Several people wrote to tell me that my results from Excel were correct and
the Bluman textbook had the wrong answer. I'm continuing to experiment with
ANOVA and I'm hoping that the list can help me answer two additional
questions I have.
First, in experimenting around with a 3x3 ANOVA without replication, I
filled the cells with =RAND() to generate random numbers and I happened to
get the following:
SS df MS F
Rows -2.105562167 2 -1.052781083 -1.336268078
Columns -0.226230909 2 -0.113115455 -0.143574551
Error 3.151406819 4 0.787851705
Total 0.819613743 8
Unfortunately, I did not convert the random number functions to numbers so
the specific data I used to generate this table was lost when the worksheet
recalculated.
I've repeated this experiment several times and sometimes the sums of
squares are positive and sometimes one or more of them are negative. Is
this a bug in Excel? I don't see how a sum of *square* can be negative even
though I am using random numbers so all of the SS should be in error.
Second, Excel offers a two-factor ANOVA with and without replication. None
of the examples of two-way ANOVA in any of the textbooks that I own shows a
two-way ANOVA without replication. It seems counter intuitive to me to
perform ANOVA with a sample size of one in each cell. Am I missing
something here?
The output from with and without replication is significantly different as
well. Using made up data, without replication, I get the following:
SS df MS F P-value F crit
Rows 56.25 1 56.25 25 0.1257 161.4462
Columns 110.25 1 110.25 49 0.0903 161.4462
Error 2.25 1 2.25
Total 168.75 3
Using made up data, with replication, I get the following
SS df MS F P-value F crit
Sample 234.38 1 234.38 130.8140 0.0000 4.3513
Columns 693.38 1 693.38 387.0000 0.0000 4.3513
Interaction 0.38 1 0.38 0.2093 0.6522 4.3513
Within 35.83 20 1.79
Total 963.96 23
It makes no sense to me to call them "rows and columns" without replication
and "sample and columns" with replication. It sort of makes sense not to
get interaction without replication. I am assuming that with a sample size
of one in each cell that there is not enough information to compute
interactions.
Ronny Richardson
.
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