A sample of size [tex]$n=57$[/tex] is drawn from a normal population. The sample mean is [tex]$\bar{x}=58.5$[/tex], and the sample standard deviation is [tex]$s=9.5$[/tex].

(a) Construct a [tex]$99.8 \%$[/tex] confidence interval for the population mean, [tex]$\mu$[/tex]. Round the answers to one decimal place.

The [tex]$99.8 \%$[/tex] confidence interval is [tex]\square \ < \ \mu \ < \ \square[/tex].

To construct a 99.8% confidence interval for the population mean [tex]$ \mu $[/tex] based on the information provided, follow these steps:

1. Identify the given information:
- Sample size ([tex]$ n $[/tex]) = 57
- Sample mean ([tex]$ \bar{x} $[/tex]) = 58.5
- Sample standard deviation ([tex]$ s $[/tex]) = 9.5
- Confidence level = 99.8%

2. Calculate the z-score:
- Since the confidence level is 99.8%, the level of significance ([tex]$ \alpha $[/tex]) is 0.2% or 0.002 in decimal.
- Find the z-score that corresponds to [tex]$(1 - \alpha/2) = 0.999$[/tex] using a standard normal distribution table. For a 99.8% confidence interval, the z-score is approximately 3.090.

3. Calculate the margin of error:
- The margin of error (ME) is calculated using the formula:
[tex]\[
\text{ME} = z \times \left(\frac{s}{\sqrt{n}}\right)
\][/tex]
- Substituting the values:
[tex]\[
\text{ME} = 3.090 \times \left(\frac{9.5}{\sqrt{57}}\right)
\][/tex]
- The margin of error is approximately 3.9.

4. Construct the confidence interval:
- The confidence interval is given by:
[tex]\[
\bar{x} - \text{ME} < \mu < \bar{x} + \text{ME}
\][/tex]
- Use the sample mean and margin of error calculated:
[tex]\[
58.5 - 3.9 < \mu < 58.5 + 3.9
\][/tex]
- Simplifying this, the confidence interval is approximately:
[tex]\[
54.6 < \mu < 62.4
\][/tex]

So, the 99.8% confidence interval for the population mean [tex]$ \mu $[/tex] is [tex]$ 54.6 < \mu < 62.4 $[/tex].