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mont4e_sm_ch11_supplemental

# mont4e_sm_ch11_supplemental - Supplemental Exercises 11-78...

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11-49 Supplemental Exercises 11-78 a) = = = = n i i n i i n i i i y y y y 1 1 1 ˆ ) ˆ ( and + = i i x n y 1 0 ˆ ˆ β β from the normal equations Then, 0 ˆ ˆ ˆ ˆ ) ˆ ˆ ( ˆ ˆ ˆ ) ˆ ˆ ( 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 = + = + + = + = = = = = n i i n i i n i i n i i i n i i x n x n x x n y x n β β β β β β β β β β b) i n i i i n i i i n i i i x y x y x y y = = = = 1 1 1 ˆ ) ˆ ( and = = = + = n i i n i i i n i i x x x y 1 2 1 1 0 1 ˆ ˆ β β from the normal equations. Then, 0 ˆ ˆ ˆ ˆ ) ˆ ˆ ( ˆ ˆ 1 2 1 1 0 1 2 1 1 0 1 0 1 1 2 1 1 0 = + = + + = = = = = = = n i i n i i n i i n i i i i n i n i i n i i x x x x x x x x β β β β β β β β c) y y n n i i = = 1 ˆ 1 ) ˆ ˆ ( ˆ 1 0 x y β β + = y x x y x x n y n n x x y n n x n n x n y n i i i i n i i = + = + = + = + = + = = 1 1 1 1 1 1 0 1 0 1 ˆ ˆ ) ˆ ˆ ( 1 ) ˆ ) ˆ ( ( 1 ) ˆ ˆ ( 1 ) ˆ ˆ ( 1 ˆ 1 β β β β β β β β β β

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11-50 11-79 a) 1.1 1.3 1.5 1.7 1.9 2.1 x 0.7 1 1.3 1.6 1.9 2.2 y Plot of y vs x Yes, a linear relationship seems plausible. b) Model fitting results for: y Independent variable coefficient std. error t-value sig.level CONSTANT -0.966824 0.004845 -199.5413 0.0000 x 1.543758 0.003074 502.2588 0.0000 -------------------------------------------------------------------------------- R-SQ. (ADJ.) = 1.0000 SE= 0.002792 MAE= 0.002063 DurbWat= 2.843 Previously: 0.0000 0.000000 0.000000 0.000 10 observations fitted, forecast(s) computed for 0 missing val. of dep. var. ± . . y x = − + 0 966824 154376 c) Analysis of Variance for the Full Regression Source Sum of Squares DF Mean Square F-Ratio P-value Model 1.96613 1 1.96613 252264. .0000 Error 0.0000623515 8 0.00000779394 -------------------------------------------------------------------------------- Total (Corr.) 1.96619 9 R-squared = 0.999968 Stnd. error of est. = 2.79176E-3 R-squared (Adj. for d.f.) = 0.999964 Durbin-Watson statistic = 2.84309 2) H 0 1 0 : β = 3) H 1 1 0 : β 4) α = 0.05 5) The test statistic is ) /( / 0 p n SS k SS f E R = 6) Reject H 0 if f 0 > f α ,1,8 where f 0.05,1,8 = 5.32 7) Using the results from the ANOVA table 9 . 255263 8 / 0000623515 . 0 1 / 96613 . 1 0 = = f 8) Because 2552639 > 5.32 reject H 0 and conclude that the regression model is significant at α = 0.05. P-value < 0.000001 d) 95 percent confidence intervals for coefficient estimates -------------------------------------------------------------------------------- Estimate Standard error Lower Limit Upper Limit CONSTANT -0.96682 0.00485 -0.97800 -0.95565 x 1.54376 0.00307 1.53667 1.55085 -------------------------------------------------------------------------------- 1 1.53667 1.55085 β e) 2) H 0 0 0 : β =