New_Linear Regression Analysis- 2

8/14/2019 New_Linear Regression Analysis- 2

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11 2009 Thomson South-Western. All Rights Reserved 2009 Thomson South-Western. All Rights Reserved

s HypothesesHypotheses

s Test StatisticTest Statistic

Testing for Significance:Testing for Significance: tt TestTest

0 1: 0H =

1: 0aH

1

1

b

bt

s= wherewhere

1 2

( )

b

i

ss

x x

=


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s Rejection RuleRejection Rule


where:where:

tt// 22is based on ais based on a tt distributiondistribution

withwith nn - 2 degrees of freedom- 2 degrees of freedom

RejectReject HH00 ififpp-value-value


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1. Determine the hypotheses.1. Determine the hypotheses.

2. Specify the level of significance.2. Specify the level of significance.

3. Select the test statistic.3. Select the test statistic.

= .05= .05

4. State the rejection rule.4. State the rejection rule.RejectReject HH00 ififpp-value-value 3.182 (with> 3.182 (with

3 degrees of freedom)3 degrees of freedom)


0 1: 0H =1: 0aH

1

1

b

bt

s=


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5. Compute the value of the test statistic.5. Compute the value of the test statistic.

6. Determine whether to reject6. Determine whether to reject HH00..

tt= 4.541 provides an area of .01 in the upper= 4.541 provides an area of .01 in the uppertail. Hence, thetail. Hence, thepp-value is less than .02. (Also,-value is less than .02. (Also,tt= 4.63 > 3.182.) We can reject= 4.63 > 3.182.) We can reject HH

00..

1

1 5 4.631.08b

bt

s= = =


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Confidence Interval forConfidence Interval for 11

HH00 is rejected if the hypothesized value ofis rejected if the hypothesized value of 11 is notis not

included in the confidence interval forincluded in the confidence interval for 11..

We can use a 95% confidence interval forWe can use a 95% confidence interval for 11 to testto test

the hypotheses just used in thethe hypotheses just used in the tttest.test.


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s The form of a confidence interval forThe form of a confidence interval for 11is:is:


11 / 2 bb t s

wherewhere is theis the ttvalue providing an areavalue providing an area

ofof /2 in the upper tail of a/2 in the upper tail of a tt distributiondistributionwithwith nn - 2 degrees of freedom- 2 degrees of freedom

2/t

bb11 is theis the

pointpoint

estimatestimatoror

is theis themarginmarginof errorof error

1/ 2 bt s


7/2677 2009 Thomson South-Western. All Rights Reserved 2009 Thomson South-Western. All Rights Reserved


RejectReject HH00 if 0 is not included inif 0 is not included in

the confidence interval forthe confidence interval for 11..

0 is not included in the confidence interval.0 is not included in the confidence interval.RejectReject HH00

= 5 +/- 3.182(1.08) = 5 +/- 3.44= 5 +/- 3.182(1.08) = 5 +/- 3.4412/1 bstb

or 1.56 to 8.44or 1.56 to 8.44


s 95% Confidence Interval for95% Confidence Interval for 11

s ConclusionConclusion



s HypothesesHypotheses

s

Test StatisticTest Statistic

Testing for Significance:Testing for Significance: FF TestTest

FF= MSR/MSE= MSR/MSE

0 1: 0H =

1: 0aH





where:where:

FF is based on anis based on an FFdistribution withdistribution with

1 degree of freedom in the numerator and1 degree of freedom in the numerator and

nn - 2 degrees of freedom in the denominator- 2 degrees of freedom in the denominator

RejectReject HH00 ifif

pp-value-value FF



1. Determine the hypotheses.1. Determine the hypotheses.

2. Specify the level of significance.2. Specify the level of significance.

3. Select the test statistic.3. Select the test statistic.

= .05= .05

4. State the rejection rule.4. State the rejection rule.RejectReject HH00 ififpp-value-value > 10.13 (with10.13 (with 1 d.f.1 d.f.in numerator andin numerator and

3 d.f. in denominator)3 d.f. in denominator)


0 1: 0H =1: 0aH

FF= MSR/MSE= MSR/MSE




5. Compute the value of the test statistic.5. Compute the value of the test statistic.

6. Determine whether to reject6. Determine whether to reject HH00..

FF= 17.44 provides an area of .025 in= 17.44 provides an area of .025 inthe upper tail. Thus, thethe upper tail. Thus, thepp-value-valuecorresponding tocorresponding to FF= 21.43 is less than= 21.43 is less than2(.025) = .05. Hence, we reject2(.025) = .05. Hence, we reject HH00..

FF= MSR/MSE = 100/4.667 = 21.43= MSR/MSE = 100/4.667 = 21.43

The statistical evidence is sufficient toThe statistical evidence is sufficient to

concludeconcludethat we have a significant relationshipthat we have a significant relationshipbetween thebetween thenumber of TV ads aired and the number ofnumber of TV ads aired and the number ofcars sold.cars sold.



Some Cautions about theSome Cautions about theInterpretation of Significance TestsInterpretation of Significance Tests

Just because we are able to rejectJust because we are able to reject HH00:: 11 = 0 and= 0 anddemonstrate statistical significance does not enabledemonstrate statistical significance does not enableus to conclude that there is aus to conclude that there is a linear relationshiplinear relationshipbetweenbetweenxxandandyy..

RejectingRejecting HH00:: 11 = 0 and concluding that= 0 and concluding that

thetherelationship betweenrelationship betweenxxandandyyis significantis significant

doesdoes not enable us to conclude that anot enable us to conclude that a cause-cause-and-effectand-effect

relationshiprelationship is present betweenis present betweenxxandandyy..



Using the Estimated Regression EquationUsing the Estimated Regression Equationfor Estimation and Predictionfor Estimation and Prediction

/ y t sp yp 2

where:where:

confidence coefficient is 1 -confidence coefficient is 1 - andand tt /2/2 is based on ais based on a ttdistributiondistribution

withwith nn - 2 degrees of freedom- 2 degrees of freedom

/ 2 indp y t s

s Confidence Interval Estimate ofConfidence Interval Estimate of

EE((yypp))

s Prediction Interval Estimate ofPrediction Interval Estimate ofyypp



If 3 TV ads are run prior to a sale, weIf 3 TV ads are run prior to a sale, weexpectexpect

the mean number of cars sold to be:the mean number of cars sold to be:

Point EstimationPoint Estimation

^yy= 10 + 5(3) = 25 cars= 10 + 5(3) = 25 cars



2

2

( )1

( )pp

y

i

x xs s

n x x

= +

s Estimate of the Standard Deviation ofEstimate of the Standard Deviation ofpy

Confidence Interval forConfidence Interval for EE((yypp))

2

2 2 2 2 2

(3 2)12.16025

5 (1 2) (3 2) (2 2) (1 2) (3 2)pys

= +

+ + + +

1 12.16025 1.44915 4p

ys = + =



The 95% confidence interval estimate of theThe 95% confidence interval estimate of themean number of cars sold when 3 TV adsmean number of cars sold when 3 TV ads

are run is:are run is:

Confidence Interval forConfidence Interval for EE((yypp))

2525 ++ 4.614.61

/ y t sp yp 2

2525 ++ 3.1824(1.4491)3.1824(1.4491)

20.39 to 29.61 cars20.39 to 29.61 cars


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2

ind 2

( )11

( )

p

i

x xs s

n x x

= + +

s Estimate of the StandardEstimate of the Standard

DeviationDeviation

of an Individual Value ofof an Individual Value ofyypp

1 12.16025 1

5 4pys = + +

2.16025(1.20416) 2.6013py

s = =

Prediction Interval forPrediction Interval foryypp


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The 95% prediction interval estimate of theThe 95% prediction interval estimate of thenumber of cars sold in one particular weeknumber of cars sold in one particular weekwhen 3 TV ads are run is:when 3 TV ads are run is:

Prediction Interval forPrediction Interval foryypp

2525 ++ 8.288.28

2525 ++ 3.1824(2.6013)3.1824(2.6013)

/ 2 indp y t s

16.72 to 33.28 cars16.72 to 33.28 cars


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Residual AnalysisResidual Analysis

i iy y

Much of the residual analysis is based on anMuch of the residual analysis is based on anexamination of graphical plots.examination of graphical plots.

Residual for ObservationResidual for Observation ii

The residuals provide the best information aboutThe residuals provide the best information about ..

If the assumptions about the error termIf the assumptions about the error term appearappear

questionable, the hypothesis tests about thequestionable, the hypothesis tests about thesignificance of the regression relationship and thesignificance of the regression relationship and theinterval estimation results may not be valid.interval estimation results may not be valid.


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Residual Plot AgainstResidual Plot Againstxx

s If the assumption that the variance ofIf the assumption that the variance of is theis the

same for all values ofsame for all values ofxxis valid, and theis valid, and theassumed regression model is an adequateassumed regression model is an adequaterepresentation of the relationship between therepresentation of the relationship between thevariables, thenvariables, then

The residual plot should give an overallThe residual plot should give an overallimpression of a horizontal band of pointsimpression of a horizontal band of points


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xx

y y

00

Good PatternGood Pattern

Res

idu

al

Res

idu

al



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xx

y y

00

Res

idu

al

Res

idu

al

Nonconstant VarianceNonconstant Variance


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xx

y y

00

Res

idu

al

Res

idu

al

Model Form Not AdequateModel Form Not Adequate


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s ResidualsResiduals


Observation Predicted Cars Sold Residuals

1 15 -1

2 25 -1

3 20 -2

4 15 2

5 25 2


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TV Ads Residual Plot

-3

-2

-1

0

1

2

3

0 1 2 3 4

TV Ads

Residual

New_Linear Regression Analysis- 2

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