ANalysis Of VAriance = ANOVA
There are actually several different kinds of
Analysis of Variance. We shall discuss the one that
is most like the generalization of Independent t-
tests.
In Independent t-tests, of course, we tested
whether 2 sample means were equal or unequal.
In ANOVA we will test whether 2 or more sample
means are all equal or whether one or more of them is
not equal to the others.
One way to show the formal hypotheses is:
H Null: Mean1=Mean2=Mean3=Mean4
H1: H Null is false
It may seem that we could solve the problem about
equality of means by doing several t-tests. However,
doing multiple t-tests changes the value of alpha away
from what intend.
For example, if we are comparing 4 Independent
sample means, and set the value for alpha at .05 and
then proceed to do the six t-tests, our actual,
experiment-wise alpha-level is:
1-(1-.05)4 = 1-(.95)4 =
1-(.95)(.95)(.95)(.95) =
1 - .815 = .185
So, we set out to have alpha at .05 but have
ended up with a very different (& much higher) alpha
of .185
Sir Ronald Fisher figured out how to avoid this
difficulty, when he invented ANOVA
Fisher says, first do an ANOVA. IF and only IF
the ANOVA rejects the null H do we proceed to do t-
tests.
If ANOVA says to accept the null H, then the
computations are over. We are through: do not do t-
tests.
If ANOVA says Reject H0 and Accept H1 then we
know that everything is NOT equal. Something is not
equal --maybe only one among the many. Then we have
to find what isn’t equal, using t-tests.
General:
When we have several related data sets, we can
temporarily collapse them into one, and compute an
overall sum, an overall (Grand) Mean, an overall sum
of Xsquares, and an overall variance.
The essence of ANOVA is that this overall
variance can be broken down into parts. The parts
represent the variance of the separate Sample Means
around the Grand Mean.
(drawing of group means, around grand mean)
_____________________________________
We will practice with some ANOVA calculations
shortly. ANOVA can analyze for differences among 2,
3, 4 or more samples.
So, when we have two Independent Samples, we CAN
use ANOVA if we want to, instead of using the t-test.
For 2 groups, t and ANOVA are equivalent.
For many ANOVA problems, it is convenient to use
a double-subscript for the individual scores, as in
Xij This is because the data usually form a matrix,
with rows and columns, as
Col-1 Col-2 Col-3 Col-4
Row-1 67 92 101 66
Row-2 65 93 89 51
Row-3 63 85 91 46
Row-5 51 89 92 55
Row-6 64 91 105 41
Row-7 68 88 99 51
Row-8 66 85 97 46
When reading the double sub-script, we always
read the row first, then the column (mnemonic 'Roman
Catholic')
so X51 would be the score 71
X23 would be the score 89
X84 would be the score 46
Quite often, we use the Columns to represent
different experimental treatments (as drug amounts;
practice time, etc.) and the rows to represent the
subjects who had that particular treatment.
So our matrix could represent 0 grams, 1 gram, 2
grams or 4 grams of some experi-mental drug we
administered. The 0-drug group would be the CONTROL
Group -- i.e., controlling for handling, injecting,
etc.
And each of the Groups here had 8 subjects who
had that particular drug treatment.
Let's compute the Group (Column) sums and means:
Col-1 Col-2 Col-3 Col-4
(g drug) 0 1 2 4
Row-1 67 92 101 66
Row-2 65 93 89 51
Row-3 59 77 94 62
Row-4 63 85 91 46
Row-5 51 89 92 55
Row-6 64 91 105 41
Row-7 68 88 99 51
Row-8 66 85 97 46
Sums 503 700 768 418
Means 62.88 87.50 96.00 52.25
By-the-Way: Sum of Xij2 = 189557
The null H says these means are all the same
(within sampling). The alternative H says they are
not.
Let's compute the Grand Sum = 503 + 700 + 768 +
418 = 2,389
Grand Mean = Grand Sum / Total n
= 2,389 / 32 = 74.66
Note: We can see that the sample means are
fluctuating around the Grand Mean.
Is this fluctuation just sampling error?
Or are the drug-group means statistically different
from each other?
This is what ANOVA helps us answer.
PROCEDURES:
Compute the Correction Factor, C.
C =(Grand Sum)(Grand Sum) / Total n
C = (2389)(2389)/ 32 = 178,354
C is a quantity that ‘centers’ the distribution of
scores on 0. In the defining equation for variance,
the numerator shows (X-X)2
The ‘-X‘ bit did the centering on 0. C does the
same thing.
Let’s compute the sum of X's squared, i.e.,
square each X, and then sum the squared X's
Xij = 189557
Total Sum of Squares = SS
SS = Xij - C = 189557 - 178354 = 11203
Variance = SS /df = 11203/31 = 361.39
This over-all variance is what ANOVA will be
breaking down into parts --including the parts
ascribable to the different drug conditions.
The ANOVA procedure will assess whether the
variation of the drug-group means, about the Grand
Mean is, or is not, statistically significant.
This means that the individual scores within a
drug-group vary about their individual group-mean.
This is just sampling variation.
Then, in addition, the group means vary about the
Grand Mean. This is the effect that we are testing
for,
ANOVA assesses whether the variation of the Group
Means about the Grand Mean is just random, or if it is
statistically significant.
ANOVA accomplishes this by forming a ratio of the
variance due to Group-means fluctuating around the
Grand Mean, to that average variance (sampling
variation) that occurred within the Groups themselves.
The test statistic for ANOVA is called F
(honoring Fisher)
To use a table of F, you need two things: the df
for the numerator of the F-ratio (i.e., # of Groups
minus 1), and the df for the denominator. The df for
the denominator is essentially the n of the data
points that estimate the sampling variation -- the
random or unexplained part.
When doing calculations for ANOVA, we calculate
one or more special variances, called mean squares.
The variance due to the fluctuation of Group
means about the Grand Mean is called mean square -
between or msb Some authors are more
grammatical, and call this component mean square -
among or msa when dealing with more than two
groups.
The average variance within groups, around each
separate group mean, is called mean square - within
or msw It is a measure of the random variation
present.
We then compute an F-ratio:
F = msb / msw
-And then look it up in the F-table.
CALCULATION
Calculate Sum of Squares Between = SSB
SSB = (Tj2/nj) -C
** the Tjs are the column sums **
SSB = (503)(503)/8 + (700)(700)/8 + (768) (768)/8 +
(418)(418)/8 - 178,354 =
= 31,626 + 61,250 + 73,728 + 21,840 - 178,354
SSB = 188,444 - 178,354 = 10,090
There are j=4 Groups, so dfB = 3
msB = SSB / df = 10090 / 3 = 3363.3
** Now we calculate msW **
TSS = SSB + SSW
SSW = TSS - SSB = 11203 - 10090 = 1113
Calculate df
There are 32 data points
We used up 3 df for the Groups
We used up 1 df for the Grand Mean
dfW = 32 - 3 -1 = 28
msW = SSW / dfW = 1113 / 28 = 39.75
We are ready to calculate F3,28
F3,28 = msB / msW = 3363.3 / 39.75 = 84.61
Examine Table A.4 to get the critical F-value. For
alpha = .05, the critical value is at 3 df x 28 df =
2.95
Our F obtained is far greater. We reject H0 and
accept HA We conclude that the drug Groups are not
equivalent. We would now have to search further,
using t-tests.
If we had accepted H0, we would end right here --
no t-tests.
Since we have a significant result, we now want
to know which Groups differ.
Since we have done several of these Independent Groups
t-tests in class, I did these by computer, to save
class time:
1 vs 2 t = -9.18 SED=2.682 p < .001
1 vs 3 t = -12.03 SED=2.754 p < .001
1 vs 4 t = 2.97 SED=3.572 p < .02
2 vs 3 t = -3.19 SED=2.666 p < .01
2 vs 4 t = 10.06 SED=3.504 p < .001
3 vs 4 t = 12.29 SED=3.559 p < .001
If we plot these Group means, we get a good
picture of what is happening:
800| X
700| X
600|
500| X
400| X
300|
200|
100|______________________________
0 2 3 4
Group
Drug groups 2 and 3 are increasing the score
significantly, but drug group 4 score decreases
significantly below that of the Controls. This is not
an unusual finding, with drug studies.