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Homework 11 Solutions

# Homework 11 Solutions - 481.666 s.e(effect =2(4.505 = 9.01...

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Problem 1 7-4. a) Predictor Coef StDev T P Constant 793.167 4.505 176.05 0.000 Position -24.167 4.505 -5.36 0.001 Temp 240.833 4.505 53.45 0.000 Pos*Temp -3.833 4.505 -0.85 0.420 b) The interaction plot does not indicate a significant interaction between temperature and position. c) The t-ratios are given in the output shown in part a. The t-ratios indicate that both Position and Temperature are significant at the α = 0.05 level. d) The 95% confidence intervals are given by effect estimate ± 2(s.e.(effect)) where effect = 2(coefficient). The coefficient is given in the Minitab output of part a. s.e.(effect) = 2[s.e.(coefficient)]. The s.e.(coefficient) is given in the Minitab output of part a. Position: effect = 2(-24.167) = -48.334 s.e.(effect) = 2(4.505) = 9.01 Approximate 95% confidence interval on the effect of Position: 48.334 ± 2(9.01) ( 66.354, 30.314) Temperature: effect = 2(240.833)=
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Unformatted text preview: 481.666 s.e.(effect) =2(4.505) = 9.01 Approximate 95% confidence interval for the effect of Temperature: 481.666 ± 2(9.01) (463.646, 499.686) Position*Temperature: effect = 2(-3.833) = -7.666 s.e.(effect) =2(4.505) = 9.01 Approximate 95% confidence interval for the effect of Position*Temperature:-7.666 ± 2(9.01) ( − 25.686, 10.354) e) Density = 793 - 24.2 Position + 241 Temp - 3.83 Pos*Temp Predictor Coef StDev T P Constant 793.167 4.505 176.05 0.000 Position -24.167 4.505 -5.36 0.001 Temp 240.833 4.505 53.45 0.000 Pos*Temp -3.833 4.505 -0.85 0.420 S = 15.61 R-Sq = 99.7% R-Sq(adj) = 99.6% Based on the regression analysis, both Position and Temperature appear to be the significant factors. This result is equivalent to that obtained in part c. Density = 793 - 24.2 Position + 241 Temp Predictor Coef StDev T P Constant 793.167 4.436 178.81 0.000...
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