A recent study of 28 city residents showed that the mean of the time they had lived at their present address was 9.3 years. The standard deviation of the population was 2 years. Find the 90% confidence interval of the true mean? Assume that the variable is approximately normally distributed. Show all your steps. (10 points) Hint use the formula given on page 312 for E and then follow the guidelines given on page 312 from the textbook. 

To find the 90% confidence interval for the true mean \(\mu\) of the time residents have lived at their current address, we'll use the following formula:

\[
\text{Confidence Interval} = \bar{x} \pm E
\]

Where:
- \(\bar{x}\) is the sample mean
- \(E\) is the margin of error, calculated as:

\[
E = Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}
\]

Given:
- \(\bar{x} = 9.3\) years
- \(\sigma = 2\) years (population standard deviation)
- \(n = 28\) (sample size)
- Confidence level = 90%

### Step 1: Determine \(Z_{\alpha/2}\)
For a 90% confidence level, \(\alpha = 1 - 0.90 = 0.10\). Since this is a two-tailed test, we split \(\alpha\) into two tails:

\[
\alpha/2 = 0.05
\]

We look up the critical value \(Z_{\alpha/2}\) for \(\alpha/2 = 0.05\) in the standard normal (Z) distribution table, which gives:

\[
Z_{\alpha/2} \approx 1.645
\]

### Step 2: Calculate the margin of error \(E\)

\[
E = 1.645 \cdot \frac{2}{\sqrt{28}}
\]

First, calculate the standard error \(\frac{\sigma}{\sqrt{n}}\):

\[
\frac{2}{\sqrt{28}} \approx \frac{2}{5.2915} \approx 0.378
\]

Now calculate \(E\):

\[
E = 1.645 \cdot 0.378 \approx 0.621
\]

### Step 3: Calculate the confidence interval

The confidence interval is given by:

\[
\text{Confidence Interval} = \bar{x} \pm E = 9.3 \pm 0.621
\]

This results in:

\[
\text{Lower Limit} = 9.3 - 0.621 = 8.679
\]
\[
\text{Upper Limit} = 9.3 + 0.621 = 9.921
\]

### Final Answer:
The 90% confidence interval for the true mean \(\mu\) of the time residents have lived at their present address is approximately:

\[
(8.679 \text{ years}, 9.921 \text{ years})
\]

Would you like further details or have any questions?

**Related Questions:**
1. What would the confidence interval be if the confidence level was 95%?
2. How would the confidence interval change if the sample size was larger?
3. Why is it important to assume that the variable is approximately normally distributed in this context?
4. How would the interval change if the standard deviation was larger?
5. What if the population standard deviation was unknown? How would you calculate the confidence interval?

**Tip:** When calculating confidence intervals, ensure you are using the correct Z-value associated with your desired confidence level.

Learn how to calculate a 90% confidence interval for the mean with a known population standard deviation using a step-by-step approach. This example involves a statistical study with sample mean, population standard deviation, and sample size.

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