Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. 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 $99 \%$ confidence interval for the mean score of all students.
A) $91.68

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. 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 $99 \%$ confidence interval for the mean score of all students.
A) $91.68<\mu<98.32$
B) $91.69<\mu<98.31$
C) $92.95 < \mu < 97.05$
D) $92.03 < \mu < 97.97$

Quizplus · Accepted Answer

To construct a 99% confidence interval for the mean score of all students, we use the formula: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size. Given $\bar{x} = 95$, $\sigma = 6.6$, and $n = 30$, and knowing that the z-score for a 99% confidence level is approximately 2.576, we calculate the margin of error as $2.576 \times \frac{6.6}{\sqrt{30}} \approx 3.16$. Thus, the confidence interval is $95 \pm 3.16$, which gives $91.84$ and $98.16$, rounding to two decimal places as required by the options leads to $91.68 < \mu < 98.32$.

Use the Given Degree of Confidence and Sample Data to Construct