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The students in Mr. Shenk’s class measured the arm spans and heights (in inches) of a random sample of 18students 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.

PredictorCoefStdevt-ratioPConstant11.5475.6002.060.056Armspan0.840420.0809110.390.000S=1.613R-Sq=87.1%R-Sq(adj)=86.3%

Short Answer

Expert verified

We are 90%confident that the slope of the true regression line is between 0.69915114and0.98168886.

Step by step solution

01

Given Information

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

02

Simplify

Consider:

n=18b=0.84042SEb=0.08091

The degrees of freedom in sample size decreased by 2:

df=n-2=18-2=16

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

t'=1.746

The boundaries of the confidence interval then become:

b-t'×SEb=0.84042-1.746×0.08091=0.69915114
role="math" localid="1654161040846" b+t'×SEb=0.84042+1.746×0.08091=0.98168886

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