Linear Correlation in Contingency Tables

To calculate the linear correlation coefficient $ r $ between two variables in a contingency table

Y / X	y₁	y₂	...	yⱼ	...	Total
x₁	f₁₁	f₁₂	...	f₁ⱼ	...	R₁
x₂	f₂₁	f₂₂	...	f₂ⱼ	...	R₂
...	...	...	...	...	...	...
xᵢ	fᵢ₁	fᵢ₂	...	fᵢⱼ	...	Rᵢ
...	...	...	...	...	...	...
Total	C₁	C₂	...	Cⱼ	...	N

Here’s the process:

Calculate the means $ \bar{x} $ and $ \bar{y} $ of variables $ X $ and $ Y $ using marginal frequencies.
Find the deviations of each value from the mean ($ x_i - \bar{x} $ and $ y_j - \bar{y} $).
Multiply each deviation of $ X $ and $ Y $ by their joint frequencies $ f_{ij} $ and sum these products.
Compute the squared deviations weighted by the marginal frequencies $ R_i $ and $ C_j $ for $ X $ and $ Y $, summing the results.
Finally, use the formula to calculate the linear coefficient $$ r = \frac{\sum (x_i - \bar{x})(y_j - \bar{y}) f_{ij}}{\sqrt{\sum (x_i - \bar{x})^2 R_i \cdot \sum (y_j - \bar{y})^2 C_j}} $$ where the double summation includes all values $ i $ and $ j $ across rows and columns. This formula gives $ r $, which reflects the strength and direction of the correlation between $ X $ and $ Y $.

This method provides $ r $, a measure of the linear correlation between the two variables.

Note: In the contingency table used to represent the joint frequency distribution of two variables, $ X $ and $ Y $:

$ x_i $ and $ y_j $ are the values of the variables $ X $ and $ Y $.
$ f_{ij} $ represents the joint frequency associated with the pair $ (x_i, y_j) $.
$ C_j $ is the column sum, representing marginal frequencies for $ Y $.
$ R_i $ is the row sum, representing marginal frequencies for $ X $.
$ N $ is the grand total, or the sum of all joint frequencies.

Overall, this structure enables the calculation of statistics like the correlation coefficient using the joint frequencies $ f_{ij} $, marginal frequencies of $ X $ and $ Y $, and the means of their marginal distributions.

A Practical Example

Let’s go through an example to calculate the correlation coefficient $ r $ using a contingency table.

Suppose we want to study the correlation between study hours ($ X $) and final grades ($ Y $) in a group of students.

	$ Y = 6 $	$ Y = 7 $	$ Y = 8 $	Total (R_i)
$ X = 4 $	3	2	1	6
$ X = 5 $	1	3	1	5
$ X = 6 $	1	1	2	4
Total (C_j)	5	6	4	15

This table shows the joint frequencies of students for each combination of study hours and final grades:

Calculate the marginal means of $ X $ and $ Y $:

$$ \bar{x} = \frac{4 \times 6 + 5 \times 5 + 6 \times 4}{15} = \frac{24 + 25 + 24}{15} = \frac{73}{15} \approx 4.87 $$

$$ \bar{y} = \frac{6 \times 5 + 7 \times 6 + 8 \times 4}{15} = \frac{30 + 42 + 32}{15} = \frac{104}{15} \approx 6.93 $$

Then calculate the deviations from the mean for each value of $ X $ and $ Y $.

The deviations for variable $ X $ are:

$ 4 - \bar{x} = 4 - 4.87 = -0.87 $
$ 5 - \bar{x} = 5 - 4.87 = 0.13 $
$ 6 - \bar{x} = 6 - 4.87 = 1.13 $

The deviations for variable $ Y $ are:

$ 6 - \bar{y} = 6 - 6.93 = -0.93 $
$ 7 - \bar{y} = 7 - 6.93 = 0.07 $
$ 8 - \bar{y} = 8 - 6.93 = 1.07 $

For clarity, I’ve listed the deviations directly above rows and alongside columns in the table.

Next, compute the product of deviations for each cell and multiply by the joint frequency $ f_{ij} $:

		$ y_j - \bar{y} $
		$ -0.93 $	$ 0.07 $	$ 1.07 $
$ x_i - \bar{x} $	$ -0.87 $	$ (-0.87)(-0.93) \times 3 $	$ (-0.87)(0.07) \times 2 $	$ (-0.87)(1.07) \times 1 $
	$ 0.13 $	$ (0.13)(-0.93) \times 1$	$ (0.13)(0.07) \times 3 $	$ (0.13)(1.07) \times 1 $
	$ 1.13 $	$ (1.13)(-0.93) \times 1$	$ (1.13)(0.07) \times 1 $	$ (1.13)(1.07) \times 2 $

For instance, in the top-left cell, $ X=4 $ deviates by $ -0.87 $ from $ \bar{x} $, while $ Y=6 $ deviates by $ -0.93 $ from $ \bar{y} $. Their joint frequency is $ 3 $, representing the number of students who studied $ X=4 $ hours and scored $ Y=6 $. The final result here is $ 2.43 $ $$ (-0.87)(-0.93) \times 3 = 2.43 $$

At this stage, we calculate each cell’s values in the table and add up the partial totals for each row.

		$ y_j - \bar{y} $
		$ -0.93 $	$ 0.07 $	$ 1.07 $	$ (x_i - \bar{x})(y_j - \bar{y}) f_{ij} $
$ x_i - \bar{x} $	$ -0.87 $	2.43	-0.12	-0.93	1.38
	$ 0.13 $	-0.12	0.03	0.14	0.05
	$ 1.13 $	-1.05	0.08	2.42	1.45

Now calculate each cell’s value and sum the partial totals for each row:

$$ \sum (x_i - \bar{x})(y_j - \bar{y}) f_{ij} = 1.38 + 0.05 + 1.45 = 2.88 $$

Calculate the squared deviations $ (x_i - \bar{x})^2 $ for each row and multiply by the row’s marginal frequency $ R_i $:

$$ \sum (x_i - \bar{x})^2 R_i = (-0.87)^2 \times 6 + (0.13)^2 \times 5 + (1.13)^2 \times 4 $$

$$ \sum (x_i - \bar{x})^2 R_i = 4.54 + 0.08 + 5.11 $$

$$ \sum (x_i - \bar{x})^2 R_i = 9.73 $$

		$ y_j - \bar{y} $
		$ -0.93 $	$ 0.07 $	$ 1.07 $	$ (x_i - \bar{x})(y_j - \bar{y}) f_{ij} $	$ (x_i - \bar{x})^2 R_i $
$ x_i - \bar{x} $	$ -0.87 $	2.43	-0.12	-0.93	1.38	4.54
	$ 0.13 $	-0.12	0.03	0.14	0.05	0.08
	$ 1.13 $	-1.05	0.08	2.42	1.45	5.11

Then, calculate the squared deviations $ (y_j - \bar{y})^2 $ for each column and multiply by the column’s marginal frequency $ C_j $:

$$ \sum (y_j - \bar{y})^2 C_j = (-0.93)^2 \times 5 + (0.07)^2 \times 6 + (1.07)^2 \times 4 $$

$$ \sum (y_j - \bar{y})^2 C_j = 4.32 + 0.03 + 4.58 $$

$$ \sum (y_j - \bar{y})^2 C_j = 8.93 $$

		$ y_j - \bar{y} $
		$ -0.93 $	$ 0.07 $	$ 1.07 $	$ (x_i - \bar{x})(y_j - \bar{y}) f_{ij} $	$ (x_i - \bar{x})^2 R_i $
$ x_i - \bar{x} $	$ -0.87 $	2.43	-0.12	-0.93	1.38	4.54
	$ 0.13 $	-0.12	0.03	0.14	0.05	0.08
	$ 1.13 $	-1.05	0.08	2.42	1.45	5.11
$ (y_i - \bar{y})^2 C_i $		4.32	0.03	4.58

Finally, apply the formula:

$$ r = \frac{\sum (x_i - \bar{x})(y_j - \bar{y}) f_{ij}}{\sqrt{\sum (x_i - \bar{x})^2 R_i \cdot \sum (y_j - \bar{y})^2 C_j}} $$

$$ r = \frac{2.88}{9.33} \approx 0.31 $$

The correlation coefficient $ r \approx 0.31 $ suggests a moderate positive correlation between study hours and final grades.

And so forth.

	\( Y = 6 \)	\( Y = 7 \)	\( Y = 8 \)	Total (R_i)
\( X = 4 \)	3	2	1	6
\( X = 5 \)	1	3	1	5
\( X = 6 \)	1	1	2	4
Total (C_j)	5	6	4	15

		\( y_j - \bar{y} \)
		\( -0.93 \)	\( 0.07 \)	\( 1.07 \)
\( x_i - \bar{x} \)	\( -0.87 \)	\( (-0.87)(-0.93) \times 3 \)	\( (-0.87)(0.07) \times 2 \)	\( (-0.87)(1.07) \times 1 \)
	\( 0.13 \)	\( (0.13)(-0.93) \times 1\)	\( (0.13)(0.07) \times 3 \)	\( (0.13)(1.07) \times 1 \)
	\( 1.13 \)	\( (1.13)(-0.93) \times 1\)	\( (1.13)(0.07) \times 1 \)	\( (1.13)(1.07) \times 2 \)