To obtain the ANOVA, we select
ANALYZE → GENERAL LINEAR MODEL → UNIVARIATE We move ac over to the Dependent Variable box and teach to the Fixed Factor(s) box.
We can see down the column ac are the achievement scores and down the column teach is the assigned teacher (1 through 4).
To get an initial feel for these data, we can get some descriptives via EXPLORE by levels of our teach factor:
teach Statistic Std.Error
ac 1.00 Mean
Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 5% Trimmed Mean
71.0000 1.80739
.845 1.741 95% Confidence
Interval for MeanLower Bound 66.3540 Upper Bound 75.6460 71.0556 71.5000 19.600 4.42719 65.00 76.00 11.00 8.75 –.290 –1.786
2.00 Mean
Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 5% Trimmed Mean 95% Confidence Interval
for Mean Lower Bound
Upper Bound
72.5000 1.60728
.845 1.741 68.3684 76.6316 72.5000 72.5000 15.500 3.93700 68.00 77.00 9.00 7.50 .000 –2.758
Mean
Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 5% Trimmed Mean 95% Confidence Interval
for Mean Lower Bound
Upper Bound
3.00 80.0000 2.42212
.845 1.741 73.7737 86.2263 80.0556 80.5000 35.200 5.93296 73.00 86.00 13.00 10.75 –.095 –2.957
4.00 Mean
Median Variance Std. Deviation Minimum Maximum Range Interquartile Range Skewness Kurtosis 5% Trimmed Mean 95% Confidence Interval
for Mean Lower Bound
Upper Bound 92.6667 90.3045 95.0289 92.7407 93.5000 5.067 2.25093 89.00 95.00 6.00 3.75 –.959 –.130
.91894
.845 1.741
The descriptives above give us a sense of each distribution of ac for the different levels of teach. See Chapter 3 for a description of these statistics.
We will select a few features for the ANOVA. We click on Plots, move teach over under Horizontal Axis, and then click on Add:
We will also select Post Hoc so that we may “snoop the data”
afterward to learn where there may be mean differences given a rejection of the overall null hypothesis for the ANOVA.
Next, we will select some Options:
We move teach over to Display Means for. We also select Estimates of effect size and Homogeneity tests. The homo- geneity tests option will provide us with Levene’s test that will evaluate whether the assumption of equal population variances is tenable. Click Continue.
We move teach over to Post Hoc Tests for and select Tukey under Equal Variances Assumed. The tests under Equal Variances Assumed are performed under the assumption that between populations on the independent variable, vari- ances within distributions are assumed to be the same (we will select a test to evaluate this assumption in a moment).
Click Continue.
The following is the syntax that will reproduce the above window commands (should you choose to use it instead of the GUI):
We obtain the following output:
Univariate Analysis of Variance
Between-Subjects Factors teach
N
1.00 6
6 6 6 2.00 3.00 4.00
Tests of Between-Subjects Effects Dependent Variable: ac
a. R Squared = .824 (Adjusted R Squared = .798) Source Type III Sum
of Squares df Mean
Square F Sig. Partial Eta Squared Corrected
Model Intercept teach Error Total Corrected Total
1764.125a 149942.042 1764.125 376.833 152083.000 2140.958
3 1 3 20 24 23
588.042 149942.042 588.042 18.842
31.210 7958.003 31.210
.000 .824
.997 .824 .000 .000
Robust Tests of Equality of Means ac
Welch Statistica
57.318
df1 df2 Sig.
3 10.419 .000 a. Asymptotically F distributed.
SPSS first confirms for us that there are N = 6 observations in each level of the teach factor.
Since we requested Homogeneity Tests, SPSS generates for us Levene’s Test of Equality of Error Variances.
This test evaluates the null hypothesis that variances in each population, as rep- resented by levels of the teach factor, are equal. That is, the null evaluated is the following: H0 12
22 32
42
: . If the null hypothesis is rejected, it suggests that somewhere among the variances, there is an inequality. The p‐value for the test is equal to 0.001, which is statistically significant, suggesting that some- where among the variances in the population, there is an inequality. However, for the purpose of demonstration, and since ANOVA is rather robust against a violation of this assumption (especially for equal N per group), we will push forth with the ANOVA and compare it with an ANOVA performed under the assumption of an inequality of variances, to see if there is a difference in the overall decision on the null hypothesis (we will conduct the Welch procedure).
Levene’s Test of Equality of Error Variancesa Dependent Variable: ac
a. Design: Intercept + teach Tests the null hypothesis that the error variance of the dependent variable is equal across groups.
F df1 df2 Sig.
7.671 3 20 .001
A one‐way fixed effects between‐
subjects analysis of variance (ANOVA) was conducted to evaluate the null hypothesis that achievement population means were equal across four experimenter‐selected teachers. A statistically significant difference was found (F = 31.210 on 3 and 20 df, p < 0.001), with an estimated effect size of 0.82 (Eta squared), sug- gesting that approximately 82% of the variance in achievement can be explained or accounted for by teacher differences featured in the experiment.
Because the assumption of equality of variances was suspect (Levene’s test indicated a violation), a more robust F‐test was also performed (Welch), for which the null hypothesis was also easily rejected (p < 0.001).
UNIANOVA ac BY teach /METHOD=SSTYPE(3) /INTERCEPT=INCLUDE /POSTHOC=teach(TUKEY) /PLOT=PROFILE(teach) /EMMEANS=TABLES(teach) /PRINT=ETASQ HOMOGENEITY /CRITERIA=ALPHẶ05) /DESIGN=teach.
Above is the ANOVA generated by SPSS. We interpret the essential elements of what is known as the ANOVA Summary Table:
● The first two rows, those of Corrected Model and Intercept, are not important for interpretation purposes, so we ignore those.
● We see that teach has a Sums of Squares equal to 1764.125. Loosely, this number represents the amount of variation due to having different teach groups. Ideally, we would like this number to be rather large, because it would suggest there are mean differences between teachers.
● The Error sum of squares is equal to 376.833. This number represents the amount of variation not due to teach, and hence “left over” after consideration of teach. It represents variation within each group of the teach factor that is not due to the grouping factor. Hence, it is unwanted variation. The bigger this number is, the more it means that within groups across all teachers, there is quite a bit of unexplained variability. That is, we want SS teach to be rather large and SS Error to be much smaller. That would be ideal under the condition of teach differences.
● The Total SS is computed to include the intercept term, and hence it is not of interest to us. We are more interested in the Corrected Total number of 2140.958. How was this number calculated? It was computed by:
SS Corrected Total = SS teach + SS Error
The above is actually one of the fundamental identities of the ANOVA, in that each ANOVA parti‑
tions SS total into two parts, that due to “between‐group” differences (as represented by teach, for our data) and “within‐group” differences, as represented by SS error. As mentioned, as a researcher, we are hoping that SS teach is much larger than SS error. Such would suggest, at least noninferen‑
tially so far, that there are mean differences on teach.
● The next column contains df or “degrees of freedom.” We divide each SS by its corresponding degrees of freedom to obtain what are known as Mean Squares. Mean Squares are a kind of “aver‑
age SS,” but unlike a normal arithmetic average where we divide the sum by N, when computing Mean Squares, we will divide SS by df. The df for teach are equal to the number of levels on the factor minus 1. If we designate the number of levels as J, then the degrees of freedom are equal to J − 1. For our data, this is equal to 4 – 1 = 3. The df for Error are computed as the total number of observations minus the number of groups (or levels). That is, they are computed as N – J. For our data, this is equal to 24 – 4 = 20.
● The Mean Squares for teach are computed as 1764.125/3 = 588.042.
● The Mean Squares for Error are computed as 376.833/20 = 18.842.
● Because the assumption of equal variances was suspect, we also conducted a Welch test (Robust Tests of Equality of Means), which can be used when the assumption of homogeneity of vari‑
ances is not met. (You can get the Welch via ANALYZE → Compare Means → One‐Way ANOVA and then select under Options). As we can see, the null hypothesis was easily rejected for this test as well.