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Weeds among the corn Refer to Exercise 13.

(a) Construct and interpret a 90% confidence interval for the slope of the true regression line. Explain how your results are consistent with the significance test in Exercise 13.

(b) Interpret each of the following in context:

(i) 8

(ii) r2

(iii) The standard error of the slope

Short Answer

Expert verified

a) (-2.1046,-0.0928)

b) (i) The expected error when making predictions using the regression line is about 7.97665bushels.

(ii) 20.9%of the variation between the variables has been explained by the linear regression line.

(iii) The slope of the least-squares regression line is expected to vary on average by 0.5823over all possible samples.

Step by step solution

01

Part(a) Step 1: Given Information

Given:

n=16

b=-1.0987

SEb=0.5712

02

Part(a) Step 2: Explanation

The degrees of freedom is the sample size decreased by 2:

df=n-2=16-2=14

The critical t-value can be found in table B in the row of df=14and in the column of c=90%:

t*=1.761

The boundaries of the confidence interval then become:

localid="1650694151582" b-t*×SEb=-1.0987-1.761×0.5712=-2.1046

localid="1650694166451" b+t*×SEb=-1.0987+1.761×0.5712=-0.0928

03

Part(b) Step 1: Given Information

Given:

s=7.97665

r2=20.9%=0.209

SEb=0.5823

04

Part(b) Step 2: Explanation

Given:

s=7.97665

r2=20.9%=0.209

SEb=0.5823

(i) s=7.97665indicates that the expected error when making predictions using the regression line is about 7.97665bushels.

(ii) r2=20.9%indicates that 20.9%of the variation between the variables has been explained by the linear regression line.

(iii) SEb=0.5823indicates that the estimation of the slope of the least-squares regression line is expected to vary on average by 0.5823over all possible samples.

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