Use the given degree of confidence and sample data to construct a confidence interval for the population mean [tex]\mu[/tex]. Assume that the population has a normal distribution.

Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a [tex]99 \%[/tex] confidence interval for the mean score of all students.

A. [tex]92.03 \ \textless \ \mu \ \textless \ 97.97[/tex]
B. [tex]91.69 \ \textless \ \mu \ \textless \ 98.31[/tex]
C. [tex]92.95 \ \textless \ \mu \ \textless \ 97.05[/tex]
D. [tex]91.68 \ \textless \ \mu \ \textless \ 98.32[/tex]



Answer :

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]