ISM Chapter 15

ISM Chapter 15 - Chapter 15 15.1a x SST = SSE = 5(10) 5(15)...

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Chapter 15 15.1a 5 5 5 ) 20 ( 5 ) 15 ( 5 ) 10 ( 5 x + + + + = = 15 SST = = - 2 j j ) x x ( n 5(10 – 15) 2 + 5(15 – 15) 2 + 5(20 – 15) 2 = 250 SSE = = - 2 j j s ) 1 n ( (5 –1)(50) + (5 – 1)(50) + (5 – 1)(50) = 600 ANOVA Table Source Degrees of Freedom Sum of Squares Mean Squares F Treatments 2 1 k = - SST = 250 2 250 1 k SST = - = 125 50 125 MSE MST = = 2.50 Error 12 k n = - SSE = 600 12 600 k n SSE = - = 50 __________________________________________ Total 14 k n = - SS(Total) = 850 b 5 5 5 ) 20 ( 10 ) 15 ( 10 ) 10 ( 10 x + + + + = = 15 SST = = - 2 j j ) x x ( n 10(10 – 15) 2 + 10(15 – 15) 2 + 10(20 – 15) 2 = 500 SSE = = - 2 j j s ) 1 n ( (10 –1)(50) + (10 – 1)(50) + (10 – 1)(50) = 1350 ANOVA Table Source Degrees of Freedom Sum of Squares Mean Squares F . Treatments 2 1 k = - SST = 500 2 500 1 k SST = - = 250 50 250 MSE MST = = 5.00 Error 27 k n = - SSE = 1350 50 27 1350 k n SSE = = - __________________________________________ Total 29 k n = - SS(Total) = 1850 c The F statistic increases. 15.2 a 4 4 4 ) 25 ( 4 ) 22 ( 4 ) 20 ( 4 x + + + + = = 22.33 SST = = - 2 j j ) x x ( n 4(20 – 22.33) 2 + 4(22 – 22.33) 2 + 4(25 – 22.33) 2 = 50.67 SSE = = - 2 j j s ) 1 n ( (4 –1)(10) + (4 – 1)(10) + (4 – 1)(10) = 90 ANOVA Table Source Degrees of Freedom Sum of Squares Mean Squares F Treatments 2 1 k = - SST = 50.67 33 . 25 2 67 . 50 1 k SST = = - 10 33 . 25 MSE MST = = 2.53 87

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Error 9 k n = - SSE = 90 9 90 1 k n SSE = - = 10 b SSE = = - 2 j j s ) 1 n ( (4 –1)(25) + (4 – 1)(25) + (4 – 1)(25) = 225 ANOVA Table Source Degrees of Freedom Sum of Squares Mean Squares F Treatments 2 1 k = - SST = 50.67 33 . 25 2 67 . 50 1 k SST = = - 0 . 25 33 . 25 MSE MST = = 1.01 Error 9 k n = - SSE = 225 0 . 25 9 225 k n SSE = = - c The F statistic decreases. 15.3 a 34 . 35 18 11 14 10 ) 4 ( 18 ) 33 ( 11 ) 35 ( 14 ) 30 ( 10 x = + + + + + + = SST = = - 2 j j ) x x ( n 10(30– 35.34) 2 +14(35 – 35.34) 2 + 11(33 –35.34) 2 + 18(40 –35.34) 2 = 737.9 SSE = = - 2 j j s ) 1 n ( (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 3 1 k = - SST = 737.9 0 . 246 3 9 . 737 1 k SST = = - 60 . 24 00 . 10 0 . 246 MSE MST = = Error 49 k n = - SSE = 490.0 00 . 10 49 0 . 490 k n SSE = = - b 34 . 135 18 11 14 10 ) 14 ( 18 ) 133 ( 11 ) 135 ( 14 ) 130 ( 10 x = + + + + + + = SST = - 2 j j ) x x ( n = 10(130– 135.34) 2 +14(135 – 135.34) 2 + 11(133 –135.34) 2 + 18(140 –135.34) 2 = 737.9 SSE = = - 2 j j s ) 1 n ( (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 3 1 k = - SST = 737.9 0 . 246 3 9 . 737 1 k SST = = - 60 . 24 00 . 10 0 . 246 MSE MST = = Error 49 k n = - SSE = 490.0 00 . 10 49 0 . 490 k n SSE = = - c No change 15.4 = μ = μ 2 1 0 : H 3 μ 88
: H 1 At least two means differ.

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This homework help was uploaded on 04/18/2008 for the course STAT 212 taught by Professor Holt during the Spring '08 term at UVA.

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ISM Chapter 15 - Chapter 15 15.1a x SST = SSE = 5(10) 5(15)...

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