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.
\[
90 \% \text { C.I. }=
\]
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Answer
5. Confidence Interval: \(\bar{x} \pm ME\)
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: \(\bar{x} \pm ME\)
