Explanation Given info: The design variable is the well type and the response is the ammonium concentration. The table provides the mean and standard deviation of the ammonium concentration corresponding to each well type. Calculation: State the hypotheses: Null hypothesis: H 0 : μ 1 = μ 2 = μ 3 = μ 4 = μ 5 Alternative hypothesis: H a : At least two of the μ i ′ s differ from each other. The ANOVA table can be obtained as follows: There are seven angles, therefore I = 5 . The total number of observations is, N = 53 + 57 + 66 + 67 + 51 = 294 . The degrees of freedom corresponding to the well can be obtained as follows: Degrees of freedom = I − 1 = 5 − 1 = 4 The degrees of freedom corresponding to the total can be obtained as follows: Degrees of freedom = N − 1 = 294 − 1 = 293 The degrees of freedom corresponding to the error can be obtained as follows: Degrees of freedom = ( N − 1 ) − ( I − 2 ) = 293 − 4 = 289 Total mean can be obtained as follows: X ¯ = X ¯ 1 + X ¯ 2 + X ¯ 3 + X ¯ 4 + X ¯ 5 I Substitute X ¯ 1 = 0.75 , X ¯ 2 = 0.90 , X ¯ 3 = 0.68 , X ¯ 4 = 0.50 , X ¯ 5 = 0.82 and I = 5 X ¯ = 0.75 + 0.90 + 0.68 + 0.50 + 0.82 5 = 3.65 5 = 0.73 The treatment sum of squares (SSTr) can be obtained as follows: SSTr = ∑ i = 1 4 J i ( X ¯ i − X ¯ ) 2 Substitute X ¯ 1 = 0.75 , X ¯ 2 = 0.90 , X ¯ 3 = 0.68 , X ¯ 4 = 0.50 , X ¯ 5 = 0.82 and J 1 = 53 , J 2 = 57 , J 3 = 66 , J 4 = 67 , J 5 = 51 SSTr = ( 53 ( 0.75 − 0.73 ) 2 + 57 ( 0.90 − 0.73 ) 2 + 66 ( 0.68 − 0.73 ) 2 + 67 ( 0.50 − 0.73 ) 2 + 51 ( 0.82 − 0.73 ) 2 ) = ( 53 ( 0.02 ) 2 + 57 ( 0.17 ) 2 + 66 ( − 0.05 ) 2 + 67 ( − 0.23 ) 2 + 51 ( 0.09 ) 2 ) = ( 53 ( 0.0004 ) + 57 ( 0.0289 ) + 66 ( 0.0025 ) + 67 ( 0.0529 ) + 51 ( 0.0081 ) ) = 0.0212 + 1

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ISBN: 9780073401331

Author: William Navidi Prof.

Publisher: McGraw-Hill Education

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