Conducting ANOVA in SPSS
We want to study the effectiveness of different treatments on anxiety. We collect a sample of 75 subjects in the following categories:
- No treatment (\(n_1\) = 27).
- Biofeedback (\(n_2\) = 24).
- Cognitive-behavioral Treatment (\(n_3\) = 24).
The dependent variable is anxiety levels. In this analysis, we will include both planned contrasts and post hoc comparisons.
Steps to doing an ANOVA
The steps to doing an ANOVA in SPSS are as follows:
1. Assert the null hypothesis: all means are equal.
- \(H_0\) : \(\mu_1 = \mu_2 = \mu_3\)
- \(H_A\) : At least two \(\mu_i\)’s are different
If even two of the means are significantly different, we reject the null hypothesis.
2. Visualize the data.
We can visualize the data by box plots to check for normality and equality of variances across groups.
We will do this in SPSS by going to
Then we select
Simple and ’Summaries for groups of cases
We will select
anxiety as our
Treatment as the
Category Axis, then click OK.
An output window with the boxplot will open.
3. Run the ANOVA.
The dialogue box looks like this:
Select the dependent variable and factor:
Click on Post Hoc.
Select Tukey, as shown. Then click Continue.
Click on Options.
Select Descriptive, as shown. Then click Continue.
Add any planned comparisons.
Contrasts, then assign weights depending on the contrasts you are conducting. You can add multiple contrasts at a time by clicking
Next. When finished click
Then click OK.
4. View Summary tables
This table gives the descriptive statistics for each treatment group, and for the combination of the three groups.
- N is the number of subjects in each group. We see that group 1 (no treatment) has 27 subjects, group 2 (biofeedback) has 24 subjects, and group 3 (CBT) has 24 subjects.
- Mean is the mean anxiety value of each group. Group 1 has a mean of 27.15, group 2 has a mean of 22.88 and group 3 has a mean of 16.88.
- Std. Deviation is the standard deviation of anxiety for each group. Group 1 has a std. deviation of 9.334, group 2 is 6.829 and group 3 is 9.786.
- Std. Error gives the standard error of the mean for each group. For group 1 it is 1.796, for group 2 it is 1.394, and for group 3 it’s 1.998.
- Lower bound gives the lower end point of the 95% Confidence Interval of the mean and upper bound give the upper end point of the 95% Confidence Interval of the mean. The interval around group 1’s mean ranges from 23.46 to 30.84. The interval around group 2’s mean ranges from 19.99 to 25.76. The interval around group 3’s mean ranges from 12.74 to 21.01.
- Group 1’s values range from a minimum of 6 to a maximum of 49. Group 2’s values range from a minimum of 11 to a maximum of 37. Group 3’s values range from a minimum of 0 to a maximum of 34.
This table gives the results of the ANOVA. The mean square values are equal to the sum of square values divided by degree of freedom. The F statistic is the ratio of the two mean square values. Our F value, from the table, is 8.746. We will compare this against an F distribution with \(df_1\) = 2, and \(df_2\) = 72 to accept or reject the null hypothesis.
This table gives the result of the planned contrast comparing group 1 to group 2. It has two rows, one where equal variances are assumed for the two groups and one where they are not.
- Value of Contrast gives the mean difference
- Std. Error gives the standard error of the mean difference
- t is the t-statistic
- df is the degrees of freedom of the test
- Sig. (2-tailed) gives the p-value
The Post Hoc Tests table gives the results of the Post Hoc test that we selected (Tukey). It compares each group to every other group and gives the results for each.
- Mean difference is the difference in means of the two groups
- Standard error is the standard error between the two groups
- Sig gives the p-value of the test
- Lower bound and Upper Bound give the end points for the 95% Confidence Interval between the two groups.
5. Compare to F distribution with \(df_B\) and \(df_W\).
We find the \(F\) statistic to be 8.746. Compare to an \(F\) distribution with \(df_1\) = 2 and \(df_2\) = 72. The cut-off for significance given these \(df\) is 3.124, so we have a significant result.
We can reject the null hypothesis that all \(\mu_i\)’s are equal.