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Q. 14

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Found in: Page 790

### The Practice of Statistics for AP Examination

Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339

# The students in Mr. Shenk’s class measured the arm spans and heights (in inches) of a random sample of $18$ students from their large high school. Here is computer output from a least-squares regression analysis of these data. Construct and interpret a $90%$ confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met.$\mathrm{Predictor}\mathrm{Coef}\mathrm{Stdev}\mathrm{t}-\mathrm{ratio}\mathrm{P}\phantom{\rule{0ex}{0ex}}\mathrm{Constant}11.5475.6002.060.056\phantom{\rule{0ex}{0ex}}\mathrm{Armspan}0.840420.0809110.390.000\phantom{\rule{0ex}{0ex}}\mathrm{S}=1.613\mathrm{R}-\mathrm{Sq}=87.1%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=86.3%$

We are $90%$ confident that the slope of the true regression line is between $0.69915114\mathrm{and}0.98168886$.

See the step by step solution

## Step 1: Given Information

We need to construct and interpret a $90%$ confidence interval for the slope of the population regression line.

## Step 2: Simplify

Consider:

$\mathrm{n}=18\phantom{\rule{0ex}{0ex}}\mathrm{b}=0.84042\phantom{\rule{0ex}{0ex}}{\mathrm{SE}}_{\mathrm{b}}=0.08091$

The degrees of freedom in sample size decreased by $2$:

$\mathrm{df}=\mathrm{n}-2=18-2=16$

The critical $\mathrm{t}$-value can be found in table B in the row of $\mathrm{df}=16$ and the column of $\mathrm{c}=90%$

$\mathrm{t}\text{'}=1.746$

The boundaries of the confidence interval then become:

$b-t\text{'}×S{E}_{b}=0.84042-1.746×0.08091=0.69915114$role="math" localid="1654161040846" $b+t\text{'}×S{E}_{b}=0.84042+1.746×0.08091=0.98168886$

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