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PSU STAT 501 - Lack of Fit Test

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Lack of Fit (LOF) TestSlide 2Slide 3Slide 4Lack of fit test for a linear functionThe full modelThe reduced modelAssumptions and requirementsNotationThe Full ModelThe Reduced ModelDecomposing the errorSlide 13Error sum of squares decompositionThe general linear testThe test (intuitively)The formal LOF testLOF Test summarized in an ANOVA TableLOF Test in MinitabSlide 20Example 1: Mortality and LatitudeSlide 22Example 2: Alligator length and weightSlide 24Example 3: Iron and corrosionSlide 26Example 4: Tread wearLack of Fit (LOF) Test An F test for checking whether a specific type of regression function adequately fits the data504030200150100LatitudeMortalityS = 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 not adequate in describing the relationship between skin cancer mortality and latitude?Example 2Do the data suggest that a linear function is not adequate 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 not adequate in describing the relationship between iron content and weight loss due to corrosion?210130120110100 90 80ironwgtlossS = 3.05778 R-Sq = 97.0 % R-Sq(adj) = 96.7 %wgtloss = 129.787 - 24.0199 ironRegression PlotLack of fit test for a linear function•Use general linear test approach.•Full model is most general model with no restrictions on the means μj at each Xj level.•Reduced model assumes that the μj are a linear function of the Xj, i.e., μj = β0+ β1Xj.The full model543213020100High school gpaCollege entrance test score12345ijjijYThe reduced model54321221814106High school gpaCollege entrance test score xYEY 10 iixY10Assumptions and requirements•The errors, εi, and hence the responses Yi, are independent.•The errors, εi, and hence the responses Yi, are normally distributed.•The errors, εi, and hence the responses Yi, have equal variances (σ2) for all x values. •The LOF test requires repeat observations, called replicates, for at least one of the X values.Notationiron wgtloss0.01 127.60.01 130.10.01 128.00.48 124.00.48 122.00.71 110.80.71 113.10.95 103.91.19 101.51.44 92.31.44 91.41.96 83.71.96 86.2• c different levels of X (c=7 with X1=0.01, X2=0.48, …, X7=1.96)• nj = number of replicates for jth level of X (Xj) (n1=3, n2=2, …, n7=2) for a total of n = n1 + … + nc observations.• Yij = observed value of the response variable for the ith replicate of Xj (Y11=127.6, Y21=130.1, …, Y27=86.2)The Full ModelAssume nothing about (or “put no structure on”) the means of the responses, μj, at the jth level of X:ijjijYLeast squares estimates of μj are sample means,jjYˆof responses at Xj level.,“Pure error sum of squares” SSPEYYFSSEcjnijijj 21 1)(The Reduced ModelAssume the means of the responses, μj, are a linear function of the jth level of X:ijjijXY10Least squares estimates of μj are as usual:jijXbbY10ˆ“Error sum of squares” SSEYYRSSEcjniijijj 21 1ˆ)(Decomposing the error2001501001501401301201101009080Size of minimum depositNumber of new accountsxy 49.07.50ˆDecomposing the error20015010015010050Size of minimum depositNumber of new accountsxy 49.07.50ˆError sum of squares decomposition   ijjjijiji jYYYYYYˆˆerror deviation pure error deviation lack of fit deviation   j iijjj ijijj iijijYYYYYY222ˆˆSS LFSSPESS E The general linear testFFRdfFSSEdfdfFSSERSSEF)()()(*2ndfRcndfFSSERSSE )(SSPEFSSE )(       MSPEMSLFcnSSPEcSSLFcnSSPEcnnSSPESSEF 22*The test (intuitively)•If the largest portion of the error sum of squares is due to lack of fit, the F test should be large.•A large F* statistic leads to a small P-value (determined by F(c-2, n-2) distribution).•If the P-value is small, reject the null and conclude significant lack of (linear) fit.The formal LOF test Null hypothesis H0: (Reduced) Alternative hypothesis HA: (Full)Test 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? (One-tailed test!)The P-value is determined by comparing F* to an F distribution with c-2 numerator degrees of freedom and n-2 denominator degrees of freedom.ijjijxY10ijjijYLOF Test summarized in an ANOVA TableLOF 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.504030200150100LatitudeMortalityS = 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 not adequate in describing the relationship between skin cancer mortality and latitude?Example 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 0.128 Pure Error 17 4310 254Total 48 5363719 rows with no replicatesExample 2Do the data suggest that a linear function is not adequate 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 2: Alligator length and weightAnalysis of VarianceSource DF SS MS F PRegression 1 342350 342350 117.35 0.000Residual Error 23 67096 2917 Lack of Fit 17 66567 3916 44.36 0.000 Pure Error 6 530 88Total 24 40944614 rows with no replicatesExample 3Do the data suggest that a linear function is not adequate in


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PSU STAT 501 - Lack of Fit Test

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