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I intend to determine Cramer's $\phi$ for two contingency tables of the following form:

Table 1 (expected empirical)
 A   B   C | sum
95  31  20 | 146

Table 2 (observed empirical)
 A   B   C | sum
70  29  18 | 117

Cramer's $\phi = \sqrt{\frac{\chi^2}{n\cdot(\min(r,~c)-1)}}$. Here, $n$ is the total number of observations, $r$ is the number of rows, and $c $ the number of columns. The value $\chi^2$ is the sum of quadratic differences between expectation and observation, normalized by the expected value. As I understand it, however, there are two ways of determining $\chi^2$ of two tables. (Note that I am aware of the conditions that need to apply for calculating $\chi^2$. Please ignore the fact that some conditions might be violated in the examples.)

On the one hand, you can determine $\chi^2$ by calculating the probabilities from one of the tables and use them to predict the number of observations in each category of the other table, given its total number of observations.

Assuming Table 1 gives the expected values and Table 2 describes the observations, observation probabilities rounded to two decimals would look like this.

Table 1a (expected theoretical)
   A    B    C
0.65 0.21 0.14

By multiplying each probability with the row sum of the observed values in Table 2 we get the following expected empirical observations.

Table 2a (expected empirical, fitted)
    A     B     C | sum
76.05 24.57 16.38 | 117

With Table 2a $\chi^2 = \frac{(70-76.05)^2}{76.05} + \frac{(29-24.57)^2}{24.57} + \frac{(18-16.38)^2}{16.38} = 0.48 + 0.80 + 0.16 = 1.44$.

On the other hand, you can determine $\chi^2$ by combining the two tables to perform a homogeneity test.

Table 3
      A   B   C | sum
(1)  95  31  20 | 146
(2)  70  29  18 | 117
sum 165  60  38 | 263

Here you get the expected values by multiplying the respective column and row sums and dividing the product with the total number of observations. As an example, for cell (1, A) this is $146 \cdot 165 / 263 = 91.60$.

Table 3a
        A     B     C
(1) 91.60 33.31 21.10
(2) 73.40 26.70 16.90

With Table 3a $\chi^2 = \frac{(95-91.60)^2}{91.60} + \frac{(31-33.31)^2}{33.31} + \frac{(20-21.10)^2}{21.10} + \frac{(70-73.40)^2}{73.40} + \frac{(29-26.70)^2}{26.70} + \frac{(18-16.90)^2}{16.90}$

$\chi^2 = 0.13 + 0.16 + 0.06 + 0.16 + 0.20 + 0.07 = 0.78$

Which $\chi^2 $-value is the right one for Cramer's $\phi$? If the first method is correct, do I take $n$ from the observed or from the expected values? If the second method is correct, how would I generate Table 3 and Table 3a from multidimensional data?

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This is actually straightforward. If you use your first strategy, you are assuming the expected frequencies are known a-priori and without error. If you use your second strategy, you are estimating the expected frequencies from the data. Your data, however, being finite, have some uncertainty with regard to the expected proportions, and this uncertainty becomes incorporated into the calculation. The second approach will thus be less powerful, but is probably more honest.

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  • $\begingroup$ Great, thanks! I guess I'll be going with the second strategy then. If you could also direct me to information on how to compare two tables that consist of several rows, instead of just one, I'll accept your answer. $\endgroup$ – wehnsdaefflae Apr 21 '15 at 20:53

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