EXST7034 – Regression Techniques Fall 2005 Geaghan Homework 6 Answer sheet Page 31 The program for the following sections follows. 8.15a KNNL) – The model fitted was 011 22iiiiYXXβββ ε=++ +. The estimate of β0 is the intercept for the group receiving the value of “0” for the indicator variable (large machined in this example). The value of β1 is the common slope, since this model has only a single slope. The value β2 is the intercept difference between the two categories. dm'log;clear;output;clear'; *************************************************************; *** EXST7034 Homework Example 1 ***; *** Problem from Neter, Wasserman & Kuttner 1989, #2.18 ***; *************************************************************; OPTIONS LS=132 PS=256 NOCENTER NODATE NONUMBER nolabel; filename copier 'C:\Geaghan\Current\EXST7034\Fall2005\Datasets (KNNL)\CH08PR15.csv'; ODS HTML style=minimal rs=none body='C:\Geaghan\Current\EXST7034\Fall2005\SAS\CH01PR20A6.html' ; Title1 'Assignment 6 : Copier maintenance example'; DATA copier; infile copier missover DSD dlm="," firstobs=2; LABEL machines = 'Number of machines serviced' minutes = 'Minutes to service machines' model = 'Machine model (small=1 or large=0)'; INPUT minutes machines model; interaction = machines * model; small = model; large = abs(model - 1); IntSmall = machines * small; IntLarge = machines * large; datalines; RUN; ; proc print data=copier; run; OPTIONS LS=99 PS=256; PROC REG DATA=copier lineprinter; id interaction; title2 'Initial regression model (parallel lines)'; MODEL minutes = machines model / clb; output out=next1 p=yhat r=e; run; OPTIONS LS=99 PS=56; proc plot data=next1; title2 'Plots from the initial regression model'; plot e*interaction / vref=0; plot e*machines=model / vref=0; run; OPTIONS LS=99 PS=256; PROC REG DATA=copier lineprinter; id interaction; title2 'Full Analysis of Covariance'; MODEL minutes = machines model interaction / clb; output out=next1 p=yhat r=e; run;EXST7034 – Regression Techniques Fall 2005 Geaghan Homework 6 Answer sheet Page 32 8.15b KNNL) – The regression coefficients for the regression model fitted in the first part were estimated as 12-0.92247 15.04614 0.75872iiiiYXXε=+ + +. The SAS output for the fitted model is given below. 8.15c KNNL) – The estimated value was not significantly different from zero, so there does not appear to be a difference in the time required to repair the two machine types. The point estimate for this parameter was 0.75872, suggesting that the small machines took about 46 seconds longer to repair than the larger machines. Of course, this was not significantly different from zero and the 95% confidence interval for this term includes zero, 2P( -4.85125 6.36870 ) = 0.95β≤≤ . 8.15d KNNL) – Clearly there is a relationship between the number of machines and the time needed to repair the machines. This test of interest for this model fits the extra SS for model after machine number (SSX2 | X0, X1). If the term X1 was omitted, and the extra SS (SSX2 | X0) was tested the large SS for X1 would be included in the error term and likely no significance could be detected even if a significant difference existed. 8.15e KNNL) – The requested residuals are plotted below. The first shows the group equal zero bunched at zero. The other points are for the groups assigned a one, which appears to show a slight increasing trend. The plot that I suggested can potentially show differing trends for the two categories. In this case there are not obvious trends. Assignment 6 : Copier maintenance example Initial regression model The REG Procedure Model: MODEL1 Dependent Variable: minutes Number of Observations Read 45 Number of Observations Used 45 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 2 76966 38483 473.94 <.0001 Error 42 3410.32825 81.19829 Corrected Total 44 80377 Root MSE 9.01101 R-Square 0.9576 Dependent Mean 76.26667 Adj R-Sq 0.9556 Coeff Var 11.81513 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| 95% Confidence Limits Intercept 1 -0.92247 3.09969 -0.30 0.7675 -7.17789 5.33294 machines 1 15.04614 0.49000 30.71 <.0001 14.05728 16.03500 model 1 0.75872 2.77986 0.27 0.7862 -4.85125 6.36870EXST7034 – Regression Techniques Fall 2005 Geaghan Homework 6 Answer sheet Page 33 Assignment 6 : Copier maintenance example Plots from the initial regression model Plot of e*interaction. Legend: A = 1 obs, B = 2 obs, etc. e | | 15 + A | A | A | C A | 10 + | A | B | B | A 5 + | A A A | A A | B A | B 0 +-A-------------------------------------------------------------------------------------------- | B A | D | A | A A -5 + | | A | B | -10 + B | A A | A | A | -15 + | | | | A -20 + | | | A | -25 + | --+---------+---------+---------+---------+---------+---------+---------+---------+---------+-- 0 1 2 3 4 5 6 7 8 9 interaction