: State the null and alternative hypotheses
We are testing if the average income is different from $24,500, so:

- **Null hypothesis ($H_0$)**: $\mu = 24,500$ (The average income is $24,500)
- **Alternative hypothesis ($H_a$)**: $\mu \neq 24,500$ (The average income is not $24,500)

This is a **two-tailed test** because the alternative hypothesis uses "$\neq$".

### Step 2: Collect the sample information
From the problem, we have the following:

- Sample mean ($\bar{x}$) = 24,350
- Population mean ($\mu_0$) = 24,500 (claimed population mean)
- Population standard deviation ($\sigma$) = 810
- Sample size ($n$) = 100
- Significance level ($\alpha$) = 0.05

### Step 3: Calculate the test statistic
Since the population standard deviation is known, we use the **Z-test** formula for the test statistic:

\[
Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
\]

Substitute the known values:

\[
Z = \frac{24,350 - 24,500}{\frac{810}{\sqrt{100}}} = \frac{-150}{\frac{810}{10}} = \frac{-150}{81} \approx -1.85
\]

### Step 4: Find the p-value
For a two-tailed test, we find the p-value corresponding to the absolute value of the Z-statistic.

Using the Z-value of $-1.85$, we can find the p-value from Z-tables or using statistical software. The p-value for $Z = -1.85$ is approximately 0.0322 for one tail. Since this is a two-tailed test, we multiply by 2:

\[
p\text{-value} \approx 2 \times 0.0322 = 0.0644
\]

### Step 5: Compare the p-value with the significance level
- If the p-value $\leq \alpha$, we reject the null hypothesis.
- If the p-value $> \alpha$, we fail to reject the null hypothesis.

In this case:

\[
p\text{-value} = 0.0644 > \alpha = 0.05
\]

### Step 6: Conclusion
Since the p-value is greater than the significance level, we **fail to reject the null hypothesis**. This means there is not enough evidence to conclude that the average income is different from $24,500.

---

Would you like further details or clarification on any part of this solution?

Here are some related questions you might consider:

1. What is the difference between a one-tailed and two-tailed test?
2. How is the p-value related to the confidence level of a hypothesis test?
3. What is the impact of the sample size on the hypothesis test?
4. How do you interpret the standard deviation in the context of this problem?
5. Why do we use the Z-test instead of the t-test here?

**Tip**: The p-value approach gives you the probability of obtaining test results at least as extreme as the observed data, under the assumption that the null hypothesis is true. Always ensure you're comparing it against the correct significance level for valid conclusions.

In a certain community, a claim is made that the average income of all employed individuals is $24,500. A group of citizens suspects this value is incorrect and gathers a random sample of 100 employed individuals in hopes of showing that $24,500 is not the correct average. The mean of the sample is $24,350 with a population standard deviation of $810. Test at α = 0.05. Show all steps, use a p-value approach.

Perform hypothesis testing on the average income claim of $24,500. A sample of 100 individuals shows a mean income of $24,350. Learn how to apply the Z-test and calculate the p-value at α = 0.05 using step-by-step methods.

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