5 Chi-Square Cross-Tabulation
5.1 The Worked Example
Here is our data:
| Groups | Yes | No |
|---|---|---|
| A | 88 | 93 |
| B | 160 | 59 |
We are going to need to know the Column Totals and Row Totals and the Total number of participants (N), so lets calculate them add them to our tables:
- Group A Row Total = 88 + 93 = 181
- Group B Row Total = 160 + 59 = 219
- Yes Column Total = 88 + 160 = 248
- No Column Total = 93 + 59 = 152
- N = 88 + 93 +160 + 59 = 400
And if we add those to our table we see:
| Groups | Yes | No | Totals |
|---|---|---|---|
| A | 88 | 93 | 181 |
| B | 160 | 59 | 219 |
| Totals | 248 | 152 | 400 |
Now the formula for the chi-square is:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
The Expected values for each condition, in the cross-tabulation in a few ways but the one we will use here is probably the easiest to use and it is:
\[Expected = \frac{Total_{row} \times Total_{column}}{N_{total}}\]
This is the same as other versions you might have seen such as:
\[Expected = \frac{Total_{row}}{N_{total}} \times \frac{Total_{column}}{N_{total}} \times N_{total}\]
The will both give the same result. So using the first approach we would see that the Expected values are:
For Group A people that said Yes:
\[Expected_{A-Yes} = \frac{181 \times 248}{400} = \frac{44888}{400} = 112.22\]
For Group A people that said No:
\[Expected_{A-No} = \frac{181 \times 152}{400} = \frac{27512}{400} = 68.78\]
For Group B people that said Yes:
\[Expected_{B-Yes} = \frac{219 \times 248}{400} = \frac{54312}{400} = 135.78\]
For Group B people that said No:
\[Expected_{B-No} = \frac{219 \times 152}{400} = \frac{33288}{400} = 83.22\]
We now have our data, let's start putting it into the formula, which we said was:
\[\chi^2 = \sum\frac{(Observed - Expected)^2}{Expected}\]
Which really means:
\[\chi^2 = \frac{(Observed_{A-Yes} - Expected_{A-Yes})^2}{Expected_{A-Yes}} + \frac{(Observed_{A-No} - Expected_{A-No})^2}{Expected_{A-No}} + \\ \frac{(Observed_{B-Yes} - Expected_{B-Yes})^2}{Expected_{B-Yes}} + \frac{(Observed_{B-No} - Expected_{B-No})^2}{Expected_{B-No}}\]
And if we start putting in the values, becomes:
\[\chi^2 = \frac{(88 - 112.22)^2}{112.22}+ \frac{(93 - 68.78)^2}{68.78}+\frac{(160 - 135.78)^2}{135.78}+\frac{(59 - 83.22)^2}{83.22}\]
And if we start to tidy those top halves up a little it becomes:
\[\chi^2 = \frac{(-24.22)^2}{112.22}+ \frac{(24.22)^2}{68.78}+\frac{(24.22)^2}{135.78}+\frac{(-24.22)^2}{83.22}\]
And now we square those top halves to give:
\[\chi^2 = \frac{586.6084}{112.22} + \frac{586.6084}{68.78} + \frac{586.6084}{135.78} + \frac{586.6084}{83.22} \]
And then divide the top halves by the bottom halves
\[\chi^2 = {5.2273071} + {8.5287642} + {4.3202858} + {7.0488873}\]
And then we sum them altogether to find:
\[\chi^2 = 25.1252443 \]
Meaning that, rounded to two decimal places, we find \(\chi^2 = 25.13\)
Degrees of Freedom
The degrees of freedom for the cross-tabulation is calculated as:
\[df = (Rows - 1) \times (Columns - 1)\]
Which is read as the number of Rows minus 1 times the number of Columns minus 1. If we look at our original data again:
| Groups | Yes | No |
|---|---|---|
| A | 88 | 93 |
| B | 160 | 59 |
Looking at the table we see we have 2 rows and 2 columns of actual observed data (not looking at the titles and group names), so:
\[Rows - 1 = 2 - 1 = 1\]
And
\[Columns - 1 = 2 - 1 = 1\]
Meaning that:
\[df = (Rows - 1) \times (Columns - 1)\]
which becomes:
\[df = (2 - 1) \times (2 - 1)\]
And reduces to:
\[df = (1) \times (1)\]
leaving us with:
\[df = 1\]
so we see that \(df = 1\)
The effect size
One of the common effect sizes for a cross-tabulation chi-square test is Cramer's \(V\) and is calculated as:
\[V = \sqrt\frac{\chi^2}{N \times \min(C-1, R-1)}\]
The key thing to note is \(\min(C-1, R-1)\) which is read as the minimum of EITHER the number of Columns (C) minus 1 OR the number of Rows (R) minus 1; whichever of those two values is smallest.
And the minimum of Columns minus 1 or Rows minus 1 is:
\[\min(C-1, R-1) = \min( 2 - 1, 2 - 1)\] \[\min(C-1, R-1) = \min( 1, 1)\] which gives us:
\[\min(C-1, R-1) = 1\]
And now we can start completing the Cramer's \(V\) formula as we know:
- \(\chi^2 = 25.13\)
- \(N = 400\)
- \(\min(C-1, R-1) = 1\)
Giving us:
\[V = \sqrt\frac{25.13}{400 \times 1}\]
And if we deal with the bottom half first we get:
\[V = \sqrt\frac{25.13}{400}\]
Then if we divide the top by the bottom we get
\[V = \sqrt{0.062825}\]
Giving us:
\[V = 0.2506492\]
So we see that, rounded to two decimal places, the effect size is \(V = 0.25\)
The write-up
If we were to look at a critical value look-up table, we would see that the critical value associated with a \(df = 1\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 3.841\). As the chi-square value of this test (i.e. \(\chi^2 = 25.13\)) is larger than \(\chi^2_{crit}\) then we can say that our test is significant, and as such would be written up as \(\chi^2(df = 1, N = 400) = 25.13,p < .05, V = 0.25\).
Following up a significant finding
If our test was significant then we would go on to carry out the relevant one-sample chi-squares to breakdown how the association manifests itself. The key thing to remember however is that in these one-sample chi-squares, following the cross-tabulation, you MUST use the expected values from the cross-tabulation and do not calculate new expected values.
For instance if you were looking at the one-sample chi-square within Group A, then you would use the expected values of \(Expected_{A-Yes} = 112.22\) and \(Expected_{A-No} = 68.78\) to compare against the observed values of \(Observed_{A-Yes} = 88\) and \(Observed_{A-No} = 93\) as such:
\[\chi^2 = \frac{(88 - 112.22)^2}{112.22}+ \frac{(93 - 68.78)^2}{68.78}\]
Which if you walk the process through you find:
\[\chi^2 = 13.7560713\]
Meaning that, rounded to two decimal places, we find \(\chi^2 = 13.76\). The degrees of freedom, based on \(df = k - 1\) (because it is a one-sample chi-square now), and that we have two groups (YES and NO, so \(k = 2\)), would be:
\[df = k - 1 = 2 - 1 = 1\]
The effect size is \(\phi\) again because it is a one-sample chi-square and, to remind us, has the formula:
\[\phi = \sqrt\frac{\chi^2}{N}\]
And we know:
- \(\chi^2 = 13.76\)
- and N = 88 + 93 = 181
and if we put those into the formula we get:
\[\phi = \sqrt\frac{13.76}{181}\]
Which becomes:
\[\phi = \sqrt{0.0760221} = 0.2757211\] And rounded to two decimal places would show \(\phi = 0.28\)
If we were to thinking about writing this one-sample up, and looking at the critical value table, we would see that the critical value associated with a \(df = 1\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 3.841\). As the chi-square value of this test (i.e. \(\chi^2 = 13.76\)) is larger than \(\chi^2_{crit}\) then we can say that our test is significant, and as such would be written up as \(\chi^2(df = 1, N = 181) = 13.76,p < .05, \phi = 0.28\).
Likewise, for the one-sample chi-square within Group B you would use the expected values of \(Expected_{B-Yes} = 135.78\) and \(Expected_{B-No} = 83.22\) to compare against the observed values of \(Observed_{B-Yes} = 160\) and \(Observed_{B-No} = 59\) as such:
\[\chi^2 = \frac{(160 - 135.78)^2}{135.78}+ \frac{(59 - 83.22)^2}{83.22}\]
Which if you walk the process through you find:
\[\chi^2 = 11.369173\]
Meaning that, rounded to two decimal places, we find \(\chi^2 = 11.37\). Again the degrees of freedom, based on \(df = k - 1\) (because it is a one-sample chi-square now), and that we have two groups (YES and NO, so \(k = 2\)), would be:
\[df = k - 1 = 2 - 1 = 1\]
The effect size is \(\phi\) again because it is a one-sample chi-square and, to remind us, has the formula:
\[\phi = \sqrt\frac{\chi^2}{N}\]
And we know:
- \(\chi^2 = 11.37\)
- and N = 160 + 59 = 219
and if we put those into the formula we get:
\[\phi = \sqrt\frac{12.85}{253}\]
Which becomes:
\[\phi = \sqrt{0.0519178} = 0.2278548\] And rounded to two decimal places would show \(\phi = 0.23\)
If we were to thinking about writing this one-sample up, and looking at the critical value table, we would see that the critical value associated with a \(df = 1\) at \(\alpha = .05\), to three decimal places, is \(\chi^2_{crit} = 3.841\). As the chi-square value of this test (i.e. \(\chi^2 = 11.37\)) is larger than \(\chi^2_{crit}\) then we can sa that our test is significant, and as such would be written up as \(\chi^2(df = 1, N = 219) = 11.37,p < .05, \phi = 0.23\).
And finally we can round everything off by stating that the most common answer for group A was No and the most common answer for group B was Yes.