Part (a): Find a 90% Confidence Interval for the True Mean Annual Salary

Given the following information:

• Population standard deviation, $\sigma = 2,023$
• Sample mean, $\bar{x} = 37,669$
• Sample size, $n = 88$
• Confidence level = 90%

The formula to compute the confidence interval when the population standard deviation is known is:

$\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean.
• $\sigma$ is the population standard deviation.
• $n$ is the sample size.
• $z_{\alpha/2}$ is the z-value corresponding to the desired confidence level (90% confidence level corresponds to a z-value of approximately 1.645).

Steps:

1. Calculate the standard error (SE):

$SE = \frac{\sigma}{\sqrt{n}} = \frac{2,023}{\sqrt{88}} \approx \frac{2,023}{9.38} \approx 215.85$

2. Calculate the margin of error (ME):

$ME = z_{\alpha/2} \cdot SE = 1.645 \cdot 215.85 \approx 355.71$

3. Construct the confidence interval:

The confidence interval is:

$\bar{x} \pm ME = 37,669 \pm 355.71$

This gives the interval:

$[37,313.29, 38,024.71]$

Part (b): Right Endpoint of the Interval

The right endpoint of the interval is $38,025$ when rounded to the nearest dollar.

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Would you like further details or have any questions?

Here are 5 related questions for you to explore:

1. How do you interpret the 90% confidence interval in the context of this survey?
2. How would the confidence interval change if the sample size were increased?
3. What would happen to the confidence interval if the population standard deviation were unknown?
4. How does the choice of confidence level affect the width of the confidence interval?
5. How can you verify the validity of using a z-distribution for this confidence interval?

Tip: The width of the confidence interval depends both on the sample size and the standard deviation, so larger samples give more precise estimates.