## Null Hypothesis Chi-Square Independence Test

A chi-square independence test evaluates if two categorical variables are associated in some population. We'll therefore try to refute the null hypothesis that
two categorical variables are (perfectly) independent in some population.
If this is true and we draw a sample from this population, then we may see *some* association between these variables in our sample. This is because **samples tend to differ** somewhat from the populations from which they're drawn.

However, a ** strong association between variables is unlikely** to occur in a sample if the variables are independent in the entire population. If we do observe this anyway, we'll conclude that the variables probably aren't independent in our population after all. That is, we'll reject the null hypothesis of independence.

## Example

A sample of 183 students evaluated some course. Apart from their evaluations, we also have their genders and study majors. The data are in course_evaluation.sav, part of which is shown below.

We'd now like to know: **is study major associated with gender?** And -if so- how? Since study major and gender are nominal variables, we'll run a chi-square test to find out.

## Assumptions Chi-Square Independence Test

Conclusions from a chi-square independence test can be trusted if two assumptions are met:

**independent observations**. This usually -not always- holds if each case in SPSS holds a unique person or other statistical unit. Since this is that case for our data, we'll assume this has been met.- For a 2 by 2 table, all
**expected frequencies > 5**.If you've no idea what that means, you may consult Chi-Square Test - What Is It?. For a larger table, no more than 20% of all cells may have an expected frequency < 5 and all expected frequencies > 1.

**SPSS will test this assumption**for us when we'll run our test. We'll get to it later.

## Chi-Square Independence Test in SPSS

In SPSS, the chi-square independence test is part of the CROSSTABS procedure which we can run as shown below.

In the main dialog, we'll enter one variable into the sex has only 2 categories (male or female), using it as our column variable results in a table that's rather narrow and high. **It will fit more easily** into our final report than a wider table resulting from using major as our column variable. Anyway, both options yield identical test results.

Under we'll just select . Clicking results in the syntax below.

## SPSS Chi-Square Independence Test Syntax

***Crosstabs with Chi-Square test as pasted from menu.**

CROSSTABS

/TABLES=major BY sex

/FORMAT=AVALUE TABLES

/STATISTICS=CHISQ

/CELLS=COUNT

/COUNT ROUND CELL.

You can use this syntax if you like but I personally prefer a shorter version shown below. I simply type it into the Syntax Editor window, which for me is much faster than clicking through the menu. Both versions yield **identical results**.

***Crosstabs with Chi-Square test - short version.**

crosstabs major by sex

/statistics chisq.

## Output Chi-Square Independence Test

First off, we take a quick look at the **Case Processing Summary** to see if any cases have been excluded due to missing values. That's not the case here. With other data, if many cases are excluded, we'd like to know why and if it makes sense.

## Contingency Table

Next, we inspect our contingency table. Note that its **marginal frequencies** -the frequencies reported in the *margins* of our table- show the frequency distributions of either variable separately.

Both distributions look plausible and since there's no “no answer” categories, there's no need to specify any user missing values.

## Significance Test

First off, **our data meet the assumption** of all expected frequencies > 5 that we mentioned earlier. Since this holds, we can rely on our significance test for which we use Pearson Chi-Square.

Right, we usually say that the association between two variables is statistically significant if **Asymptotic Significance (2-sided) < 0.05** which is clearly the case here.

Significance is often referred to as “**p**”, short for probability; it is the probability of observing our sample outcome if our variables are independent in the entire population. This probability is 0.000 in our case. **Conclusion: we reject the null hypothesis** that our variables are independent in the entire population.

## Understanding the Association Between Variables

We conclude that our variables are associated but what does this association look like? Well, one way to find out is inspecting either column or **row percentages**. I'll compute them by adding a line to my syntax as shown below.

***Show only variable/value labels in output.**

set tvars labels tnumbers labels.

***Crosstabs with frequencies and row percentages.**

crosstabs major by sex

/cells count row

/statistics chisq.

## Adjusting Our Table

Since I'm not too happy with the format of my newly run table, I'll right-click it and select

We select

and then drag and drop right underneath “What's your gender?”. We'll close the pivot table editor.## Result

Roughly half of our sample if female. Within psychology, however, a whopping 87% is female. That is, females are highly overrepresented among psychology students. Like so, **study major “says something” about gender**: if I know somebody studies psychology, I know she's probably female.

The opposite pattern holds for economy students: some 80% of them are male. In short, our **row percenages describe the association** we established with our chi-square test.

We could quantify the **strength of the association** by adding Cramér’s V to our test but we'll leave that for another day.

## Reporting a Chi-Square Independence Test

We report the significance test with something like
“an association between gender and study major was observed, χ^{2}(4) = 54.50, p = 0.000.
Further, I suggest including our final contingency table (with frequencies and row percentages) in the report as well as it gives a lot of insight into the nature of the association.

So that's about it for now. Thanks for reading!

## This tutorial has 61 comments

## By Ruben Geert van den Berg on January 18th, 2016

Dear Elaine,

The Fisher exact test can only be reported for 2 * 2 tables which is why you don't see it for a 6 * 2 table.

Keep in mind that "all expected frequencies < 5" is nothing more than a rough rule of thumb; if some cells have an expected frequency of like 4.8 or something, I wouldn't bother too much.

However, if some cells have a

verysmall frequency, you may consider merging some of the categories of the categorical variable. This will result in a smaller crosstab and hence reduces the chance of finding many (nearly) empty cells.## By Elaine Tennant on January 18th, 2016

Thank you for this helpful tutorial. I am analysing how speciality affects whether a certain test is done or not. So I have a categorical dependent variable and a categorical nominal independent variable with six levels. When I've done chi squared tests in the past and the independent variable has two levels (y/n), I have used the fishers exact test as the statistic if any cells have the expected count < 5. My output for this particular test shows that six cells have expected count less than five but a result for fishers is not reported. Can I use the test statistic for the pearson chi square? Many thanks for your help.

## By Ruben Geert van den Berg on December 30th, 2015

Short answer: yes.

Long answer: it's the highest

statisticalsignificance. It only means there's a (roughly) 0 probability that the association between variables iszeroin the population (variables perfectly independent).However, it does not mean that the variables are strongly associated; because the p value depends on the sample size, a minor association with a huge sample size can result in a p value of 0.000. A table (or better: a chart) gives a much better idea of how (strongly) variables are associated.

We therefore recommend you always add it to the p value. Also see this tutorial.

## By sher bahadar khan on December 30th, 2015

As we know that if p value is less than 0.05 it is considered as significant, so if the p value = .000 it means the highest significance?

## By Aliu on August 18th, 2015

Beautiful tutorials, quite helpful & easy to understand. Good job.