To construct a 99% confidence interval for the population mean [tex]\(\mu\)[/tex] based on the given sample data, follow these steps:
1. Identify the sample statistics:
- Sample size ([tex]\(n\)[/tex]): 30
- Sample mean ([tex]\(\bar{x}\)[/tex]): 95
- Sample standard deviation ([tex]\(s\)[/tex]): 6.6
2. Determine the confidence level and the corresponding z-score:
- Confidence level: 99%
- The corresponding z-score for a 99% confidence level is approximately 2.576 (this is something typically found in z-score tables or statistical software).
3. Calculate the standard error of the mean (SE):
[tex]\[
SE = \frac{s}{\sqrt{n}} = \frac{6.6}{\sqrt{30}} \approx 1.205
\][/tex]
4. Compute the margin of error (ME):
[tex]\[
ME = z \times SE = 2.576 \times 1.205 \approx 3.1038
\][/tex]
5. Calculate the confidence interval:
[tex]\[
\text{Lower limit} = \bar{x} - ME = 95 - 3.1038 \approx 91.896
\][/tex]
[tex]\[
\text{Upper limit} = \bar{x} + ME = 95 + 3.1038 \approx 98.104
\][/tex]
Therefore, the 99% confidence interval for the mean score of all students is approximately:
[tex]\[
91.896 < \mu < 98.104
\][/tex]
From the provided options, the interval [tex]\(91.68 < \mu < 98.32\)[/tex] closely resembles the calculated result due to rounding differences. However, the closest match to calculated bounds [tex]\(91.896\)[/tex] and [tex]\(98.104\)[/tex] is:
[tex]\[
91.69 < \mu < 98.31
\][/tex]
Hence, the correct answer is:
[tex]\[
91.69 < \mu < 98.31
\][/tex]
