Thank you for the clarification. Let's recalculate the confidence interval without rounding off intermediate values and only round the final result to two decimal places.

Step 1: Determine $Z_{\alpha/2}$

As before, for a 90% confidence level, $\alpha/2 = 0.05$, and the critical value $Z_{\alpha/2} \approx 1.645$.

Step 2: Calculate the margin of error $E$

$E = Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} = 1.645 \cdot \frac{2}{\sqrt{28}}$

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

$\frac{2}{\sqrt{28}} = \frac{2}{5.291502622129181} \approx 0.378$

Now, calculate $E$ without rounding:

$E = 1.645 \cdot 0.378 \approx 0.621231926$

Step 3: Calculate the confidence interval

The confidence interval is given by:

$\text{Confidence Interval} = \bar{x} \pm E = 9.3 \pm 0.621231926$

This results in:

$\text{Lower Limit} = 9.3 - 0.621231926 = 8.678768074$ $\text{Upper Limit} = 9.3 + 0.621231926 = 9.921231926$

Final Answer (Rounded to Two Decimal Places):

The 90% confidence interval for the true mean $\mu$ of the time residents have lived at their present address is:

$(8.68 \text{ years}, 9.92 \text{ years})$

Would you like any further details or have additional questions?

Related Questions:

1. How would the confidence interval change with a 95% confidence level?
2. What effect does increasing the sample size have on the margin of error?
3. Why is the standard error important in calculating confidence intervals?
4. How would using a t-distribution instead of the Z-distribution affect the interval?
5. What role does the population standard deviation play in determining the interval?

Tip: Avoid rounding during intermediate calculations to maintain accuracy, only rounding the final result as required.