A simple linear regression analysis for n = 20 data points produced the followin

Question

A simple linear regression analysis for n = 20 data points produced the following results: y = 2.1 + 3.4x x = 2.5 y = 10.6 SS_xx = 4.77 SS_yy = 59.21 SS_xy = 16.22 Find SSE and s^2 Find a 95% confidence interval for E(y) when x = 2.5. Interpret this interval. Find a 95% interval for E (y) when x = 2.0. Interpret this interval Find a 95% confidence interval for E(y) when x = 3.0. Interpret this interval. Examine the widths of the confidence obtained in parts b, c, and d. What happens to the width of the confidence interval for E(y) as the value of x moves away from the value of x? Find a 95% prediction interval for a value of y to be observed in the future when x = 3.0. Interpret its value.

Explanation / Answer

ans)here SST =SSyy =59.21

and SSR =(SSxy)2/Sxx =55.155

hence SSE =SST-SSR =4.055

therefore s2 =SSE/(n-2) =0.225

b) for xo =2.5 ;

Yreg =2.1+3.4*2.5 =10.6

std error =(s2(1/n+(x0-xbar)2/Sxx)1/2 =0.106

for 95% CI and (n-2) 18 df, t =2.1009

hence confidence interval =10.6 +/- t*std error =10.377 ; 10.823

c)

for xo =2.;

Yreg =2.1+3.4*2.0 =8.9

std error =(s2(1/n+(x0-xbar)2/Sxx)1/2 =0.152

for 95% CI and (n-2) 18 df, t =2.1009

hence confidence interval =8.9 +/- t*std error =8.581 ; 9.219

d)similarly

for xo =3.;

Yreg =2.1+3.4*3.0 =12.3

std error =(s2(1/n+(x0-xbar)2/Sxx)1/2 =0.152

for 95% CI and (n-2) 18 df, t =2.1009

hence confidence interval =12.3 +/- t*std error =11.981 ;12.619

e) as it is observed from above as x moves away from mean, confidence interval increases due to higher standard error.

f)for 95% prediction interval

std errorr =(s2(1 +1/n+(x0-xbar)2/Sxx)1/2 =0.498

hence confidence interval =12.3 +/- t*std error =11.253 ; 13.347

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