New_Linear Regression Analysis- 2
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Transcript of 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
Testing for Significance:Testing for Significance: tt TestTest
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)
Testing for Significance:Testing for Significance: tt TestTest
0 1: 0H =1: 0aH
1
1
b
bt
s=
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Testing for Significance:Testing for Significance: tt TestTest
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:
Confidence Interval forConfidence Interval for 11
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
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Confidence Interval forConfidence Interval for 11
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 Rejection RuleRejection Rule
s 95% Confidence Interval for95% Confidence Interval for 11
s ConclusionConclusion
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s HypothesesHypotheses
s
Test StatisticTest Statistic
Testing for Significance:Testing for Significance: FF TestTest
FF= MSR/MSE= MSR/MSE
0 1: 0H =
1: 0aH
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s Rejection RuleRejection Rule
Testing for Significance:Testing for Significance: FF TestTest
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
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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 > 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)
Testing for Significance:Testing for Significance: FF TestTest
0 1: 0H =1: 0aH
FF= MSR/MSE= MSR/MSE
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Testing for Significance:Testing for Significance: FF TestTest
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.
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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..
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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
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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
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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 = + =
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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
Residual Plot AgainstResidual Plot Againstxx
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Residual Plot AgainstResidual Plot Againstxx
xx
y y
00
Res
idu
al
Res
idu
al
Nonconstant VarianceNonconstant Variance
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Residual Plot AgainstResidual Plot Againstxx
xx
y y
00
Res
idu
al
Res
idu
al
Model Form Not AdequateModel Form Not Adequate
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s ResidualsResiduals
Residual Plot AgainstResidual Plot Againstxx
Observation Predicted Cars Sold Residuals
1 15 -1
2 25 -1
3 20 -2
4 15 2
5 25 2
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Residual Plot AgainstResidual Plot Againstxx
TV Ads Residual Plot
-3
-2
-1
0
1
2
3
0 1 2 3 4
TV Ads
Residual
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End of Chapter 12End of Chapter 12