Explanation Given info: The design variable is the location and the response is the pH measurements. The table provides the mean and standard deviation of the pH measurements. Calculation: State the hypotheses: Null hypothesis: H 0 : μ 1 = μ 2 = μ 3 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 three locations, therefore I = 3 . The total number of observations is, N = 49 + 31 + 30 = 110 . The degrees of freedom corresponding to the location can be obtained as follows: Degrees of freedom = I − 1 = 3 − 1 = 2 The degrees of freedom corresponding to the total can be obtained as follows: Degrees of freedom = N − 1 = 110 − 1 = 109 The degrees of freedom corresponding to the error can be obtained as follows: Degrees of freedom = ( N − 1 ) − ( I − 2 ) = 109 − 2 = 107 Total mean can be obtained as follows: X ¯ = X ¯ 1 + X ¯ 2 + X ¯ 3 + X ¯ 4 I Substitute X ¯ 1 = 6 , X ¯ 2 = 6.2 , X ¯ 3 = 6.4 and I = 3 X ¯ = 6 + 6.2 + 6.4 3 = 18.6 3 = 6.2 The treatment sum of squares (SSTr) can be obtained as follows: SSTr = ∑ i = 1 4 J i ( X ¯ i − X ¯ ) 2 Substitute X ¯ 1 = 6 , X ¯ 2 = 6.2 , X ¯ 3 = 6.4 , J 1 = 49 , J 2 = 31 and J 3 = 30 in the above equation. SSTr = 49 ( 6 − 6.2 ) 2 + 31 ( 6.2 − 6.2 ) 2 + 30 ( 6.4 − 6.2 ) 2 = 49 ( − 0.2 ) 2 + 31 ( 0 ) 2 + 30 ( 0.2 ) 2 = 49 ( 0.04 ) + 0 + 30 ( 0.04 ) = 1.96 + 1.2 = 3.16 The error sum of squares (SSE) can be obtained as follows: S S E = ∑ i = 1 I ( J i − 1 ) s i 2 Substitute s 1 = 0.2 , s 2 = 0

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Solution Summary: The author states that the ANOVA table provides the mean and standard deviation of the pH measurements.

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