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# 420Hw10ans - STAT 420 Spring 2010 Homework#10(due Friday...

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Unformatted text preview: STAT 420 Spring 2010 Homework #10 (due Friday, April 23, by 4:00 p.m.) 1. Do NOT use a computer for this problem. Consider the following salary data for the full professors in the Department of Philosophical Engineering at Anytown State University in thousands of dollars per month ( Y ) and x 1 ( years in rank ). Females Males x 1 2 3 3 4 x 1 4 5 5 6 Y 2 3 5 6 Y 8 6 12 10 Consider the model Y = β + β 1 x 1 + β 2 x 2 + e , where the dummy variable x 2 is the indicator for males ( i.e., x 2 = 0 for females, x 2 = 1 for males ). Then X T X = 4 20 4 20 140 32 4 32 8 , ( X T X ) – 1 = ---- 5 . 1 5 . 25 . 1 5 . 25 . 75 . 25 . 1 75 . 5 . 2 , X T Y = 36 234 52 , and β ˆ = - 2 5 . 1 5 . . a) Complete the ANOVA table for the “significance of the regression” test. State the null hypothesis. Is the regression significant at α = 0.05? n = 8, p = 2. Source SS df MS F Regression 59 p = 2 29.5 7.0238 Residual (Error) 21 n – p – 1 = 5 4.2 Total 80 n – 1 = 7 H : β 1 = β 2 = vs H 1 : at least one of β 1 , β 2 is not zero. Critical Value: F 0.05 ( 2 , 5 ) = 5.79 . Decision: Reject H . ( p-value ≈ 0.0353 ) b) Does there seem to be discrimination against females? Test H : β 2 = 0 vs. H 1 : β 2 > 0 at a 5% level of significance. Var( 2 β ˆ ) = C 22 × s 2 = 1.5 × 4.2 = 6.3. Test Statistic: t = 3 . 6 2- ≈ 0.79682 . Critical Value: t 0.05 ( 5 ) = 2.015. Decision: Do NOT Reject H . 1.476 = t 0.10 ( 5 ) > t > t 0.25 ( 5 ) = 0.727 P-value = Area to the right of t. 0.10 < p-value < 0.25 . ( p-value ≈ 0.23085 ) c) Find the partial correlation coefficient of Y and x 1 , with the effects of gender removed. Var( 1 β ˆ ) = C 11 × s 2 = 0.25 × 4.2 = 1.05. Test Statistic for H : β 1 = 0: t = 05 . 1 5 . 1 ≈ 1.46385. r Y X 1 • X 2 = ( 29 5 46385 . 1 46385 . 1 1 2 2 t t + =-- + p n ≈ 0.5477 . d) Construct a 90% prediction interval for the monthly salary of a female full professor with 5 years in rank. female professor with 5 years in rank ⇒ X T = [ 1 5 0 ] Y ˆ = 1 × ( – 0.5 ) + 5 × 1.5 + 0 × 2 = 7. X T C X = [ ] ---- ⋅ ⋅ 5 1 5 . 1 5 . 25 . 1 5 . 25 . 75 . 25 . 1 75 . 5 . 2 5 1 = [ ] -- ⋅ 25 . 1 50 . 25 . 1 5 1 = 1.25. r a ˆ V ( Y ˆ ) = [ 1 + X T C X ] s 2 = ( 1 + 1.25 ) × 4.2 = 9.45. t 0.05 ( 5 ) = 2.015. 7 ± 2.015 × 9.45 7 ± 6.194 e) Find the values of the Multiple R-squared, Adjusted R-squared, and Akaike’s Information Criterion (AIC). Multiple R-squared = 80 59 SYY SSRegr 80 21 1 SYY SSResid 1 = =- =- = 0.7375 ....
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420Hw10ans - STAT 420 Spring 2010 Homework#10(due Friday...

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