Problem
Constructing a Confidence Interval for a Population Mean You intend to estimate a population mean \( \mu \) with the following sample. \begin{tabular}{|l|l|l|l|} \hline 14 & 12 & 18 & 14 \\ \hline 14 & 16 & 17 & 11 \\ \hline 12 & 15 & 12 & 12 \\ \hline \end{tabular} You believe the population is normally distributed. Find the \( 90 \% \) confidence interval for the population mean. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places. \[ 90 \% \text { C.I. }= \] Question Help: \( \square \) Message instructor Submit Question Submit Question Applying for the \( 0 \ldots . . \mathrm{pdf} \wedge \) O Type here to search
Solution
Step 1 ：1. Calculate sample mean \(\bar{x}\): \(\frac{14+12+18+14+14+16+17+11+12+15+12+12}{12}\)
Step 2 ：2. Calculate sample standard deviation (s): \(\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\)
Step 3 ：3. Find the critical t-value (t*) for a 90% confidence interval from a t-table with df = n-1 = 11
Step 4 ：4. Calculate margin of error (ME): \(t* \cdot \frac{s}{\sqrt{n}}\)
Step 5 ：5. Confidence Interval: \(\bar{x} \pm ME\)
