relation, and that it will become 1 when they are mutually exclusive.

In general, if we have a population arranged with reference to the presence or absence of two characters, P and Q, we arrange the population as in the accompanying diagram with respect to the characters in question.

A formula for correlation which satisfies the conditions above is the following:

If care be taken to arrange the table so that be shall be numerically less than ad, then the correlation is positive. Whenever ad and be become equal the formula gives 0/(ad + bc) = 0, and there is no correlation. Whenever b or c becomes zero, then the formula becomes ad/ad 1, and there is perfect positive correlation. Whenever a or d becomes zero, then the formula becomes

=

- bc/bc

=

1, and there is perfect negative

correlation.

EXERCISES

1. Compute the correlation between deafness and white color in cats, from the following table:

2. Find the measure of the association of bay color and speed from

Such a low correlation cannot be relied upon as a proof of any distinct connection.

237. A Correlation Table. The preceding formula gives us no aid if we desire to find the correlation between characters present in varying degrees as weight, length, and circumference of ears of corn. If we examine the length and circumference of ears we will find that, in general, long ears are also large ears, but many can be found that are long and slender, many that are short and small, and still others that are short and large. What now is the correlation? To answer this question we construct a correlation table. One character is recorded in columns and the other in rows.

For example, if the characters are weight of ears in ounces and length of ears in inches, a large number of ears are taken at random, weighed and measured, and the data arranged in tabular form, appearing as on page 315.

In this table each ear is recorded in the proper square to represent both its weight and its length.. That is, all the ears of the same weight that are also of the same length are recorded together in the same square. This means that the various rows are frequency distributions of weight with respect to length (as 1, 6, 11, 26, 11, 8, 6, 1, the frequency distribution corresponding to the length 6.5), and all the columns are frequency distributions of length with respect to weight. Such frequency distributions with respect to a correlated character are technically known as arrays. The entire table, therefore, may be looked upon as made up of two systems of parallel arrays

with

Length of Ear in Inches.

respect to the two characters in question. They are in no respect different from any other frequency distribution; and their means, standard deviations, variability, and other determinations are calculated by the same methods as for ordinary frequency distributions.

CORRELATION BETWEEN WEIGHT AND LENGTH OF EAR

238. The Correlation Coefficient. The method of obtaining the correlation coefficient may be explained in connection with the above table.

M1,

Find in the usual way the mean with respect to each character involved-in this case mean length of ears, and mean weight of ears, Mw. Find the deviation Dɩ of ear length from mean length, and the deviation Dw of weight from mean weight, for each ear tabulated. For each ear tabulated find the product of Dr and Dw and then add all of these products. This sum we will indicate by ZDDw. Find in the usual way the standard deviation of length of ears, σ, and the standard deviation of weight of ears, σw. Then the coefficient of correlation, r, is

where n is the number of things observed, in this case the total number of ears.

A convenient arrangement for computing D, for each ear length and Du for each ear weight is shown in the table below.

The row labeled 6.5 inches (table § 237), gives the frequency distribution of ears with respect to weight. There is one ear of weight 4 oz., 6 ears of weight oz., 11 ears of weight 6 oz., 26 ears of weight 7 oz., etc.; a total of 70 ears, fi, of length 6.5 inches.

fiVi

=

1 X 4 + 6 X 5 + 11 × 6 + 26 × 7 + 11 × 8 + 8 X 9

+6 x 10 + 1 X 11 = 455.0 The mean length of ear is obtained by adding the numbers in the column headed fiVi and dividing this sum by the total number, fi 993, of ears.

=

Mw

10,576

=

= 10.65

NOTATION:

fi class frequencies of total population with respect to length.

=

Vi = value or measurement corresponding to a given frequency with

respect to length.

M1 = mean length of ears.

Di = deviation of ear length from mean length.

standard deviation of length of ears.

class frequencies of total population with respect to weight.

value corresponding to a given frequency with respect to weight.

= mean weight of ears.

=

deviation of weight from mean weight.

standard deviation of weight of ears.

r = coefficient of correlation.

This gives Mi

=

7.85. In the row labeled 6.5 and in the column headed Di we write the difference between this mean length 7.85 and the length 6.5. This gives the number The number 306.8 in the last column is obtained as follows:

1.3 of the column headed Dr.

(− 1.3)[1(— 6.7) + 6(− 5.7) + 11(− 4.7) + 26(− 3.7)

+ 11(−2.7) + 8(− 1.7) + 6(− 0.7) + 1(0.3)]

That is, the ear of weight 4 oz. deviates from the mean weight by 6.7 oz., the 6 ears of weight 5 oz. deviate from the mean weight by 5.7 oz., the 11 ears of weight 6 oz. deviate from the mean weight by 4.7 oz., etc.

The number 306.8 represents the sum of the products of the corresponding length and weight deviations for every individual in the horizontal row to which the number belongs. To find the correlation coefficient add the numbers in the column headed DiDw, obtaining in this case 4947.2.

993,

Divide this number 4947.2 by n X or X σw. In this case n = and σ, σ have been computed to be 1.57 and 3.63 respectively. This gives the correlation coefficient