The purpose of this paper is to explain the logic and vocabulary of one-way analysis of variance (ANOVA). The null hypothesis tested by one-way ANOVA is that two or more population means are equal. The question is whether (H0) the population means may equal for all groups and that the observed differences in sample means are due to random sampling variation, or (Ha) the observed differences between sample means are due to actual differences in the population means.
The logic used in ANOVA to compare means of multiple groups is similar to that used with the t-test to compare means of two independent groups. When one-way ANOVA is applied to the special case of two groups, one-way ANOVA gives identical results as the t-test.
Not surprisingly, the assumptions needed for the t-test are also needed for ANOVA. We need to assume: 1)random, independent sampling from the k populations; 2)normal population distributions;
3)equal variances within the k populations.
Assumption 1 is crucial for any inferential statistic. As with the t-test, Assumptions 2 and 3 can be relaxed when large samples are used, and Assumption 3 can be relaxed when the sample sizes are roughly the same for each group even for small samples. (If there are extreme outliers or errors in the data, we need to deal with them first.) As a first step, we will review the t-test for two independent groups, to prepare for an extension to ANOVA.
Review of the t-test for independent groups
Let us start with a small example. Suppose we wish to compare two training programs in terms of performance scores for people who have completed the training course. The table below shows the scores for six randomly selected graduates from each of two training programs. These (artificially) small samples show somewhat lower scores from the first program than from the second program.
But, can these fluctuations be attributed to chance in the sampling process or is this compelling evidence of a real difference in the populations? The t-test for independent groups is designed to address just this question by testing the null hypothesis H0: (1 = (2. We will conduct a standard t-test for two independent groups, but will develop the logic in a way that can be extended easily to more than two groups.
Program 1 Program 210210090108971049411198105101102Mean[pic] 97 [pic]105Variance s12 = 20 s22 = 16The mean of all 12 scores = Grand mean = [pic] 101The first step is to check the data to make sure that the raw data are correctly assembled and that assumptions have not been violated in a way that makes the test inappropriate. In our example, a plot of the data shows that the sample distributions have roughly the same shape, and neither sample has extreme scores or extreme skew. The sample sizes are equal, so equality of population variances is of little concern. Note that in practice you would usually have much larger samples.
We assume that the variance is the same within the two populations (Assumption 3). An unbiased estimate of this common population variance can be calculated separately from each sample. The numerator of the variance formula is the sum of squared deviations around the sample mean, or simply the sum of squares for sample j (abbreviated as SSj). The denominator is the degrees of freedom for the population variance estimate from sample j (abbreviated as dfj).
Unbiased estimateof (j2 =[pic] [Formula 1]
For the first sample, SS1 = (102-97)2 + … + (101-97)2 = 100, and for the second sample, SS2 = 80. This leads to [pic] = 100/5 = 20, and [pic] = 80/5 = 16.
To pool two or more sample estimates of a single population variance, each sample variance is weighted by its degrees of freedom. This is equivalent to adding together the sums of squares for the separate estimates, and dividing by the sum of the degrees of freedom for the separate estimates.
Pooled estimate of [pic] = [pic] [Formula 2]
Thus, for our example
sy2 = (6-1)(20) + (6-1)(16) = 100 + 80 = 180 = 18 (6 + 6 – 2) 5 + 5 10
A t-test can be conducted to assess the statistical significance of the difference between the sample means. The null hypothesis is that the population means are equal (H0: (1 = (2).
df = (n1 + n2 – 2) = (6 + 6 – 2) = 10.
For a two-tailed t-test with alpha set at .01 and df=10, the tabled critical value is 3.169. Because the absolute value of the observed t just exceeds the critical value we can reject the null hypothesis (Ho: (1 = (2) at the .01 level of significance. The exact p=.0085. It is unlikely that the population means are equal, or that the population mean for Group 1 is larger than the population mean for Group 2.
An equivalent test of the null hypothesis can be calculated with the F distribution, because t2 with df = ( is equal to F (df = 1,(). (The Greek letter ( “nu”is often used to represent df for the t-test.) For our example, t2 = (-3.266)2 = 10.67. From
the F table, F(1,10; .01) = 10.04, so we find that the null hypothesis can just be rejected at the .01 level of significance (p = .0085). This test result is identical to the result of the t test.
ANOVA as a comparison of two estimates of the population variance
In this section we examine a second approach to testing two means for equality. The logic of this approach extends directly to one-way analysis of variance with k groups. We can use our data to calculate two independent estimates of the population variance: one is the pooled variance of scores within groups, and the other is based on the observed variance between group means.
These two estimates are expected to be equal if the population means are equal for all k groups (H0: (1 = (2 = …= ( k), but the estimates are expected to differ if the population means are not all the same.
Within-groups estimate. Our single best estimate of the population variance is the pooled within groups variance, sy2 from Formula 2. In our example sy2 = 18, with df = 10. In ANOVA terminology, the numerator of Formula 2 is called the Sum of Squares Within Groups, or SSWG, and the denominator is called the degrees of freedom Within Groups, or dfWG. The estimate of the population variance from Formula 2, SSWG/dfWG, is called the Mean Square Within Groups, or MSWG. Formula 3 is an equivalent way to express and compute MSWG.
Within-groups estimate of (y2 = [pic] [Formula 3]
Between-groups estimate. If the null hypothesis ((1 = (2) is true and the assumptions are valid (random, independent sampling from normally distributed populations with equal variances), then a second independent estimate of the population variance can be calculated.
As is stated by the Central Limit Theorem, if independent samples of size n are drawn from some population with variance = (y2, then the variance of all possible such sample means [pic] is (y2/n. We can use our observed sample means to calculate an unbiased estimate of the variance for the distribution of all possible sample means (for samples of size n). Our estimate of the variance of means is not very stable because it is based on only two scores, [pic]=97 and [pic]=105, but nonetheless it is an unbiased estimate of [pic]. With our data, [pic] and df = 1, as calculated with Formula 4.
= (97-101)2 + (105-101)2 = (-4)2 + (4)2 = 16 + 16 = 32.(2 – 1) 1Because [pic] = (y2/n, it follows that (y2 = n[pic]. Now we can estimate the variance of the population based on the observed variance of 32 for the sample means. With our data, where n=6 for each sample, we find sy2 =(n)([pic]) = (6)(32) = 192. This tells us that if we draw samples of size nj = 6 from a population where (y2 = 192, the expected variance of sample means is [pic] = (y2/n = 192/6 = 32. Thus, if the groups in the population have equal means and variances, we can estimate this common population variance to be 192, because that would account for the observed variance of 32 between our sample means.
Calculation of this second estimate of the population variance using ANOVA notation is shown in Formula 5. The MSBG is our best estimate of the population variance based only on knowledge of the variance among the sample means. Formula 5 allows for unequal sample sizes.
Between-groups estimate of (y2 = [pic] [Formula 5]
Comparing the two estimates of population variance. The estimate of the population variance based on the variability between sample means (MSBG = 192) is considerably larger than the estimate based on variability within samples (MSWG = 18). We should like to know how likely it is that two estimates of the same population variance would differ so widely if all of our assumptions are valid and ((1 = (2). The F ratio is designed to test this question. (Ho: (12 = (22 )
[pic] [Formula 6]
F(1,10) = 192 = 10.67 (p = .0085) 18
The degrees of freedom for the two estimates of variance in Formula 6 are dfBG = k-1 = 2-1 = 1, and dfWG = (n1 + n2 – k) = (6 + 6 – 2) = 10. Notice that these are exactly the same F ratio and degrees of freedom that we calculated earlier when we converted the t-test to an F-test.
If the null hypothesis and assumptions were true, such that independent random samples were drawn from two normally distributed populations with equal means and equal variances, then it would be very surprising indeed (p