Explanation Given info: An experiment was carried out to test the drilling speed A , feed rate B and material C on the thrust force. The sum of squares of factor A is 19,149.73, due to factor B is 2, 89,047.62, due to factor C is 157,437.52, sum of squares due to error and total sum of squares is 56,819.50 and 2,983,164.81. Calculation: The sum of squares due to the interaction of factor A , B and C is calculated as follows: SST=SSA+SSB+SSC+SSAB+SSBC+SSAC+SSABC+SSE 2 , 983 , 164.81 = [ 19 , 149.73 + 2 , 589 , 047.62 + 157 , 437.52 + 53 , 238.21 + 91 , 880.04 + 9 , 033.73 + SSABC + 56 , 819.50 ] SSAB = 2 , 983 , 164.81 − [ 19 , 149.73 + 2 , 589 , 047.62 + 157 , 437.52 + 53 , 238.21 + 91 , 880.04 + 9 , 033.73 + 56 , 819.50 ] SSAB = 2 , 983 , 164.81 − 2 , 976 , 605.62 = 6 , 559.2 The general ANOVA table is given below: Source Degrees of freedom Sum of squares Mean sum of squares F -ratio Factor A I − 1 SSA = ∑ i ∑ j ∑ k ( X ¯ i ⋅ ⋅ − X ⋅ ⋅ ⋅ ¯ ) 2 MSA = SSA ( I − 1 ) f A = MSA MSE Factor B J − 1 SSB = ∑ i ∑ j ∑ k ( X ¯ ⋅ j ⋅ − X ⋅ ⋅ ⋅ ¯ ) 2 MSB = SSB ( J − 1 ) f B = MSB MSE Factor C L − 1 SSC = ∑ i ∑ j ∑ k ( X ¯ ⋅ ⋅ L − X ⋅ ⋅ ⋅ ¯ ) 2 Interaction AB ( I − 1 ) ( J − 1 ) ( L − 1 ) SSAB = K L ∑ i ∑ j ( X ¯ i j ⋅ ⋅ − X ¯ i ⋅ ⋅ ⋅ − X ¯ ⋅ j ⋅ ⋅ + X ⋅ ⋅ ⋅ ⋅ ¯ ) 2 MSAB = SSAB ( I − 1 ) ( J − 1 ) f A B = MSAB MSE Error I J K ( L − 1 ) SSE = ∑ i ∑ j ∑ k ∑ l ( X i j k l − X ¯ i j k ⋅ ) 2 MSE = SSE I J ( K − 1 ) Total I J K L − 1 SST = ∑ i ∑ j ∑ k ∑ l ( X i j k l − X ¯ ⋅ ⋅ ⋅ ⋅ ) 2 The ANOVA for the given data is shown below: Source Degrees of freedom Sum of squares Mean sum of squares F -ratio Drilling speed A 4 − 1 = 3 SSA = 19 , 149.73 MSA = 19 , 149.73 3 = 6 , 383.24 4.72 Feed rate B 3 − 1 = 2 SSB = 2 , 589 , 047.62 MSB = 2 , 589 , 047.62 2 = 1 , 294 , 523.81 5.61 Material C 2 − 1 = 1 SSC = 157 , 437.52 MSB = 157 , 437.52 1 = 157 , 437.52 7.82 Interaction AB ( 4 − 1 ) ( 3 − 1 ) = 6 SSAB = 53 , 238.21 MSAB = 53,238 .21 6 = 8 , 873.04 3.67 Interaction BC ( 3 − 1 ) ( 2 − 1 ) = 2 SSBC = 91 , 880.04 MSBC = 91,880 .04 2 = 45 , 940.02 5.61 Interaction AC ( 4 − 1 ) ( 2 − 1 ) = 3 SSAC = 9 , 033.73 MSAC = 9,033 .73 3 = 3 , 011.24 4.72 Interaction ABC ( 4 − 1 ) ( 3 − 1 ) ( 2 − 1 ) = 6 SSAC = 9 , 033.73 MSABC = 6,558 .46 6 = 1 , 093.08 3.67 Error ( 4 ) ( 3 ) ( 2 ) ( 2 − 1 ) = 24 SSABC = 6 , 559.2 MSE = SSE I J ( K − 1 ) Total ( 4 × 3 × 2 × 2 − 1 ) = 48 − 1 = 47 Testing the Hypothesis for the fixed effect A: Null hypothesis: H 0 A : α 1 = α 2 = α 3 = α 4 = 0 That is, there is no significant difference in the thrust force due to the four levels of drilling speed

Probability and Statistics for Engineering and the Sciences

9th Edition

ISBN: 9781305251809

Author: Jay L. Devore

Publisher: Cengage Learning

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Chapter 11.3, Problem 28E

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Probability and Statistics for Engineering and the Sciences

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