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This might be a stupid question. I know how to calculate them but mine not sure what they are telling me about my data set. What does it mean if $\sum $$(SSB)^2$ $\ge \sum (SSW)^2$, or vise versa, or does $SSB$ and $SSW$ only have meaning when you sum them to find Total $\sum $$(SST)^2$.

pink floyd
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toom
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  • I am not sure what to make of the notation $\sum(SSB)^2$, etc. SSB and SSW are standard notation and I use them in my Answer below. – BruceET Apr 26 '15 at 23:22

1 Answers1

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I assume you are talking about a one-way (or one-factor) ANOVA. Below is an example with fake data in which there are 30 observations: r = 10 replications within each of g = 3 treatment groups.

 A:   9.0   4.4   1.6   4.4   0.4  23.8   1.5  22.6   6.0  18.3 
 B:  24.5  26.9  21.4  17.5  28.6  24.8  21.9  23.7   4.2   8.4
 C:  19.0  33.4  23.9  21.2  24.7  23.6  30.0  21.2  25.2  26.2

The null hypothesis in such an ANOVA is that the population means of all three treatment groups are equal, that is $H_0: \mu_A = \mu_B = \mu_C$, and the alternative is that there are some differences among these three population means.

Just looking at the data we see that the observations in Group A are generally smaller than the others. So we suspect that the null hypothesis is not true. The sample means and standard deviations of the three groups are as follows:

 Group   N    Mean  StDev
 A      10   9.200  8.984
 B      10  20.190  7.988
 C      10  24.840  4.281

We see that the sample mean for Group A is indeed much smaller than the other two sample means. Also, the mean for Group C is a little bigger than the mean for Group B. The question is whether these differences in sample means based on just 10 observations in each group are large enough that they are unlikely to have occurred by chance alone.

The standard deviations are a key to judging this. In this ANOVA we assume that the three populations have the same variance. The sample variances (squares of sample standard deviations) are somewhat different, but no by enough to contradict our assumption. Averaging the three sample variances we have 54.3, one good estimate of the common variance of the three populations. This number is shown in the ANOVA table below as SS(Within), which is your SSW. [The square root of 54.3 is the 'pooled' SD 7.37, which lies among the three group SDs, but is not their average.]

 Source   DF      SS     MS      F      P
 Between   2  1290.0  645.0  11.88  0.000
 Within   27  1465.5   54.3
 Total    29  2755.6

Provided that the null hypothesis is true, there is another good estimate of the common population variance, which is based on the three sample means. If $H_0$ is not true this second estimate tends to be too big. For our data that estimate is shown in the table above as SS(Between) = 645.0. This is your SSB. This estimate is enormously larger than SSW. Because these two estimates of the common population variance agree so poorly, we conclude that the null hypothesis is not true.

The ratio SSB/SSW = F = 11.88 is used as the criterion whether to reject. If $H_0$ is true the F-statistic should be around 1 (the ratio of two good estimates of the same thing). For the dimensions of this problem (10 observations on each of 3 groups), we would reject the null hypothesis if F > 3.36. And 11.88 is certainly greater than 3.36. (The DF column of the ANOVA table tells you how to look up the 'critical value', here 3.36, that separates low non-rejection values of F from high rejection values.)

Thus to answer your question, here is the intuitive interpretation of SSB and SSW: No matter whether $H_0$ is true, SSW is a good estimate of the common population variance $\sigma^2$. This estimate is based on the sample variances (or SDs shown above). A second estimate of $\sigma^2$ is SSB is based on the sample means. It is a good estimate if $H_0$ is true and too high if not.

The Analysis of Variance (ANOVA) procedure compares these two variance estimates. If they agree pretty well, we do not reject $H_0$. If they don't agree, so that F = SSB/SSW is too large, then we do reject $H_0$.

Notes: (1) SSB is based on the degree to which the g group means differ. (2) It is conceivable, but rare, to have SSB much smaller than SSW. In this case statisticians begin to worry whether there is something wrong with the way the data were collected, or whether the assumptions of the ANOVA model are correct: normally distributed populations with the same variance $\sigma^2.$ (A bit smaller is OK, they are only estimates after all.) (3) If you reject $H_0$, then you have to try to figure out what pattern of differences there is among the population means. For our data, it is pretty clear that $\mu_A$ is smaller than the other two population means. There is no strong evidence that $\mu_B$ differs from $\mu_C.$ Procedures to check for such patterns, after rejecting $H_0$, are called multiple comparison procedures. (4) The one-factor ANOVA is a direct generalization of the pooled 2-sample T test. In that test the difference between two means in the numerator is compared with the pooled sample variance in the denominator; the square of the T-statistic is an F-statistic with numerator DF = 1. (5) For normal data sample means and variances are independent random variables. That turns out to be important in doing the computations of critical values given in F tables. B.T.

BruceET
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