Find the confidence interval specified. Assume that the population is normally distributed.
-Thirty randomly selected students took the calculus final. If the sample mean was 93 and the standard deviation was 13.3, construct a 99% confidence interval for the mean score of all students.
A)88.87 to 97.13
B)86.31 to 99.69
C)87.02 to 98.98
D)86.33 to 99.67

Question

Find the confidence interval specified. Assume that the population is normally distributed.
-Thirty randomly selected students took the calculus final. If the sample mean was 93 and the standard deviation was 13.3, construct a 99% confidence interval for the mean score of all students. 
A)88.87 to 97.13
B)86.31 to 99.69
C)87.02 to 98.98
D)86.33 to 99.67

Quizplus · Accepted Answer

To construct a 99% confidence interval for the mean score of all students, we use the formula for the confidence interval for the mean, which is $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, 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 $\bar{x} = 93$, $\sigma = 13.3$, and $n = 30$, and knowing that the z-score for a 99% confidence interval is approximately 2.576, we calculate the margin of error as $2.576 \times \frac{13.3}{\sqrt{30}}$, which gives us a margin of error of approximately 6.69. Therefore, the confidence interval is $93 \pm 6.69$, which gives us $86.31$ to $99.69$.

Find the Confidence Interval Specified