Chapter 15

# Chapter 15 - 272 Chapter 15 15.1 5 5 5 20 5 15 5 10 5 = x =...

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272 Chapter 15 15.1 5 5 5 ) 20 ( 5 ) 15 ( 5 ) 10 ( 5 + + + + = x = 15 SST = å = - 2 ) ( x x n j j 5(10 L 15) 2 + 5(15 — 15) 2 + 5(15 & 15) 2 = 250 SSE = å = - 2 ) 1 ( j j s n (5 &1)(50) + (5 “ 1)(50) + (5 p 1)(50) = 600 ANOVA table Degrees of Sum of Mean Source Freedom Squares Squares F Treatments k - 1=3- 1=2 SST = 250 2 250 1 = - k SST = 125 50 125 = MSE MST = 2.50 Error n - k=15- 3=12 SSE = 600 12 600 1 = - k n SSE = 50 __________________________________________ Total n - 1=15 - 1=14 SS(Total) = 850 15.2 5 5 5 ) 20 ( 10 ) 15 ( 10 ) 10 ( 10 + + + + = x = 15 SST = å = - 2 ) ( x x n j j 10(10 p 15) 2 + 10(15 & 15) 2 + 10(15 Y 15) 2 = 500 SSE = å = - 2 ) 1 ( j j s n (10 ±1)(50) + (10 & 1)(50) + (10 F 1)(50) = 1350 ANOVA table Degrees of Sum of Mean Source Freedom Squares Squares F Treatments k- 1=3- 1=2 SST = 500 2 500 1 = - k SST = 250 50 125 = MSE MST = 5.00 Error n - k=30- 3=27 SSE = 1350

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27 1350 1 = - k n SSE = 50 __________________________________________ Total n - 1=30- 1=29 SS(Total) = 1850 15.3 The F statistic increases. 273 15.4 5 5 5 ) 20 ( 5 ) 15 ( 5 ) 10 ( 5 + + + + = x = 15 SST = å = - 2 ) ( x x n j j 5(10 W 15) 2 + 5(15 — 15) 2 + 5(15 & 15) 2 = 250 SSE = å = - 2 ) 1 ( j j s n (5 &1)(25) + (5 “ 1)(25) + (5 p 1)(25) = 300 ANOVA table Degrees of Sum of Mean Source Freedom Squares Squares F Treatments k - 1=3- 1=2 SST = 250 2 250 1 = - k SST = 125 25 125 = MSE MST = 5.00 Error n - k=15- 3=12 SSE = 300 12 300 1 = - k n SSE = 25 __________________________________________ Total n - 1=15 - 1=14 SS(Total) = 550 15.5 The F statistic increases. 15.6 5 5 5 ) 120 ( 5 ) 115 ( 5 ) 110 ( 5 + + + + = x = 115 SST = å = - 2 ) ( x x n j j 5(110 & 115) 2 + 5(115 • 115) 2 + 5(115 & 115) 2 = 250 SSE = å = - 2 ) 1 ( j j s n (5 &1)(50) + (5 “ 1)(50) + (5 p 1)(50) = 600 ANOVA table Degrees of Sum of Mean Source Freedom Squares Squares F Treatments k - 1=3- 1=2 SST = 250 2 250
1 = - k SST = 125 50 125 = MSE MST = 2.50 Error n - k=15- 3=12 SSE = 600 12 600 1 = - k n SSE = 50 __________________________________________ Total n - 1=15 - 1=14 SS(Total) = 850 15.7 The F statistic is unchanged. 15.8 5 5 5 ) 25 ( 5 ) 15 ( 5 ) 5 ( 5 + + + + = x = 15 SST = å = - 2 ) ( x x n j j 5(5 & 15) 2 + 5(15 ± 15) 2 + 5(25 ± 15) 2 = 1000 SSE = å = - 2 ) 1 ( j j s n (5 ±1)(50) + (5 “ 1)(50) + (5 p 1)(50) = 600 274 ANOVA table Degrees of Sum of Mean Source Freedom Squares Squares F Treatments k - 1=3- 1=2 SST = 1000 2 1000 1 = - k SST = 500 50 500 = MSE MST = 10.0 Error n - k=15- 3=12 SSE = 600 12 600 1 = - k n SSE = 50 __________________________________________ Total n - 1=15 - 1=14 SS(Total) = 1600 15.9 The F statistic increased fourfold. 15.10 5 5 5

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) 15 ( 5 ) 15 ( 5 ) 15 ( 5 x + + + + = = 15 SST = å = - 2 ) ( x x n j j 5(15 & 15) 2 + 5(15 ( 15) 2 + 5(15 & 15) 2 = 0 SSE = å = - 2 ) 1 ( j j s n (5 &1)(50) + (5 “ 1)(50) + (5 U 1)(50) = 600 ANOVA table Degrees of Sum of Mean Source Freedom Squares Squares F Treatments k - 1=3- 1=2 SST = 0 2 0 1 = - k SST
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Chapter 15 - 272 Chapter 15 15.1 5 5 5 20 5 15 5 10 5 = x =...

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