A sample of size n = 14 has a sample mean $\bar { x } = 11.9$ and sample standard deviation s = 2.1. It is reasonable to assume that the population is approximately normal. Construct a 99% confidence Interval for the population mean $\mu$
A) (10.7, 13.1)
B) (11.4, 12.4)
C) (10.2, 13.6)
D) (10.4, 13.4)

Question

Quizplus · Accepted Answer

To construct a 99% confidence interval for the population mean $\mu$, we use the formula: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, 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 corresponding to the desired confidence level and degrees of freedom $df = n - 1$. For a 99% confidence interval and $df = 14 - 1 = 13$, the t-score (from a t-distribution table or calculator) is approximately 3.012. Plugging in the values: $11.9 \pm 3.012 \cdot \frac{2.1}{\sqrt{14}}$, we get the interval (10.2, 13.6).

A Sample of Size N = 14 Has a Sample xˉ=11.9\bar { x } = 11.9xˉ=11.9

A Sample of Size N = 14 Has a Sample $\bar { x } = 11.9$