Faculty members at a local university complain that achieving tenure at their university takes much longer than at comparable universities.They randomly sample nine tenured faculty members at their university and find that the average time of service prior to tenure was 13.55 years, with a standard deviation of 1.98 years.What is a 95% confidence interval for the mean number of years of service to achieve tenure at this university?
A) (12.028, 15.072)
B) (12.057, 15.043)
C) (12.256, 14.844)
D) (8.984, 18.116)

Question

Quizplus · Accepted Answer

To calculate a 95% confidence interval for the mean with a sample size of nine (n=9) and a sample standard deviation of 1.98 years, we use the formula for the confidence interval for the mean, which is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score for a 95% confidence level with $n-1$ degrees of freedom. Given $\bar{x} = 13.55$, $s = 1.98$, and $n = 9$, and knowing that the t-score for 95% confidence with 8 degrees of freedom is approximately 2.306, we calculate the margin of error as $2.306 \cdot \frac{1.98}{\sqrt{9}} = 1.521$. Adding and subtracting this margin of error from the sample mean gives us the confidence interval: $13.55 \pm 1.521$, which results in (12.029, 15.071), rounded to (12.028, 15.072).

Faculty Members at a Local University Complain That Achieving Tenure