Homework 6 Solution (Due Oct 31th, Thursday)Problems:1) 6.5a) From “PROC INSIGHT, one can see that the linear relationship between Y (liking) and X1.Itseems that Y slightly increases as the level of X2increases.From “PROC CORR”, the following correlation m at ri x is produced. Based on these correlati ons , thecorrelation between Y and X1is much str onge r than the correlation between Y and X2.The CORR ProcedurePearson Correlation Coefficients, N = 16Prob > |r| under H0: Rho=0liking moisture sweetnessliking 1.00000 0.89239 0.39458<.0001 0.1304moisture 0.89239 1.00000 0.00000<.0001 1.0000sweetness 0.39458 0.00000 1.000000.1304 1.0000b) The regression function isˆY = 37.650 + 4.425X1+4.375X2. b1is interpreted as follows: The meanresponse of liking increases by 4.425 with a unit increase in X1when X2is held constant, no matterwhat th e level of X2.Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t|Intercept 1 37.65000 2.99610 12.57 <.0001moisture 1 4.42500 0.30112 14.70 <.0001sweetness 1 4.37500 0.67332 6.50 <.000116c) Usi n g the PROC UNIVARIATE function in SAS, the following stem-and-leaf plot for the residualswas obtained. For this small data set (n=16), it is not easy to see if this plot shows an approximatelynormal di s t ri b ut i on . We will also examine the normal Q-Q plot in part d) below.The UNIVARIATE ProcedureVariable: residual (Residual)Stem Leaf #42 13 24 224 11 23 20 26 2-0 1 1-1 760 3-2 60 2-3 1 1-4 4 1----+----+----+----+d) See the plots below. For all of these plots, there is no strong pattern, and the distribution of theresiduals is similar above and below the horizontal line at 0. Note that the plots of residuals v s.ˆY andresiduals vs. moisture show a possible nonlinear pattern (possibly this is c aus ed by small sample size).From the plots of residuals vs. moisture the variance seems to be nonconstant. Note t h at for the plotof residuals vs. moisture*sweetness, there is also no pattern, which indicates that no interaction effectsreflected by the model term X1X2appear to be present. The normal Q-Q plot of residuals falls approx-imately on the straight line, indicating that the assumption of normality of the residuals is appropriate.Plot of resid ual s vs.ˆY Plot of resid ual s vs. moisturePlot of resid ual s vs. sweetness Plot of residual s vs. moisture*sweetness17Normal QQ plot for residuals2) 6.9a) Using the “PROC UNIVARIATE function in SAS, the following stem-and-leaf plots are obtained.For X1(# of cases shipped), the distribution is slightly skewed to the right, with several large valuesthat could be considered outliers. However, overall this distribution is close to normal. For X2(indirectcosts of the total labor hours), there is one small value that could be considered an outlier, but thedistribution is close to a normal distri bu t i on.Variable: shipStem Leaf #46 2 144 3 142 7 140 24 238 3 136 70 234 2 132 22388 530 1233677 728 380023467 926 15688999022477 1424 668027 622 8 120 2 1----+----+----+----+Multiply Stem.Leaf by 10**+4Variable: costStem Leaf #96 190 18 566 38 00011224 87 66777888899 117 00222222344 116 577888999 96 123344 658 1546 1----+----+----+----+c) From the scatter plot matrix, Y increases when the week has a holiday(i.e., X3= 1). It seemsthat the linear relationship between Y and X1exists when X1is not large. Note that the r el at i ons hi pbetween Y (liking) and X2seems to be unclear. There does not appear to be significant correlation18among the three predic t or variables.Pearson Correlation Coefficients, N = 52Prob > |r| under H0: Rho=0hours ship cost holidayhours 1.00000 0.20766 0.06003 0.810580.1396 0.6725 <.0001ship 0.20766 1.00000 0.08490 0.045660.1396 0.5496 0.7479cost 0.06003 0.08490 1.00000 0.113370.6725 0.5496 0.4236holiday 0.81058 0.04566 0.11337 1.00000<.0001 0.7479 0.42363) 6.10a) See SAS output below. The estimated regression function isˆY = 4149.89+0.000787X1−13.166X2+623.554X3where X3=holiday. b1is interpreted as follows: The mean response increases by 0.000787with a unit increase in X1when both X2and X3are held constant. Interpretations are similar for b2and b3.Parameter EstimatesParameter StandardVariable DF Estimate Error t Value Pr > |t|Intercept 1 4149.88721 195.56541 21.22 <.0001ship 1 0.00078708 0.00036455 2.16 0.0359cost 1 -13.16602 23.09173 -0.57 0.5712holiday 1 623.55448 62.64095 9.95 <.00014) 6.11a) H0: β1= β2= β3=0vs. Ha: Not all βk= 0(k =1, 2, 3). Since MSR = 725, 535 and MSE =20.531.9, F�obs= 725, 535/20.531.9 = 35.337. Since F (0.95; 3, 48) = 2.79806. If F�obs< 2.79806,conclude H0, other wi se Ha.SinceF�obs> 2.79806, conclude Haand its P-value is 0+.19b) Since s(b1)=.000365, s(b3) = 62.6409 and B = t(.9875; 48) = 2.3139, a 95% Bonferr oni basedjoint confidence intervals for β1and β3are β1∈ .000787 ± 2.3139(.000365) = [−.000058,.00163] andβ3∈ 623.554 ± 2.3139(62.6409) = [478.6092, 768.4988]. The 95% joint confid en ce interval means th atboth intervals for β1and β3will be in the i nterval at least 95% of the time.c) SSR =2, 176, 606 and SST O =3, 162, 136, R2= .6883. This measures the proportionate re du ct i onof total variation i n Y associated with the use of X1,X2and X3in the model.Analysis of VarianceSum of MeanSource DF Squares Square F Value Pr > FModel 3 2176606 725535 35.34 <.0001Error 48 985530 20532Corrected Total 51 3162136Root MSE 143.28946 R-Square 0.6883Dependent Mean 4363.03846 Adj R-Sq 0.6689Coeff Var 3.28417Obs t11 2.313905) The following is based on partial SAS output (below) for the new cases.Output StatisticsDependent Predicted Std ErrorObs Variable Value Mean Predict 95% CL Mean Residual1 4264 4296 21.5340 4253 4339 -32.06352 4496 4327 35.6211 4255 4398 169.20513 4317 4339 67.0370 4204 4474 -21.82544 4292 4346 32.9794 4280 4412 -54.11965 4945 4869 65.0869 4738 5000 75.93376 4325 4297 23.5980 4250 4344 28.00077 4110 4267 24.2408 4218 4315 -156.68448 4111 4278 34.0151 4210 4346 -166.98349 4161 4298 28.7193 4240 4355 -136.6910---------------------------48 4993 5022 76.0618 4869 5175 -28.753349 4309 4306 22.4132 4261 4351 2.731350 4499 4273 26.4996 4220 4327 225.697851 4186 4371 45.9379 4279 4463 -184.877652 4342 4277 23.7795 4230 4325 64.516853 . 4293 21.3567 4250 4336 .54 . 4245 29.7021 4186 4305 .55 . 4279 24.4444 4230 4329 .Obs t21 2.48078From SAS, B = t(1 − .05/(2 · 3); 48) = 2.48078. Thus, the j oi nt confidence interval s are:Case number 1 : 4293 ± (2.48078)(21.3567) = (4239.98, 4346.02)Case number 2 : 4245