M5 Assignment Answer Key.docx - M5 Assignment Answer Key...

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M5 Assignment Answer Key 14.3 a ¯ ¯ x = 10 ( 30 )+ 14 ( 35 )+ 11 ( 33 )+ 18 ( 40 ) 10 + 14 + 11 + 18 = 35.34 SST = n j ( ¯ x j −¯ ¯ x ) 2 = 10(30– 35.34) 2 +14(35 – 35.34) 2 + 11(33 –35.34) 2 + 18(40 –35.34) 2 = 737.9 SSE = ( n j 1 ) s j 2 = (10 –1)(10) + (14 – 1)(10) + (11 – 1)(10) + (18−1)(10) = 490.0 ANOVA Table Source Degrees of Freedom Sum of Squares Mean Squares F Treatments k 1 = 3 SST = 737.9 SST k 1 = 737.9 3 = 246.0 MST MSE = 246.0 10.00 = 24.60 Error n k = 49 SSE = 490.0 SSE n k = 490.0 49 = 10.00 b ¯ ¯ x = 10 ( 130 )+ 14 ( 135 )+ 11 ( 133 )+ 18 ( 140 ) 10 + 14 + 11 + 18 = 135.34 SST = n j ( ¯ x j −¯ ¯ x ) 2 = 10(130– 135.34) 2 +14(135 – 135.34) 2 + 11(133 –135.34) 2 + 18(140 –135.34) 2 = 737.9 SSE = ( n j 1 ) s j 2 = (10 –1)(10) + (14 – 1)(10) + (11 – 1)(10) + (18−1)(10) = 610.0 ANOVA Table Source Degrees of Freedom Sum of Squares Mean Squares F Treatments k 1 = 3 SST = 737.9 SST k 1 = 737.9 3 = 246.0 MST MSE = 246.0 10.00 = 24.60 Error n k = 49 SSE = 490.0 SSE n k = 490.0 49 = 10.00 c No change 14.6 H 0 1 = µ 2 = µ 3 H 1 : At least two means differ 349
Rejection region: F > F α ,k 1 ,n k = F .05 , 2 , 12 = 3.89 BA BSc BBA Mean 3.94 4.78 5.76 Variance1.26 .92 1.00 Grand mean = 4.83 SST = n j ( ¯ x j −¯ ¯ x ) 2 = 5(3.94 – 4.83) 2 + 5(4.78 – 4.83) 2 + 5(5.76 – 4.83) 2 = 8.30 SSE = ( n j 1 ) s j 2 = (5 –1)(1.26) + (5 – 1)(.92) + (5 – 1)(1.00) = 12.73 ANOVA table Source Degrees of Freedom Sum of Squares Mean Squares F . Treatments k 1 = 2 SST = 8.30 SST k 1 = 8.30 2 = 4.15 MST MSE = 4.15 1.06 = 3.91 Error n k = 12 SSE = 12.73 SSE n k = 12.73 12 = 1.06 F = 3.91, p-value = .0493. There is enough evidence to conclude that students in different degree program differ in their summer earnings. 14.60 a. LSD = t α / 2 ,n k MSE ( 1 n i + 1 n j ) = 1.782 1.06 ( 1 5 + 1 5 ) = 1.16 Treatment Means Difference i = 1, j = 2 3.94 4.78 −.84 i = 1, j = 3 3.94 5.76 −1.82 i = 2, j = 3 4.78 5.76 −1.02 Means of BAs and BBAs differ. b. C = 3(2)/2 = 3, α E = .10, α = α E / C = .0333: t α / 2 ,n k = t .0167 , 12 = 2.404 (from Excel) LSD = t α / 2 ,n k MSE ( 1 n i + 1 n j ) = 2.404 1.06 ( 1 5 + 1 5 ) = 1.57 Treatment Means Difference i = 1, j = 2 3.94 4.78 −.84 i = 1, j = 3 3.94 5.76 −1.82 i = 2, j = 3 4.78 5.76 −1.02 Means of BAs and BBAs differ. 14.88 a k = 4, b = 3, Grand mean = 5.6 SS(Total) = j = 1 k i = 1 b ( x ij −¯ ¯ x ) 2 =( 6 5.6 ) 2 +( 8 5.6 ) 2 +( 7 5.6 ) 2 +( 5 5.6 ) 2 +( 5 5.6 ) 2 +( 6 5.6 ) 2 350
+( 4 5.6 ) 2 +( 5 5.6 ) 2 +( 5 5.6 ) 2 +( 4 5.6 ) 2 +( 6 5.6 ) 2 +( 6 5.6 ) 2 = 14.9 SST = j = 1 k b ( ¯ x [ T ] j −¯ ¯ x ) 2 = 3 [( 7 5.6 ) 2 +( 5.3 5.6 ) 2 +( 4.7 5.6 ) 2 +( 5.3 5.6 ) 2 ]= 8.9 SSB = i = 1 b k ( ¯ x [ B ] i −¯ ¯ x ) 2 = 4 [( 4.8

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