# $\chi^2$ of multidimensional data

I want to test if two observations of nominal data accord to the same distribution. I am using the chi squared statistics to perform a chi squared homogeneity test and normalize the result with Cramer's $\phi$.

Unfortunately, all the examples for performing a chi squared homogeneity test I could find (e.g. here) perform the test with two one-dimensional observations. The example after the link above, for example, compares boys and girls dependent on their viewing preferences. This makes two observations in the form $[x_0, \ldots, x_n]$. However, I want to test observations in the form $[[x_0, \ldots, x_n], \ldots, [z_0, ..., z_m]]$. I don't know, if "multidimensional" is the right term in this context. I figured it might be.

Could you explain to me, how a chi squared homogeneity test can be performed with multi-row observations? I know it works, because scipy provides unique values for two dimensional input lists.

import scipy.stats as sps
observation1 = [[95, 31, 20], [70, 29, 18]]
observation2 = [[21, 69, 98], [54, 35, 11]]
data = [observation1, observation2]
print sps.chi2_contingency(data)


The code above yields (159.18016188570166, 4.772222443744986e-31, 7, array([[[ 69.44008748, 47.45072645, 42.53205358], [ 45.11526642, 30.82876539, 27.63310068]], [[ 76.04085626, 51.96125177, 46.57502446], [ 49.40378984, 33.75925639, 30.25982128]]])) where the first value is chi squared. Flattening the observations yields different values, so there must be a difference.

How do you calculate $\chi^2$ for determining the homogeneity of multidimensional observations? Please note that I know how to calculate $\chi^2$ on a multi-row contingency table. Instead I want to know how to perform a for the homogeneity of two contingency tables.

Example:

Table 1                                   Table 2
outcome0 outcome1 outcome2 sum                outcome0 outcome1 outcome2
action0       95       31       20 146        action0       21       69       98
action1       70       29       18 117        action1       54       35       11
sum      165       60       38 263


Question: Are these observations following the same distribution?

Scipy allows to determine the expected values (and $\chi^2$) of multi-dimensional observations. The expected values are returned as the last element of its call:

Expected table 1                      Expected table 2
outcome0 outcome1 outcome2            outcome0 outcome1 outcome2
action0    69.44    47.45    42.53    action0    76.04    51.96    46.58
action1    45.12    30.83    27.63    action1    49.40    33.76    30.26


The sum of the squared differences, divided by the expected value, yields the $\chi^2$ value. Note, however, that the expected tables are not just the expected values from each individual table (quick proof for the first cell of table 1: $\frac{146 \cdot 165}{263} = 91.60$). Instead their calculation seems to be somehow dependent on each other.

I'd like to know how the expected values depend on each other, so that I can calculate it myself.

• @whuber Thanks for your comment. I know what a contingency table is and how to calculate $\chi^2$. The 1st link provides information on Fisher's exact test on one-dimensional observations ("behaviours"), which is not applicable to my question. Your 2nd link provides information on when to use which test. I need information on $\chi^2$, this is also not applicable. Your 3rd and 4th link provide information on how to calculate $\chi^2$ on a contingency table containing one-dim. observations. The 5th doesn't work and the 6th isn't about $\chi^2$. I'll edit to emphasize homogeneity. – wehnsdaefflae Apr 23 '15 at 14:01
• These posts cover various aspects of your problem, including dealing with multi-way tables. If you know what contingency tables are and how to calculate $\chi^2$, then there are only two other aspects to your question: (1) computing the degrees of freedom, which presumably you also know (because that's part of understanding contingency tables) and (2) using the $\chi^2$ distribution to test homogeneity. In your edit you now refer to "two" contingency tables. Could you elaborate on that and explain precisely what it means for two tables to be "homogeneous"? – whuber Apr 23 '15 at 14:10
• @whuber thanks for your response. i'll edit the question accordingly. – wehnsdaefflae Apr 23 '15 at 14:12

To analyze a multi-way contingency table, you use log-linear models. In truth, log-linear models are a special case of the Poisson generalized linear model, so you could do that, but log-linear models are more user-friendly. In Python, you may need to use the Poisson GLM, as I gather log-linear models may not be implemented. I will demonstrate the log-linear model using your data with R.

library(MASS)
tab = array(c(95, 31, 20, 70, 29, 18, 21, 69, 98, 54, 35, 11), dim=c(3,2,2))
tab = as.table(tab)
names(dimnames(tab)) = c("outcomes", "actions", "observations")
dimnames(tab)[] = c("0", "1", "2")
dimnames(tab)[] = c("0", "1")
dimnames(tab)[] = c("1", "2")
tab
# , , observations = 1
#         actions
# outcomes  0  1
#        0 95 70
#        1 31 29
#        2 20 18
#
# , , observations = 2
#         actions
# outcomes  0  1
#        0 21 54
#        1 69 35
#        2 98 11


Log-linear models are simply a series of goodness of fit tests. We can start with a (trivial) null model that assumes all cells have the same expected value:

summary(tab)
# Number of cases in table: 551
# Number of factors: 3
# Test for independence of all factors:
#  Chisq = 159.18, df = 7, p-value = 4.772e-31


The null is rejected. Next, we can fit a saturated model:

m.sat = loglm(~observations*actions*outcomes, tab)
m.sat
# Call:
# loglm(formula = ~observations * actions * outcomes, data = tab)
#
# Statistics:
#                  X^2 df P(> X^2)
# Likelihood Ratio   0  0        1
# Pearson            0  0        1


Naturally, this fits perfectly. At this point, we could build up from the null model seeing if additional terms improve the fit, or drop terms from the saturated model to see if the fit gets significantly worse. The latter is more convenient and is conventional. To see if the distribution of outcomes by actions differs as a function of the observation, we need to drop the interactions between the observations and the actions * outcomes. If we also drop the marginal effect of observations, we are testing if the mean count differs between the two levels of observations. That may or may not be of interest to you, I don't know.

m1 = loglm(~observations + actions*outcomes, tab)
sum(tab[,,1])  # 263
sum(tab[,,2])  # 288
m2 = loglm(~actions*outcomes, tab)
anova(m2, m1)
# LR tests for hierarchical log-linear models
#
# Model 1:
#   ~actions * outcomes
# Model 2:
#   ~observations + actions * outcomes
#
#           Deviance df Delta(Dev) Delta(df) P(> Delta(Dev))
# Model 1   126.4172  6
# Model 2   125.2825  5   1.134691         1         0.28678
# Saturated   0.0000  0 125.282534         5         0.00000


Model 1 has dropped a single degree of freedom from Model 2 (note that, confusingly, Model 1 $\leftrightarrow$ m2, and Model 2 $\leftrightarrow$ m1), but the decrease in model fit is very small. It is not significant. There is not enough evidence to suggest that the mean counts differ by observation. On the other hand, when Model 2 is compared to the Saturated model, the decrease in fit is highly significant. The data are inconsistent with the idea that the distribution of counts is the same in both levels of observation.

• @gung There is not enough evidence to suggest that the mean counts differ by observation. I'm a bit confused... The anova(m2, m1) above effectively amounts to testing the significance of the main-effect regressor in a loglinear model (respecting the principle of marginality). However, in CAR it is said that "the tests for the main effects are generally irrelevant in loglinear models and are usually ignored---they pertain to the marginal distributions of the factors". I'm not sure if this is in contraction with the answer above... – landroni Jun 17 '15 at 18:53
• That's a reasonable point, @landroni. There is potentially some question of whether m2 vs m1 is the right question to ask. However, I don't think they are contradictory, m2 fits a different model, it isn't a test of an observations parameter in the context of m1. – gung - Reinstate Monica Jun 17 '15 at 19:10
• Should we unpack the underlying Poisson GLM, we could specify it so as to test if there is a difference in counts by observations at the reference level of the other 2 variables. In this case however, we are testing whether the marginal counts differ by observations. Ie, the main effect of obs in that Poisson GLM would be a test of 95 vs 21, whereas the test above compares 263 vs 288. – gung - Reinstate Monica Jun 17 '15 at 19:10
• If we want to test if the contingency table is the same for each observation, which p-value would we use? It seems like you’re saying to use the p-value $\approx 0$ for “saturated”, but m1 vs m2 ($p\approx 0.29$) seems like the usual likelihood ratio test of the “observation” parameter. – Dave Mar 29 at 19:08
• @Dave, I'm not sure if I follow your question. Do you want to know if the expected value for every cell is the same? Usually, people don't care about the actual numbers, they want to know if the variables are related. – gung - Reinstate Monica Mar 30 at 1:06

I found the answer here under 5.4.1: "Mutual (Complete) Independence". Obviously, the term I was missing is three-way contingency tables. Determining the expected values in a three dimensional contingency table is actually pretty much analogue to the standard variant.

In normal contingency tables you get the expected values by multiplying the row sum with the column sum and dividing the product by the total sum. Commonly denoted as $e_{ij} = \frac{o_{i.} \cdot o_{.j}}{n}$, where $o_{i.}$ is the row sum, $o_{.j}$ is the column sum and $n$ is the total.

In three-way contingency tables, you do not multiply the sum of a line (i.e. row or column) but the sum of a plane. Accordingly you divide by the square of all the observations involved.

With the given example, the calculation for the first cell in table 1 goes as follows:

$\frac{(95+70+21+54) \cdot (95+31+20+21+69+98) \cdot (95+31+20+70+29+18)}{(263+288)^2} = \frac{240 \cdot 313 \cdot 263}{303601} = 69.44$

The calculation for the last cell in table 2:

$\frac{(11+98+20+18) \cdot (70+29+18+54+35+11) \cdot (21+69+98+54+35+11)}{(263+288)^2} = 30.26$

The sought-after $\chi^2$ value is the squared sum of differences between observed and expected, divided by expected, as usual.