As noted at the beginning of this chapter, when observed items are sorted according to two or more separate dimensions of classification concurrently, they are said to be crosscategorized. The most efficient way to represent crosscategorized frequency data is with what is known as a contingency table. Suppose, for example, that there is a certain University—we will call it University Z, or UZ for short—which has several graduate professional schools. An investigator is interested in determining whether the students at two of these entities, the Lschool and the Bschool, differ in their respective proportions of liberals and conservatives. To this end, the investigator takes a random sample of students from each of the two schools and asks them which of the two labels they prefer as a selfdescription, "liberal" or "conservative." She then crosscategorizes the subjects according to (i) their school affiliation, L or B, and (ii) their preferred selfdescription as either "liberal" or "conservative." The resulting contingency table would have the following form, except that the various verbal descriptions beginning with the phrases "number of" and "total" would of course be replaced with specific numerical values.
 Preferred SelfDescription

 Liberal
 Conservative

School
 Lschool
 number of Lschool liberals
 number of Lschool conservatives
 total Lschool subjects

Bschool
 number of Bschool liberals
 number of Bschool conservatives
 total Bschool subjects

 total liberals
 total conservatives
 total subjects

When chisquare procedures are applied to a contingency table of this sort, it is typically with the aim of determining whether the two categorical variables are associated; hence the name of this version, the
chisquare test of association. If Lschool and Bschool subjects divided themselves as "liberal" and "conservative" in equal proportions, as in the following example
 Liberal
 Conservative

 Lschool
 55% of Lschool subjects
 45% of Lschool subjects

Bschool
 55% of Bschool subjects
 45% of Bschool subjects

there would clearly be no evidence of association between the two variables. If on the other hand they divided themselves up in different proportions, for example
 Liberal
 Conservative

 Lschool
 55% of Lschool subjects
 45% of Lschool subjects

Bschool
 45% of Bschool subjects
 55% of Bschool subjects

then there would be at least the suggestion of an association between the two variables. In this particular instance, the null hypothesis would be that there is no overall difference between Lschool and Bschool students at UZ with respect to how they divide themselves up on the liberal/conservative selfdescription. The chisquare test of association permits one to determine how likely it is that random samples of Lschool and Bschool students could have produced a liberal/conservative difference this large or larger, by mere chance coincidence,
if the null hypothesis were true.
Here is a concrete example of crosscategorized frequency data, adapted from a medical study that sought to determine whether estrogen supplementation might delay or prevent the onset of Alzheimer's disease in postmenopausal women.

MX. Tang, D. Jacobs, Y. Stern, et al., "Effect of oestrogen during menopause on risk and age at onset of Alzheimer's disease." Lancet 1996; 348, 42932.

The subjects in the study were 1,124 postmenopausal women who were initially free of Alzheimer's disease, Parkinson's disease, and stroke, and who were taking part in a longitudinal study of ageing and health in a New York City community. Of these women, 156 had received estrogen supplementation following the onset of menopause, while the remaining 968 had not. The following table breaks these two groups down according to whether the women did or did not show signs of Alzheimer's disease onset during a fiveyear followup period, as judged by annual clinical assessments and criterionbased diagnoses.

Alzheimer's onset during 5year period


No

Yes

received
estrogen
 Yes
 147
 9
 156

No
 810
 158
 968

 957
 167
 1,124

In case the bare numbers in this table do not jump right out at you, here is the bottom line in percentage terms: Of the women who did not receive estrogen supplementation, 16.3% (158/968) showed signs of Alzheimer's disease onset during the fiveyear period; whereas, of the women who did receive estrogen supplementation, only 5.8% (9/156) showed signs of disease onset. It is not difficult to adduce reasons that might account for such a difference. As the authors of the study note, estrogen "promotes the growth of cholinergic neurons, stimulates the secretase metabolism of the amyloid precursor protein, and may interact with apolipoprotein E. All these factors could affect the risk of Alzheimer's disease." But of course, none of these possibilities amounts to anything at all unless we can establish that the difference between the observed percentages—16.3% versus 5.8%—reflects anything other than mere random variability. Once again it is the question of statistical significance: How likely is it that a difference this large or larger could have occurred by mere chance coincidence?
The procedures for applying chisquare to a twodimensional situation of this sort are the same as we saw for the onedimensional situation, except for two small modifications. The first has to do with what we take to be the MCE expected cell frequencies, and the second pertains to how we calculate the appropriate value for degrees of freedom. In both of these modifications the underlying logic is the same; it is only the details that differ.
In the onedimensional chisquare test the values of
E are simply stipulated in advance. In the fish example they were given as
100/100/100, because the sample was to be compared with the proportions of the three fish species that had been recorded over the preceding years. In the student survey example they were set to match the proportions of responses in the various categories that had been found in the large national survey. In the twodimensional situation, on the other hand, the expected cell frequencies are not given in advance, nor are they intuitively obvious. We will illustrate the logic of the point with a simple if somewhat fanciful example.
Two friends, A and B, believe their friendship to be so deep as to produce a remarkable pattern of correspondences. Often they find that they are thinking the same thing at the same time. Often, even when separated, they find that they are doing the same thing at the same time. To put their faith to the test, they each toss a coin 100 times in succession, recording on each occasion the head/tail outcome of A's toss and the corresponding
head/tail outcome of B's toss. Their hypothesis is that when A gets a head, B will also tend to get a head; and that when A gets a tail, B will also tend to get a tail. They of course do not suppose that even their relationship is so deep as to ensure these correspondences in 100 percent of the tosses, though they do believe that the pattern of such correspondences will significantly exceed what would be expected on the basis of mere chance coincidence.
For the sake of discussion, suppose that A and B each end up with exactly 50 heads and 50 tails. In that case the contingency table would have the following marginal totals, irrespective of how much or little the heads and tails outcomes of A and B might correspond.

Outcomes for B


Tails

Heads

Outcomes
for A
 Heads
 [cell a]
 [cell b]
 50

Tails
 [cell c]
 [cell d]
 50

 50
 50
 100

Actually, this is one of the rare scenarios for which the values of
E would be fairly intuitively obvious. If you were asked to guess the values of the MCE expected frequencies for cells a, b, c, and d, I expect you would answer
25/25/25/25—and this would be quite correct. Now, if only you can make explicit the hidden logic that leads you to this answer, you will have the procedure for figuring out the values of
E for twodimensional chisquare situations in general. I suspect the core of your implicit logic runs something like this: If A gets 50 percent heads and B gets 50 percent tails; and if nothing other than mere chance coincidence is operating in the situation; then the (conjunctive) probability that any particular one of the 100 paired tosses will include a head for A and a tail for B is
.5x.5=.25. Thus, the expected frequency for cell a is 25 percent of the total number of paired tosses:
E_{a}=.25x100=25. The same logic would also lead you to
E_{b}=25,
E_{c}=25, and
E_{d}=25.
Now see how the same logic can be extended to situations that are not intuitively obvious. When A and B perform their series of 100 paired tosses, it is actually not very likely that they would both end up with exactly 50 percent heads and 50 percent tails. It is much more likely that they each would end up with something slightly different from an exact
50/50 split. Suppose that A comes out with 46 heads and 54 tails, while B ends up with 48 heads and 52 tails. In this case the marginal totals would be distributed as follows

Outcomes for B


Tails

Heads

Outcomes
for A
 Heads
 [cell a]
 [cell b]
 46

Tails
 [cell c]
 [cell d]
 54

 52
 48
 100

and the logic would run like this. It is the same logic as outlined above, but now we will lay it out more formulaically. The proportion of paired tosses that include a head for person A is
46/100=.46; the proportion that include a tail for person B
is 52/100. If nothing other than mere chance coincidence is operating in the situation, the probability that any particular one of the 100 paired tosses will include a head for A and a tail for B is therefore
.46x.52=.2392; and the mean chance expected number of such conjunctions in 100 paired tosses is accordingly
 E_{a} =
 46 100
 x
 52 100
 x 100 = 23.92

This same structure can now be generalized to any situation at all where we have frequency data arranged in a rows by columns matrix. Defining "R" as the marginal total of the row to which the cell belongs, "C" as the marginal total of the column to which the cell belongs, and "N" as the total number of crosscategorized observations, we can write it as
which reduces algebraically to the simpler structure
 E_{cell} =
 RxC N

 That is: For each cell, multiply the marginal total for the row to which the cell belongs by the marginal total for the column to which the cell belongs, and then divide the result by the total number of crosscategorized observations.

The following illustration shows how this simple calculation works out for each of the cells of the present cointoss example.

Outcomes for B


Tails

Heads

Outcomes
for A
 Heads

E_{a} =
46x52 100
= 23.92

E_{b} =
46x48 100
= 22.08
 46

Tails

E_{c} =
54x52 100
= 28.08

E_{a} =
54x48 100
= 25.92
 54

 52
 48
 100

And here is the same procedure applied to our estrogen/Alzheimer's example.

Alzheimer's onset during 5year period


No

Yes

received
estrogen
 Yes

E_{a} =
156x957 1124
= 132.82

E_{b} =
156x167 1124
= 23.18
 156

No

E_{c} =
968x957 1124
= 824.18

E_{d} =
968x167 1124
= 143.82
 968

 957
 167
 1,124

Translation: The null hypothesis holds that estrogen supplementation in postmenopausal women makes no difference with respect to Alzheimer's onset. If this null hypothesis were true, then these are the frequencies we would have expected for the various cells of the table, given the marginal totals that appear along the right edge and the bottom of the table.
In case all this calculation and its accompanying jargon are proving a bit confusing, step back for a moment and think of it this way. On the null hypothesis, we would have expected the two categories of subjects—those who received estrogen supplementation and those who did not—to show approximately the same proportionate incidence of Alzheimer's onset during the fiveyear period. And that is exactly what the calculated values of E are specifying. Thus, of the 156 subjects who did receive estrogen supplementation, we would have expected a 23.18/156=14.9% rate of onset; and for the 968 subjects who did not receive supplementation, we would have expected an identical 143.82/968=14.9% rate. The same holds true for the expected rate of nononset: 132.82/156=85.1% for those who received supplementation, and 824.18/968=85.1% for those who did not receive supplementation.
The next table shows these values of E, now listed in red, in comparison with the original values of O. Notice, incidentally, that the values of O and E, when summed across the rows and the columns, both add up to the same marginal totals.

Alzheimer's onset during 5year period


No

Yes

received
estrogen
 Yes
 147
132.82
 9
23.18
 156

No
 810
824.18
 158
143.82
 968

 957
 167
 1,124

Beyond this point, the calculation of chisquare for the twodimensional situation depends on the particular numbers of rows and columns. If there are
more than two rows or more than two columns, it proceeds exactly as it does in the onedimensional case, in accordance with the formula
 _{} =

 (O—E)^{2} E

When there are
exactly two rows and two columns, the calculation requires a
correction for continuity analogous to the correction described in Chapter 6 for binomial probability situations:
 _{} =

 (O — E—.5)^{2} E


I.e., for each cell, subtract onehalf unit from the absolute (unsigned) difference between O and E before squaring that difference.

As our current example has exactly two rows and two columns, the appropriate procedure is the latter. The following table shows the specific calculations for each of the four cells.

Alzheimer's onset during 5year period

 No
 Yes

received estrogen
 Yes

(147—132.82.5)^{2} 132.82
= 1.41

(9—23.18.5)^{2} 23.18
= 8.07

No

(810—824.18.5)^{2} 824.18
= 0.23

(158—143.82.5)^{2} 143.82
= 1.3

 sum:
_{} = 11.01

The next step is to refer our calculated value of chisquare to the appropriate sampling distribution, which you will recall is defined by the applicable number of degrees of freedom. When applying chisquare procedures to situations in which there is only one dimension of categorization, the general principle for determining degrees of freedom is
df = (number of cells)—1
With two dimensions of classification, it is a different formula but the same basic logic. Suppose you have two rows and two columns of cells, as shown below, and you are free to plug any integer numbers that you want into them, subject only to the stipulation that the cell values summed across rows, down columns, and overall, must add up to the fixed row, column, and overall sums that appear along the right edge and across the bottom.
 cell a = ?
 cell b = ?
 50

cell c = ?
 cell d = ?
 40

48
 42
 90

Here again your "freedom" is limited by the fixed sums. Arbitrarily plug 10 into cell a and the remaining three cells are instantly fixed at
b=40, c=38, and d=2. Plugging 20 into cell d fixes the other cells at
b=22, c=20, and
a=28; and so on. For chisquare situations involving two rows and two columns of cells, degrees of freedom will in every instance be equal to one.
If you have two rows and three columns, your "freedom" is increased a bit, though still limited by the fixed sums.
 cell a = ?
 cell b = ?
 cell c = ?
 50

cell d = ?
 cell e = ?
 cell f = ?
 40

25
 40
 25
 90

If you plug 10 arbitrarily into cell a, cell d becomes fixed at 15, but the other four cells remain "free" to vary. Plug some other number into any of these remaining four cells, however, and everything else becomes instantly fixed. For chisquare situations involving two rows and three columns (or three rows and two columns), degrees of freedom is in every case equal to two. More generally, when applying chisquare procedures to situations in which there are two dimensions of categorization, the general principle for determining degrees of freedom is
 df = (r—1)(c—1)

 r = number of rows
c = number of columns

For our estrogen/Alzheimer's example, there are 2 rows and 2 columns. Hence
Figure 8.6 shows the sampling distribution of chisquare for
df=1, along with the critical values of chisquare for several levels of significance. As indicated, a chisquare value equal to or greater than 3.84 would be significant at or beyond the .05 level; a chisquare value equal to or greater than 5.02 would be significant at or beyond the .025 level; and so on.
Figure 8.6. Sampling Distribution of ChiSquare for df=1
Our calculated value of
_{}=11.01 exceeds the value of chisquare (10.83) required for significance at the .001 level; hence we can say that the observed result is significant
beyond the .001 level. That is: If the null hypothesis were true—if estrogen supplementation were unrelated to the onset of Alzheimer's disease in postmenopausal women—the likelihood of obtaining a differential onset rate as great as the one observed (16.3%
versus 5.8%), by mere chance coincidence, would be smaller than onetenth of one percent. Our investigators can therefore reject the null hypothesis with a high degree of confidence.