Statistics Homework Solutions 202

# Statistics Homework Solutions 202 - 202 CHAPTER 8 Cubic...

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Unformatted text preview: 202 CHAPTER 8 Cubic Model 15 Residual 10 5 There is no obvious pattern to the residual plot, so the cubic model appears to ﬁt well. 0 −5 −10 400 ¨ ¡¥ ¨ ¡¥ ¡¥ ¤ 237 02. ¨ 176 80. ¨ ¡¥ ¢¥ ¡¥ ¢¥ 0 164 75 ¨ ¡¥ ¤ ¨ ¨ ¤ ¥ ¥ ¤ ¨ ¤ ¥ ¨ ¨ ¢ ¨ ¤ ¥ ¡ ¨ ¢ ¨ ¤ ¨ ¡ ¨ ¢ ¥ ¡ ¨ ¢ ¤ ¡ ¨ ¨ ¤ ¡ ¡ ¨ ¢ ¥ ¡ ¨ ¢ ¡ ¡ ¨ ¤ ¡ ¨ ¨ ¤ ¤ ¡ ¤ ¢ ¥ ¡ ¨ ¤ 163 89. 180 78. ¨ ¨ ¤ 0 164 100 ¨ ¤ 0 2106 200 236 39. ¨ ¢¥ 0 3660 1400 187 56. 234 18. 0 16047 200 0 5198 75 ¤ 0 5198 100 ¡¥ ¨ 0 41641 1400 0 16047 300 0 2106 300 0 37820 1400 0 164 150 ¨ ¢ ¨ ¡ 380 1 ¨ 301 8 ¤ 320 59 Under model 2, the prediction is ¥ 0 3660 1600 301 8 (c) Under model 1, the prediction is ¡ ¡ Under model 3, the prediction is ¨ 0 37820 1600 0 41641 1600 ¤ ¨ 380 1 Under model 2, the prediction is 0 2106 400 166 55. ¡¥ ¢ 320 59 0 5198 150 182 52. ¨ ¨ 0 3660 1500 301 8 (b) Under model 1, the prediction is ¤ Under model 3, the prediction is ¨ ¡ ¡ Under model 3, the prediction is 0 41641 1500 ¥ 380 1 0 16047 400 ¡ Under model 2, the prediction is 0 37820 1500 ¨ 320 59 9. (a) Under model 1, the prediction is ¤ 200 300 Fitted Value ¡¥ 100 ¨ −15 0 (d) (iv). The output does not provide much to choose from between the two-variable models. In the three-variable model, none of the coefﬁcients are signiﬁcantly different from 0, even though they were signiﬁcant in the two-variable models. This suggest collinearity. S = 12.308 Coef 40.751 0.54013 R-sq = 4.2% Analysis of Variance StDev 5.4533 0.61141 T 7.4728 0.88341 R-sq(adj) = ¡ Linear Model Predictor Constant Hardwood 1 2% ¨ 11. (a) P 0.000 0.389 ...
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