Applied Statistics and Probability for Engineers

512 CHAPTER 14 DESIGN OF EXPERIMENTS WITH SEVERAL FACTORS

The hypotheses that we will test are as follows:

1. H 0 :  1  2 a 0 (no main effect of factor A)
   H 1 : at least one i 0


2. H 0 :  1  2 b 0 (no main effect of factor B) (14-2)
   H 1 : at least one j 0


3. H 0 : () 11 () 12 ()ab 0 (no interaction)
   H 1 : at least one ()ij 0
   As before, the ANOVA tests these hypotheses by decomposing the total variability in the data
   into component parts and then comparing the various elements in this decomposition. Total
   variability is measured by the total sum of squares of the observations

and the sum of squares decomposition is defined below.

SST a

a

i 1

a

b

j 1

a

n

k 1

1 yijky... 22

p

p

p

Equations 14-3 and 14-4 state that the total sum of squares SSTis partitioned into a sum of squares

for the row factor A(SSA), a sum of squares for the column factor B(SSB), a sum of squares for the

interaction between Aand B(SSAB), and an error sum of squares (SSE). There are abn1 total

degrees of freedom. The main effects Aand Bhave a1 and b1 degrees of freedom, while

the interaction effect ABhas (a1)(b1) degrees of freedom. Within each of the abcells in

Table 14-3, there are n1 degrees of freedom between the nreplicates, and observations in the

same cell can differ only because of random error. Therefore, there are ab(n1) degrees of free-

dom for error. Therefore, the degrees of freedom are partitioned according to

If we divide each of the sum of squares on the right-hand side of Equation 14-4 by the

corresponding number of degrees of freedom, we obtain the mean squaresfor A,B, the

abn 1  1 a 12  1 b 12  1 a 121 b 12 ab 1 n 12

Thesum of squares identity for a two-factor ANOVAis

(14-3)

or symbolically,

SSTSSASSBSSABSSE (14-4)

 a

a

i 1

(^) a
b
j 1
(^) a
n
k 1
1 yijkyij. 22
na
a
i 1
(^) a
b
j 1
1 yij.yi..y.j.y... 22
ana
b
j 1
1 y.j.yp 22
a
a
i 1 a
b
j 1 a
n
k 1
1 yijky... 22 bna
a
i 1
1 yi..y... 22
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