Main Effect of Factor B
Hypotheses:
H0: No effect of Factor B (that is average change time independent of vehicle) or j =
0 for all j
Ha: There is an effect due to Factor B or j ≠ 0 for some j
If Factor B has little effect then we expect the mean of each level to be close to the
overall mean and therefore the F statistic below should be small
2
∑ a×r× x(−•j )
j
F = MSB = b−1 ~ F
MSE MSE b−1,ab(r−1)
Example: Brake Replacement
Source Degrees Sum oMean SF-statistic
VariatiFreedoSquareof Squares -> Compare with earlier table that included the interaction.
Manufacture2 595.8 297.9 7.73 First two rows and the last row are the same but now the
Vehicle 1 6840.36840.3177.46
Total 29 8438.338.55 “interaction” sum of squares has been included in the error
sum of squares
Multiple Pairwise Comparisons
—> Comparing the means allows us to explore the nature of the differences detected by
the F-tests
When there is interaction between the factors in how they affect the response variable,
the main interest is comparing the treatment means
When there are main effects, then we may be interested in comparing row means or
column means, or comparing Factor A or Factor B means
Comparison of Treatment Means
—> For each set of comparisons, we start by determining how many means there are to
compare:
In our example, the primary concern is with the pairwise comparisons of the ab = 6
treatment means
There are a total of J = 6×5/2 = 15 pairwise comparisons
Because each comparison is effectively a hypothesis test, the multiple comparisons
demands an adjustment of the level of signiﬁcance for each individual test to ensure
that the overall level of signiﬁcance is no larger than
Comparing Treatment Means
—> To compare each pair of treatment means, we look at the observed differences and
compare them to the ± margin of error:
⎛ ⎞
⎜ ⎛1 1 ⎞ ⎟
⎜xij• ±tkl•SE ⎜ + ⎟ ⎟
⎝ ⎝ r r⎠ ⎠ —> where t* is the upper α/(2J) point of the t-distribution with degrees of freedom given
by the MSE (ab(r - 1)) and r is the number of observations that comprise each treatment
mean
Calculations
Here, if α = .05, then α/(2J) = .05/ (2×15) = .05/30 = .001667
Since the MSE has 24 d.f., the critical value of t is 3.2583:
Student's t distribution with 24 DF
P( X <= x ) x
0.998333 3.25830
The margin of error is:
3.2583×sqrt(34.53×2/5) = 3.2583×3.71645 = 12.109
The six treatment means are:
(A, Car) = 60.2, (B, Car) = 65, (C, Car) = 74.6
(A, SUV) = 97.2, (B, SUV) = 91.8, (C,SUV) = 101.4
Calculations
By inspection:
The pairs (60.2,65), (65

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