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Material Type: Assignment; Class: Applied Regression Analysis; Subject: Statistics; University: Ohio State University - Main Campus; Term: Unknown 1989;

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Pre 2010

1 / 6

Download Homework to Solutions - Applied Regression Analysis | STAT 645 and more Assignments Statistics in PDF only on Docsity! STAT 645 HOMEWORK SOLUTION FOR CHANPTER 7 AND 8 Chapter 7 Problem 7.4 a. The AVNOVA table is: Source DF Seq SS MS X1 1 136366 136366 X3|X1 1 2033565 2033565 X2|X1, X3 1 6675 6675 Error 48 985530 20531.88 Total 51 3162136 62002.67 b. H0: β2 = 0 HA: β2 ≠ 0 3251.0 48 985530 1 6675 4n )X,X,SSE(X 1 )X,X|SSR(XF* 321312 =÷= − ÷= F(0.95, 1, 48) = 4.0427 F* < F(0.95, 1, 48), we conclude H0, that is, X2 can be dropped. P-value = 0.5713 c. SSR(X1) + SSR(X2|X1) = 136366 + 5726 = 142092 SSR(X2) + SSR(X1|X2) = 11395 + 130697 = 142092 So they are equal to each other, and the value is actually SSR(X1, X2). Problem 7.7 a. The AVNOVA table is: Source DF Seq SS MS X4 1 67.775 67.775 X1|X4 1 42.275 42.275 X2|X4, X1 1 27.857 27.857 X3|X4, X1, X2 1 0.42 0.42 Error 76 98.231 1.292513 Total 80 236.557 2.956963 b. H0: β3 = 0 HA: β3 ≠ 0 3249.0 76 231.98 1 42.0 5n )X,X,X,SSE(X 1 )X,X,X|SSR(XF* 43212143 =÷= − ÷= F(0.99, 1, 76) = 6.9806 F* < F(0.99, 1, 76), we conclude H0, that is, X3 can be dropped. P-value = 0.5704 Problem 7.8 H0: β2 = 0 and β3 = 0 HA: β2 ≠ 0 or β3 ≠ 0 9388.10 76 231.98 2 110.05-138.327 5n )X,X,X,SSE(X 2 )X,X|X ,SSR(XF* 43211432 =÷= − ÷= F(0.99, 2, 76) = 4.8958 F* > F(0.99, 1, 76), we conclude HA, that is, X2 and X3 can be dropped at the same time. P-value < 0.001 Problem 7.14 a. The R values are: 4021.0 7106.4 2857.6 )X,SSE(X )X,X|SSR(XR 4579.0 0.8509 3896.0 )SSE(X )X|SSR(XR 6190.0 13369.3 8275.4 )SSTO(X )SSR(XR 32 3212 |23Y1 2 212 |2Y1 1 12 Y1 === === === Problem 7.22 The significance of the predictors depends on other variables in the model. When we add four additional variables, if they are related to the three already in the model it is possible for none of the new variables to be significant and those previously in the model can go from significant to not significant. This does not mean the model is worse. Problem 7.25 a. The regression equation is: Y = 4080 + 0.000935 X1 b. In 6.10, the regression function is: Y = 4150 + 0.000787 X1 - 13.2 X2 + 624 X3. What we can see from here is that if we add the factor X2 and X3, there will be a large impact on the coefficient for X1. c. SSR(X1) = 136366, SSR(X1|X2) = 130697, they are not equal each other. d. r12 = 0.08490 If the effects for X1 and X2 are additive, then the two lines should be parallel to each other. But from the graph, they have very different slope, so the effect are not additive. Problem 8.21 a. Hard hat: E{Y} = (β0 +β2) + β1X1 Bump cap: E{Y} = (β0 +β3) + β1X1 None: E{Y} = β0 + β1X1 b. When X1 fixed, test if wearing a bump cap reduced the expected severity of injury. H0: β3 ≥ 0 HA: β3 < 0 When X1 is fixed, test if the expected severity of injury the same when wearing a hard hat as when wearing a bump cap. H0: β2 = β3 HA: β2 ≠ β3 Problem 8.23 It means summer and winter are about the same. Problem 8.24 a. The plot looks as the following: X1 Y 80.077.575.072.570.0 100 90 80 70 60 X2 0 1 Scatterplot of Y vs X1 The regression relations seem to be different. b. The regression equation is: Y = - 127 + 2.78 X1 + 76.0 X2 - 1.11 X1X2 To test if there is any regression relation, the null and alternative hypotheses are: H0: β2 = β3 = 0 HA: β2 ≠ 0 or β3 ≠ 0 To test it, 62.18 60 909.1 2 3670.9-4237.1),,(),,|,( *F 32132132 =÷= − ÷ − = pn XXXSSE qp XXXXXSSR F(0.95; 2, 60) = 3.15041 F* > F(0.95; 2, 60), we reject H0, which implies that there is a regression relation between Y and X2. c. The estimated regression functions for two populations are: Y = −126.9052 + 2.7759X1 for noncorner lots Y = −50.8836 + 1.6684X1 for corner lots From the graph, we can see that the prices for houses not located at the corners are higher than the prices for houses located at the corner. Problem 8.34 a. The linear regression model is: Y = β0 + β1X1 + β2X2 + β3X3 b. The response functions for three types of banks are: Commercial bank: Y = (β0 + β2) + β1X1 Mutual savings: Y = (β0 + β3) + β1X1 Savings and loan: Y = (β0 – β2 – β3) + β1X1 c. β2, β3, –β2–β3 indicates the adjustments for profit or loss of the three different types of bank.