(Donnelly 8.19)
Construct a \(90\%\) confidence interval to estimate the population mean when \(\overline{x}=68\) and \(s=13.9\) for the sample sizes below.
(a)
\(n=18\)
Answer.
\(df=17,\; t_{.05}=T.INV.2T(.10,17)\approx 1.74\)
\(\hat{\sigma_{\overline{x}}}=\frac{13.9}{\sqrt{18}}\approx 3.276\)
\(ME_{\overline{x}}\approx 1.74\cdot (3.276)\approx 5.7007\)
\(\overline{x}\pm ME_{\overline{x}}\rightarrow 68\pm 5.7007\)
We are \(90\%\) confident that
\begin{equation*}
62.299\lt \mu\lt 73.701
\end{equation*}
(b)
\(n=41\)
Answer.
\(df=40,\; t_{.05}=T.INV.2T(.10,40)\approx 1.684\)
\(\hat{\sigma_{\overline{x}}}=\frac{13.9}{\sqrt{41}}\approx 2.171\)
\(ME_{\overline{x}}\approx 1.684\cdot (2.171)\approx 3.656\)
\(\overline{x}\pm ME_{\overline{x}}\rightarrow 68\pm 3.656\)
We are \(90\%\) confident that
\begin{equation*}
64.344\lt \mu\lt 71.656
\end{equation*}
(c)
\(n=64\)
Answer.
\(df=63,\; t_{.05}=T.INV.2T(.10,63)\approx 1.669\)
\(\hat{\sigma_{\overline{x}}}=\frac{13.9}{\sqrt{64}}\approx 1.7375\)
\(ME_{\overline{x}}\approx 1.669\cdot (1.7375)\approx 2.8999\)
\(\overline{x}\pm ME_{\overline{x}}\rightarrow 68\pm 2.8999\)
We are \(90\%\) confident that
\begin{equation*}
65.1001\lt \mu\lt 70.8999
\end{equation*}