To solve this problem, we will first calculate the 90% confidence interval for the true mean annual salary of adults living in Townsburg.

Part (a): Finding the Confidence Interval

The formula for a confidence interval for a mean when the population standard deviation is unknown is:

$\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean (given as $30,498),
• $t_{\alpha/2}$ is the critical value from the t-distribution corresponding to the confidence level (90%),
• $s$ is the sample standard deviation (given as $3,148),
• $n$ is the sample size (given as 19).

Step 1: Find the critical value $t_{\alpha/2}$

For a 90% confidence level, the area in each tail of the t-distribution is $\frac{1 - 0.90}{2} = 0.05$. Thus, we need to find the t-value for 18 degrees of freedom (since $n - 1 = 19 - 1 = 18$).

Using a t-table or statistical calculator, the critical t-value for 18 degrees of freedom at a 90% confidence level is approximately 1.734.

Step 2: Calculate the margin of error

The margin of error (E) is calculated using:

$E = t_{\alpha/2} \times \frac{s}{\sqrt{n}}$

Substitute the values:

$E = 1.734 \times \frac{3,148}{\sqrt{19}} = 1.734 \times \frac{3,148}{4.359} \approx 1.734 \times 722.17 \approx 1,251.63$

Step 3: Calculate the confidence interval

Now, calculate the confidence interval by adding and subtracting the margin of error from the sample mean:

$\text{Lower bound} = 30,498 - 1,251.63 \approx 29,246.37$ $\text{Upper bound} = 30,498 + 1,251.63 \approx 31,749.63$

So, the 90% confidence interval is approximately $29,246 to $31,750.

Part (b): Right endpoint of the interval

The right endpoint of the interval is approximately $31,750, which we round to the nearest dollar.

Final Answer:

• The right endpoint of the 90% confidence interval for the true mean annual salary is $31,750.

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Would you like a more detailed explanation of any part of this process? Here are some related questions for further practice:

1. How would the confidence interval change if the sample size were larger or smaller?
2. How can the sample size be determined for a specific margin of error at a certain confidence level?
3. What would happen if we used a 95% confidence level instead of 90%?
4. How does the t-distribution differ from the normal distribution when constructing confidence intervals?
5. What assumptions must be met in order to use the t-distribution for confidence intervals?

Tip: When calculating confidence intervals, always check the assumptions: random sampling, the data being approximately normally distributed, and using the t-distribution for small sample sizes (less than 30).