A student wanted to quickly construct a 99% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?
A) [17.42, 20.78]
B) [17.48, 20.72]
C) [14.23, 23.98]
D) [0.44, 3.80]
E) [17.95, 20.25]

Question

Quizplus · Accepted Answer

To construct a 99% confidence interval for the population mean with a sample size of 9, the t-distribution is used because the population standard deviation is unknown. The formula for the confidence interval is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score corresponding to the confidence level and degrees of freedom (n-1), $s$ is the sample standard deviation, and $n$ is the sample size. Given $\bar{x} = 19.1$, $s = 1.5$, and $n = 9$, and knowing that the degrees of freedom is 8, the t-score for a 99% confidence interval is approximately 3.355 (from t-distribution tables). Plugging these values into the formula gives a confidence interval of approximately [17.42, 20.78].

A Student Wanted to Quickly Construct a 99% Confidence Interval