We continue our example on diameters of inhibition zones around disks with oils (see Example 1). We will use a fragment taken from Example 1. Now we will compare inhibition zones for tea tree oil against three bacterial species – E. coli, E. faecalis and S. aureus:
Example for one-way ANOVA
The first column contains numerical identifier of a strain (for reference purposes), the second column contains grouping variable (“Species”), and the third – diameter of inhibition zone in mm, which is independent variable. So, we have one factor and three compared groups.
We can now formulate research question: Are there differences in activity of tea tree oil against different bacterial species?
Note, that in SPSS for ANOVA grouping variable should be coded with numerical values (1, 2, 3):
It is also useful to write label for numerical values (to type in the Label column of the Variable View “1 – E.coli, 2 – E.faecalis, 3 – S.aureus”); also the type of grouping variable should be changed to Numeric.
ANOVA should not be started without checking normality of distribution of variable values in each group. Therefore, first we perform descriptive statistics analysis with assessment of distribution. It is not always necessary to split a file into groups. Another way to obtain separated results for groups is to specify group variable in the Explore dialog box which appears after clicking Analyze menu, pointing Descriptive Statistics and selecting Explore… Variable “Species” should be specified in the Factor List: list box:
Results of statistical tests of normality prove that distribution of values in all groups is normal: deviation from normal distribution is not statistically significant. Therefore, we can use one-way ANOVA.
Normality tests prove that our data are normally distributed – differences from the normal distribution are not significant. Therefore, we can use parametric ANOVA tests:
SPSS gives an opportunity to calculate plenty of post-hoc tests. Among them the most popular are Tukey and Games-Howell tests.
The Tukey test is powerful and widely accepted but it also has some limitations. It assumes that both variances in population and group sizes are equal. If this conditions are not true, we should use Gabriel's procedure, or if the sizes are very different – Hochberg's GT2.
The Games-Howell test does not require any of these conditions and, therefore, represent a good alternative to Tukey test.
Apart from post-hoc tests, the Homogeneity of Variance Test, Brown-Forsythe and Welch tests should be calculated.
Assumption of ANOVA is homogeneity of variance and because of this, such test should be performed. However, if this assumption is not satisfied, Brown-Forsythe and Welch options will display alternative versions of the F statistic and results will still be appropriate for use.
To specify one-way ANOVA:
1) Click the Analyze menu, point to Compare Means, and select One-Way ANOVA… :
The One-Way ANOVA dialog box opens:
2) Select the dependent variable (“Tea Tree Diffusion”) and factor variable (“1 – E.coli, 2 – E.faec…”) and move them to Dependent List and to Factor by clicking upper and lower transfer arrow buttons , respectively.
3) Click the Post Hoc… button; One-Way ANOVA: Post Hoc Multiple Comparisons dialog box opens:
4) Select Tukey test in the Equal Variances Assumed section and Games-Howell test in the Equal Variances Not Assumed section.
5) Click the Continue button. This returns you to the One-Way ANOVA dialog box.
6) Click the Options… button; the One-Way ANOVA: Options dialog box opens:
7) Select Homogeneity of Variance Test, Brown-Forsythe and Welch.
8) Click the Continue button. This returns you to One-Way ANOVA dialog box.
9) Click the OK button. An Output Viewer window opens and displays the statistics for one-way ANOVA.
The Output Viewer contains five tables with statistical results: the Test of Homogeneity of Variances table, the ANOVA table with results for ANOVA itself, the Robust Tests of Equality of Means table, the Multiple Comparisons table with results of post hoc tests and the table with means for homogeneous subsets of groups:
From the first table we see that variance are homogenous within groups, this was assessed by the Levene test. Significance level is more than 0.05 which indicates that the null hypothesis about homogeneity of variances is accepted. Therefore, we can readily use results of all ANOVA tests as the main assumption of ANOVA is satisfied.
The ANOVA table shows that differences in means between groups are statistically significant (p = 0.000): we can reject the null hypothesis about equality of means between groups. Welch and Brown-Forsythe tests indicate the same in the next table (p = 0.000 for both tests).
From the table with post hoc results we can use Tukey test because homogeneity of variances was proved. This test shows the results of paired comparison and we can see that all groups differe significantly from each other: when the 1st and the 2nd groups are compared p = 0.004, when the 1st and the 3rd groups are compared p = 0.000, when the 2nd and the 3rd groups are compared also p = 0.000 (p<0.001).
The last table contains means for homogenous groups: 11.2 mm, 15.5 mm and 23.6 mm for the 1st, 2nd and 3rd groups, respectively.
Formal reporting results of the one-way ANOVA includes mentioning F statistics, degrees of freedom and p-value. So, in our example results is reported as:
There are significant differences between inhibition zones against E. coli, E. faecalis and S. aureus, F(2,33) = 50.96, p < 0.001.