Linear Lack of Fit (LOF) TestWhere does this topic fit in?Slide 3Slide 4Slide 5Some notationDecomposing the errorSlide 8The basic ideaA geometric decompositionSlide 11Breakdown of degrees of freedomDefinitions of Mean SquaresExpected Mean SquaresExpanded Analysis of Variance TableThe formal lack of fit F-testLOF Test in MinitabSlide 18Is there lack of linear fit?Slide 20Slide 21Slide 22Example 1: Mortality and LatitudeSlide 24Example 2: Alligator length and weightSlide 26Example 3: Iron and corrosionSlide 28Example 4: Tread wearWhen is it okay to perform the LOF Test?Linear Lack of Fit (LOF) Test An F test for checking whether a linear regression function is inadequate in describing the trend in the dataWhere does this topic fit in?•Model formulation•Model estimation•Model evaluation•Model use504030200150100LatitudeMortalityS = 19.1150 R-Sq = 68.0 % R-Sq(adj) = 67.3 %Mortality = 389.189 - 5.97764 LatitudeRegression PlotExample 1Do the data suggest that a linear function is inadequate in describing the relationship between skin cancer mortality and latitude?Example 2Do the data suggest that a linear function is inadequate in describing the relationship between the length and weight of an alligator?150140130120110100 90 80 70 60700600500400300200100 0LengthWeightS = 54.0115 R-Sq = 83.6 % R-Sq(adj) = 82.9 %Weight = -393.264 + 5.90235 LengthRegression PlotExample 3Do the data suggest that a linear function is inadequate in describing the relationship between iron content and weight loss due to corrosion?210130120110100 90 80Iron contentWeight lossS = 3.05778 R-Sq = 97.0 % R-Sq(adj) = 96.7 %wgtloss = 129.787 - 24.0199 ironRegression PlotSome notation20015010015010050Size of minimum depositNumber of new accountsxy 49.07.50ˆ2811y4212y351y12462y1146y10461y2.87ˆˆ1211yy1.148ˆˆ6261yyDecomposing the error20015010015010050Size of minimum depositNumber of new accountsxy 49.07.50ˆ 14742ˆ2i jijijyy 13594ˆ2i jijiyy 11482i jiijyyDecomposing the error100 150 2008090100110120130140150Size of minimum depositNumber of new accountsxy 49.07.50ˆ 1.45ˆ2i jijijyy 6.6ˆ2i jijiyy 5.382i jiijyyThe basic idea•Break down the residual error (“error sum of squares – SSE) into two components:–a component that is due to lack of model fit (“lack of fit sum of squares” – SSLF)–a component that is due to pure random error (“pure error sum of squares” – SSPE)•If the lack of fit sum of squares is a large component of the residual error, it suggests that a linear function is inadequate.A geometric decomposition20015010015010050Size of minimum depositNumber of new accounts iijijiijijyyyyyy ˆˆijyijijxbby10ˆiyThe decomposition holds for the sum of the squared deviations, too: cinjiijcinjijicinjijijiiiyyyyyy1 121 121 12ˆˆError sum of squares (SSE)Lack of fit sum of squares (SSLF)Pure error sum of squares (SSPE)SSPESSLFSSE Breakdown of degrees of freedom cncn 22Degrees of freedom associated with SSEDegrees of freedom associated with SSLFDegrees of freedom associated with SSPEDefinitions of Mean SquaresAnd, the pure error mean square (MSPE) is defined as:The lack of fit mean square (MSLF) is defined as: cnSSPEcnyyMSPEiij2 22ˆ2cSSLFcyyMSLFijiExpected Mean Squares 2)(2102cXnMSLFEiii2)(MSPEE• If μi = β0+β1Xi, we’d expect the ratio MSLF/MSPE to be … • If μi ≠ β0+β1Xi, we’d expect the ratio MSLF/MSPE to be …• Use ratio, MSLF/MSPE, to reject whether or not μi = β0+β1Xi.Expanded Analysis of Variance TableSource DF SS MS FRegression 1Residual errorn-2Lack of fit c-2Pure error n-cTotal n-1 cinjijijiyySSE1 12ˆ cinjijiyySSTO1 12 cinjijiyySSR1 12ˆ1SSRMSR 2nSSEMSEMSEMSRF cinjijiiyySSLF1 12ˆ cinjiijiyySSPE1 122cSSLFMSLFcnSSPEMSPEMSPEMSLFF The formal lack of fit F-testNull hypothesis H0: μi = β0+β1XiAlternative hypothesis HA: μi ≠ β0+β1XiTest statisticMSPEMSLFF *P-value = What is the probability that we’d get an F* statistic as large as we did, if the null hypothesis is true? The P-value is determined by comparing F* to an F distribution with c-2 numerator degree of freedom and n-c denominator degrees of freedom.LOF Test in Minitab•Stat >> Regression >> Regression …•Specify predictor and response.•Under Options…–under Lack of Fit Tests, select the box labeled Pure error.•Select OK. Select OK.Decomposing the error20015010015010050Size of minimum depositNumber of new accountsxy 49.07.50ˆ 14742ˆ2i jijijyy 13594ˆ2i jijiyy 11482i jiijyyIs there lack of linear fit?Analysis of VarianceSource DF SS MS F PRegression 1 5141 5141 3.14 0.110Residual Error 9 14742 1638 Lack of Fit 4 13594 3398 14.80 0.006 Pure Error 5 1148 230Total 10 198831 rows with no replicatesDecomposing the error100 150 2008090100110120130140150Size of minimum depositNumber of new accountsxy 49.07.50ˆ 1.45ˆ2i jijijyy 6.6ˆ2i jijiyy 5.382i jiijyyIs there lack of linear fit?Analysis of VarianceSource DF SS MS F PRegression 1 5448.9 5448.9 1087.06 0.000Residual Error 9 45.1 5.0 Lack of Fit 4 6.6 1.7 0.21 0.919 Pure Error 5 38.5 7.7Total 10 5494.01 rows with no replicatesExample 1Do the data suggest that a linear function is not adequate in describing the relationship between skin cancer mortality and latitude?504030200150100LatitudeMortalityS = 19.1150 R-Sq = 68.0 % R-Sq(adj) = 67.3 %Mortality = 389.189 - 5.97764 LatitudeRegression PlotExample 1: Mortality and LatitudeAnalysis of VarianceSource DF SS MS F PRegression 1 36464 36464 99.80 0.000Residual Error 47 17173 365 Lack of Fit 30 12863 429 1.69
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