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A reporter for a student newspaper is writing an article on the cost of off-campus housing. A sample of 16 one-bedroom apartments within a half-mile of campus resulted in a sample mean of \(\$ 750\) per month and a sample standard deviation of \(\$ 55\).

Provide a \(95 \%\) confidence interval estimate of the mean rent per month for the population of one-bedroom apartments within a half-mile of campus. We will assume this population to be normally distributed.
in Data Science & Statistics by Platinum (131,394 points) | 280 views

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- At \(95 \%\) confidence, \(\alpha=.05\), and \(\alpha / 2=.025 .\)
- \(t_{.025}\) is based on \(n-1=16-1=15\) degrees of freedom.

 

\begin{array}{|ccccccc|}
\hline {\begin{array}{c}
\text { Degrees } \\
\text { of Freedom }
\end{array}} & {\text { Area in Upper Tail }} \\
& . \mathbf{2 0} & .10 & .05 & .025 & .01 & .005 \\
\hline 15 & .866 & 1.341 & 1.753 & 2.131 & 2.602 & 2.947 \\
\hline 16 & .865 & 1.337 & 1.746 & 2.120 & 2.583 & 2.921 \\
17 & .863 & 1.333 & 1.740 & 2.110 & 2.567 & 2.898 \\
18 & .862 & 1.330 & 1.734 & 2.101 & 2.520 & 2.878 \\
19 & .861 & 1.328 & 1.729 & 2.093 & 2.539 & 2.861 \\
. & . & . & . & . & . & . \\
\hline
\end{array}

 

\(\$ 750\) mean monthly rent \(\$ 55\) standard deviation 16 sampled one-bedroom apt

\(\bar{x} \pm t_{.025} \frac{S}{\sqrt{n}}\)

\(750_{\simeq} \pm 2.131 \frac{55}{\sqrt{16}}=750 \pm 29.30\)

We are \(95 \%\) confident that the mean rent per month for the population of one-bedroom apartments within a half-mile of campus is between \(\$ 720.70\) and \(\$ 779.30\).
by Platinum (131,394 points)

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