Most data analysts are familiar with post hoc tests for ANOVA. Oddly, post hoc tests for the chi-square independence test are not widely used. This tutorial walks you through 2 options for obtaining and interpreting them in SPSS.

- Option 1 - CROSSTABS
- CROSSTABS with Pairwise Z-Tests Output
- Option 2 - Custom Tables
- Custom Tables with Pairwise Z-Tests Output
- Can these Z-Tests be Replicated?

## Example Data

A sample of N = 300 respondents were asked about their education level and marital status. The data thus obtained are in edu-marit.sav. All examples in this tutorial use this data file.

## Chi-Square Independence Test

Right. So let's see if education level and marital status are associated in the first place: we'll run a chi-square independence test with the syntax below. This also creates a contingency table showing both frequencies and column percentages.

***Basic crosstabs with chi-square independence test.**

crosstabs marit by educ

/cells count column

/statistics chisq.

Let's first take a look at the actual test results shown below.

First off, we reject the null hypothesis of independence:
education level and marital status are associated,
χ^{2}(12) = 23.57, p = 0.023.
Note that that SPSS wrongfully reports this 1-tailed significance as a 2-tailed significance. But anyway, what we really want to know is
precisely *which* percentages differ significantly from each other?

## Option 1 - CROSSTABS

We'll answer this question by slightly modifying our syntax: adding `BPROP`

(short for “**B**onferroni **prop**ortions”) to the `/CELLS`

subcommand does the trick.

***Crosstabs with pairwise z-tests.**

crosstabs marit by educ

/cells count column

**bprop**. /*bprop = Bonferroni adjusted z-tests for column proportions.

Running this simple syntax results in the table shown below.

## CROSSTABS with Pairwise Z-Tests Output

First off, take a close look at the table footnote: **“Each subscript letter denotes a subset of Education Level categories whose column proportions do not differ significantly from each other at the .05 level.”**

These conclusions are based on z-tests for independent proportions. These also apply to the percentages shown in the table: within each row, each possible pair of percentages is compared using a z-test. If they don't differ, they get a similar subscript. Reversely, within each row, percentages that

*don't*share a subscript

are significantly different. For example, the percentage of people with middle school who never married is 46.2% and its frequency of n = 18 is labeled “a”. For those with a Master’s degree, 16.7% never married and its frequency of 9 is

*not*labeled “a”. This means that 46.2% differs significantly from 16.7%.

The frequency of people with a Bachelor’s degree who never married (n = 21 or 25.0%) is labeled both “a” and “b”. It doesn't differ significantly from any cells labeled “a”, “b” or both. Which are all cells in this table row.

Now, a Bonferroni correction is applied for the number of tests within each row. This means that for \(k\) columns,

$$P_{bonf} = P\cdot\frac{k(k - 1)}{2}$$

where

- \(P_{bonf}\) denotes a Bonferroni corrected p-value and
- \(P\) denotes a “normal” (uncorrected) p-value.

Right, now our table has 5 education levels as columns so
$$P_{bonf} = P\cdot\frac{5(5 - 1)}{2} = P \cdot 10$$

which means that each p-value is multiplied by 10 and only then compared to alpha = 0.05. Or -reversely- only z-tests yielding an uncorrected p < 0.0**0**5 are labeled “significant”. This holds for all tests reported in this table. I'll verify these claims later on.

## Option 2 - Custom Tables

A second option for obtaining “post hoc tests” for chi-square tests are Custom Tables. They're found under

but only if you have a Custom Tables license. The figure below suggests some basic steps.
You probably want to select both frequencies and column percentages for education level.

We recommend you add totals for education levels as well.

Next, our z-tests are found in the Test Statistics tab shown below.

Completing these steps results in the syntax below.

***Contingency table with pairwise z-tests via CTABLES.**

CTABLES

/VLABELS VARIABLES=marit educ DISPLAY=DEFAULT

/TABLE marit BY educ [COUNT 'N' F40.0, COLPCT.COUNT '%' PCT40.1]

/CATEGORIES VARIABLES=marit ORDER=A KEY=VALUE EMPTY=INCLUDE TOTAL=YES POSITION=AFTER

/CATEGORIES VARIABLES=educ ORDER=A KEY=VALUE EMPTY=INCLUDE

/CRITERIA CILEVEL=95

/COMPARETEST TYPE=PROP ALPHA=0.05 ADJUST=BONFERRONI ORIGIN=COLUMN INCLUDEMRSETS=YES

CATEGORIES=ALLVISIBLE MERGE=YES STYLE=SIMPLE SHOWSIG=NO.

## Custom Tables with Pairwise Z-Tests Output

Let's first try and understand what the footnote says: **“Results are based on two-sided tests. For each significant pair, the key of the category with the smaller column proportion appears in the category with the larger column proportion. Significance level for upper case letters (A, B, C): .05. Tests are adjusted for all pairwise comparisons within a row of each innermost subtable using the Bonferroni correction.”**

Now, for normal 2-way contingency tables, the “innermost subtable” is simply the entire table. Within each row, each possible pair of column proportions is compared using a z-test. If 2 proportions differ significantly, then the higher is flagged with the column letter of the lower. Somewhat confusingly, SPSS flags the frequencies instead of the percentages.

In the first row (never married),
the D in column A indicates that these 2 percentages

differ significantly:
the percentage of people who never married is significantly higher for those who only completed middle school (46.2% from n = 39) than for those who completed a Master’s degree (16.7% from n = 54).

Again, all z-tests use α = 0.05 after Bonferroni correcting their p-values for the number of *columns* in the table. For our example table with 5 columns, each p-value is multiplied by \(0.5\cdot5(5 - 1) = 10\) before evaluating if it's smaller than the chosen alpha level of 0.05.

## Can these Z-Tests be Replicated?

Yes. They can.

Custom Tables has an option to create a table containing the exact p-values for all pairwise z-tests. It's found in the Test Statistics tab. Selecting it results in the syntax below.

***Contingency table with Bonferroni corrected z-tests (exact p-values) in separate table.**

CTABLES

/VLABELS VARIABLES=marit educ DISPLAY=DEFAULT

/TABLE marit BY educ [COUNT 'N' F40.0, COLPCT.COUNT '%' PCT40.1]

/CATEGORIES VARIABLES=marit ORDER=A KEY=VALUE EMPTY=INCLUDE TOTAL=YES POSITION=AFTER

/CATEGORIES VARIABLES=educ ORDER=A KEY=VALUE EMPTY=INCLUDE

/CRITERIA CILEVEL=95

/COMPARETEST TYPE=PROP ALPHA=0.05 ADJUST=BONFERRONI ORIGIN=COLUMN INCLUDEMRSETS=YES

CATEGORIES=ALLVISIBLE MERGE=NO STYLE=SIMPLE SHOWSIG=YES.

## Exact P-Values for Z-Tests

For the first row (never married), SPSS claims that the Bonferroni corrected p-value for comparing column percentages A and D is p = 0.020. For our example table, this implies an *uncorrected* p-value of p = 0.0020.

We replicated this result with an Excel z-test calculator. Taking the Bonferroni correction into account, it comes up with the exact same p-value as SPSS.

All other p-values reported by SPSS were also *exactly* replicated by our Excel calculator.

I hope this tutorial has been helpful for obtaining and understanding pairwise z-tests for contingency tables. If you've any questions or feedback, please throw us a comment below.

Thanks for reading!

## THIS TUTORIAL HAS 7 COMMENTS:

## By Maud on April 3rd, 2021

Thanks for the tutorial, it was very useful.

However, when I want to generate p-values using your syntax it doesn't display p-values if one of the two proportions is zero (but the crosstabs do say that there is a significant difference). How can I determine p-value in this case? Is this possible with the Excel z-test calculator you mention? And if so, how does it work?

## By Ruben Geert van den Berg on April 4th, 2021

Hallo Maud, great question!

I replicated this scenario and I see the problem.

I'm not 100% sure but I think CTABLES may refuse a pairwise test because a zero proportion is a huge violation of the sample sizes assumptions required by the z-test for independent proportions:

p1 * n1 > 5,

(1−p1) * n1 > 5,

p2 * n2 > 5,

(1−p2) * n2 > 5

Because of this, our Excel tool tells you that your data don't meet the assumptions. It still computes a p-value but you can't rely on it.

I think CROSSTABS uses this exact same p-value but ignores this huge violation of assumptions.

Note that a 2 by 2 chi-square independence test doesn't help because it requires the exact same assumptions.

The quick solution is to simply report that you can't perform a significance test because you don't meet the assumptions. I'm pretty confident noboby will dispute.

Technically, however, you could isolate the 2 proportions in your data with RECODE and SELECT IF.

If you do this correctly, then a CROSSTABS should come up with a 2 by 2 table containing only the proportions you'd like to compare. You can now inspect (and test) the Phi-coefficient or (equivalently) a Pearson correlation between the 2 (now dichotomous) variables. If your observations are independent and your overall sample size > 25 or so, this would be statistically perfectly valid.

Of course, you'd have to do this for each pair of proportions you'd like to test. As this is cumbersome, I recommend just going for the quick solution.

Hope that helps!

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