Question

The following table shows the Myers-Briggs personality preferences for a random sample of 409 people in the listed professions-

Occupation	Extroverted	Introverted	Row Total
Clergy (all denominations)	65	43	108
M.D.	69	95	164
Lawyer	57	80	137
Column Total	191	218	409

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.10 level of significance

Answer 1

Note that first using the given row totals and column totals, the expected value for each of the 6 cells is computed as:

E_i = (Row Total x Column Total ) / 409

After this for each cell the chi square test statistic contribution is computed as:

The expected cell values are shown in the circular bracket and the square bracket terms contain the chi square test statistic contribution as:

Now the chi square test statistic for each cell is summed up to get the chi square test statistic value as:

Degrees of freedom here is computed as:

DF = (num of columns - 1)( num of rows - 1) = ( 2 - 1)(3 - 1) = 2

From chi square tables, we get the p-value as:

As the p-value here is 0.00468 < 0.1 which is the level of significance, therefore the test is significant and we can reject the null hypothesis here and conclude that the variables are not related and are independent.