The accompanying data resulted from an experiment to investigate whether yield f

Question

The accompanying data resulted from an experiment to investigate whether yield from a certain chemical process depended either on the formulation of a particular input or on mixer speed. a. A statistical computer package gave SS (formula) = 2253.44, SS (speed) = 230.81, SS (formula*speed) = 18.58, and SSE = 71.87. Fill the following two-way ANOVA table: b. Based on the values that you found for I, J, and K, what are the sets of hypotheses that should be considered in this experiment. c. What are the conclusion for each of these hypotheses (alpha = 5%)? Explain your conclusions.

Explanation / Answer

Here number of levels of factor A are ( r ) = 2

Number of levels of factor B are (c )= 3

Number of replication at each block are (n )= 3

Total number of observations are = n*r*c = 3*2*3 = 18

So degrees of freedom,

For Factor A = r-1 = 2-1 = 1

For factor B = c-1 = 3-1 = 2

For interaction A*B are = (r-1)*(c-1) = 1*2 = 2

For Error = r*c*(n-1) = 2*3*2 = 12

total = ncr-1 = 3*3*2-1 = 17

We are testing three null hypothesis:

There is no difference in the means of factor A
There is no difference in means of factor B
There is no interaction between factors A and B

Verses Alternative hypothesis are

There is significance difference in the means of factor A
There is significance difference in means of factor B
There is interaction between factors A and B

Here ,

MSS(A) = SS(A)/df(A) = 2253.44/ 1 = 2253.44

MSS(B) = SS(B) / df(B) = 230.81/2 = 115.405 and so on.

Fratio (Factor A) = MSS(Factor A) / MSS(Error) = 2253.44/ 5.9892 = 376.25

Fratio (Factor B) = MSS(Factor B) / MSS(Error) = 115.405/5.9892 = 19.27

Fratio (Interaction AB) = MSS(Interaction AB) / MSS(Error) = 9.29/ 5.9892 = 1.55

P value is given by excel command as FDIST( F ration, numerator DF, denominator DF) then hit on enter

For A: =FDIST(2253.44,1,12) then hit on enter.

Here pvalue for A = 1.9*10^-10 << 0.05 so we reject null hupothesis and conclude that,

There is significance difference in the means of factor A

P-value for B = 0.00018 < 0.05 So we reject null hypothesis,

and conclude that There is significance difference in the means of factor B

P-value for AB = 0.2516 > 0.05 We fail to reject null hypothesis ,

so we conclude that There is no interaction between factors A and B

Source Degrees of freedom Sum of squares Mean SS F ratio P-value Factor A 1 2253.44 2253.44 376.25 1.9*E-10 Factor B 2 230.81 115.405 19.27 0.00018 Interaction AB 2 18.58 9.29 1.55 0.2516 Error 12 71.87 5.9892 Total 17 2574.7

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