A hospital wants to know how a homeopathic medicine for depression performs in comparison to alternatives. They adminstered 4 treatments to 100 patients for 2 weeks and then measured their depression levels. The data, part of which are shown above, are in depression.sav.

## Data Inspection - Split Histogram

Before running any statistical tests, let's first just take a look at our data. In this case, a split histogram basically tells the whole story in a single chart. We don't see many SPSS users run such charts but you'll see in a minute how incredibly useful it is. The screenshots below show how to create it.

In step below, you can add a nice title to your chart. We settled for “Distribution BDI per Medicine”.

## Syntax for Split Histogram

Clicking

results in the syntax below. Running it generates our chart.***Run histogram of BDI scores for the four treatments separately.**

GRAPH

/HISTOGRAM=bdi

/PANEL ROWVAR=medicine ROWOP=CROSS

/TITLE='Distribution BDI per Medicine'.

## Result

- All
**distributions look plausible**. We don't see very low or high BDI scores that should be set as missing values and the BDI scores even look reasonably normally distributed. - The medicine
**“None” results in the highest BDI scores**, indicating the worst depressive symptoms. “Pharmaceutical” results in the lowest levels of depressive illness and the other two treatments are in between. - The four histograms are roughly equally wide, suggesting BDI scores have
**roughly equal variances**over our four medicines.

## Means Table

We'll now take a more precise look at our data by running a means table with the syntax below.

***Run basic means table.**

means bdi by medicine/cells count min max mean variance.

## Result

Unsurprisingly, our table mostly confirms what we already saw in our histogram. Note (under “N”) that each medicine has 25 observations so these two variables don't contain any missing values.

So can we conclude that “Pharmaceutical” performs best and “None” performs worst? Well, for our sample we can. For our population (all people suffering from depression) we can't.

The basic problem is that **samples differ from the populations from which they are drawn**. If our four medicines perform equally well in our population, then we may still see some differences between our sample means. However, *large* sample differences are unlikely if all medicines perform equally in our population. This basic reasoning is explained further in ANOVA - What Is It?.

The question we'll now answer is: **are the sample means different enough** to reject the hypothesis that the mean BDI scores in our population are equal?

## ANOVA Basics

We'll try to demonstrate that some medicines perform better than others by rejecting the **null hypothesis** that the mean BDI scores for our four medicines are all equal in our population. In short, our ANOVA tests whether all 4 means are equal. If they aren't then we'd like to know exactly which means are unequal with post hoc (Latin for “after that”) tests.

Our ANOVA will run fine in SPSS but in order to have confidence in its results, we need to satisfy some assumptions.

## ANOVA - Main Assumptions

**Independent observations**often holds if each case (row of cells in SPSS) represents a unique person or other statistical unit. That is, we usually don't want more than one row of data for one person, which holds for our data;**Normally distributed variables**in the population seems reasonable if we look at the histograms we inspected earlier. Besideds, violation of the normality assumption is no real issue for larger sample sizes due to the central limit theorem.**Homogeneity**means that the population variances of BDI in each medicine group are all equal, reflected in roughly equal sample variances. Again, our split histogram suggests this is the case but we'll try and confirm this by including**Levene's test**when running our ANOVA.

## Running our ANOVA in SPSS

There's many ways to run the exact same ANOVA in SPSS. Today, we'll go for partial eta squared as an estimate for the effect size of our model.

because it'll provide us withWe'll briefly jump into

and before pasting our syntax.The post hoc test we'll run is Tukey’s HSD (Honestly Significant Difference), denoted as “Tukey”. We'll explain how it works when we'll discuss the output.

“Estimates of effect size” refers to partial eta squared. “Homogeneity tests” includes Levene’s test for equal variances in our output.

## Post Hoc ANOVA Syntax

Following the previous screenshots results in the syntax below. We'll run it and explain the output.

***ANOVA syntax with Post Hoc (Tukey) test, Homoscedasticity (Levene's test) and effect size (partial eta squared).**

UNIANOVA bdi BY medicine

/METHOD=SSTYPE(3)

/INTERCEPT=INCLUDE

/POSTHOC=medicine(TUKEY)

/PRINT=ETASQ HOMOGENEITY

/CRITERIA=ALPHA(.05)

/DESIGN=medicine.

## SPSS ANOVA Output - Levene’s Test

Levene’s Test checks whether the *population* variances of BDI for the four medicine groups are all equal, which is a requirement for ANOVA. “Sig.” = 0.949 so there's a 94.9% probability of finding the slightly different variances that we see in our sample. This sample outcome is very likely under the null hypothesis of homoscedasticity; we satisfy this assumption for our ANOVA.

## SPSS ANOVA Output - Between Subjects Effects

If our population means are really equal, there's a 0% chance of finding the sample differences we observed. **We reject the null hypothesis of equal population means**.

The different medicines administered account for some 39% of the variance in the BDI scores. This is the **effect size** as indicated by partial eta squared.

Partial Eta Squared is the Sums of Squares for medicine divided by the corrected total sums of squares (2780 / 7071 = 0.39).

Sums of Squares Error represents the variance in BDI scores not accounted for by medicine. Note that + = .

## SPSS ANOVA Output - Multiple Comparisons

So far, we only concluded that all four means being equal is very unlikely. So exactly **which mean differs from which mean?** Well, the histograms and means tables we ran before our ANOVA point us in the right direction but we'll try and back that up with a more formal test: Tukey’s HSD as shown in the multiple comparisons table.

Right, now comparing 4 means results in (4 - 1) x 4 x 0.5 = 6 distinct comparisons, each of which is listed twice in this table. There's three ways for telling which means are likely to be different:

Statistically significant mean differences are **flagged** with an asterisk (*). For instance, the very first line tells us that “None” has a mean BDI score of 6.7 points higher than the placebo -which is quite a lot actually since BDI scores can range from 0 through 63.

As a rule of thumb, **“Sig.” < 0.05 indicates a statistically significant difference** between two means.

A confidence interval *not* including zero means that a zero difference between these means in the population is unlikely.

Obviously, , and result in the same conclusions.

So that's it for now. I hope this tutorial helps you to run ANOVA with post hoc tests confidently. If you have any suggestions, please let me know by leaving a comment below.

## This tutorial has 25 comments

## By abdo nazim on March 4th, 2018

how to compare multible means using superscripts after one way ANOVA and post hoc LSD?

## By Veronica Ocampo on March 25th, 2017

Thank you! A very down-to-earth tutorial. The screenshots are invaluable plus the ease in interpretation of all the numbers. A great help in completing my homework!

## By Andrea on February 10th, 2017

Hello, I wanted to know if ANOVA works out when it comes to prove independence between variables. Particularly, I am studying how several conditions regarding an industrial process (such as temperatures and flows) affect one of the chemical components of the product. I know that the dependent variable would be the amount of the chemical component and that the process conditions would be the fixed factors. But how can I interpret the interaction between this variables (and the combination of them) with the chemical component with the ANOVA results?

Thank you in advance!

## By Ruben Geert van den Berg on October 28th, 2016

Hi Ernest, thanks for the compliment and a great question!

If the sample sizes are (roughly) equal, don't bother about homogeneity being violated because the F-test is pretty robust in this case. However, if you have sharply unequal sample sizes, run a Kruskal-Wallis test besides the ANOVA since it doesn't require equal variances.

For larger samples, both tests will often lead to the same conclusion so the KW-test kinda backs up your ANOVA conclusion. However, if they lead to different conclusions, the KW-test conclusion should probably prevail.

Hope that helps!

## By Ernest on October 28th, 2016

Beautiful explanation. So what do i do if my Levene's Test gives a sig of 0.00? Does it mean there are unequal variances, and what do i do?