Explore BrainMass
Share

# Chi square test for association: Personality & Occupation

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

See attached file.

There are 3 pages to the problems.

Chi -Square Distribution to test the independence of two variables

Step 1: Set up the hypotheses:

H0: The variables are independent
H1: The variables are not independent.

Step 2: Compute the expected frequency for each cell in the contingency table by use of the formula:

E=Expected frequency= (Row total)(Column Total)
_____________________
Sample size

Step 3: Compute the statistic &#967; ²=&#931;(O-E)² / E
Where O is the observed frequency, E is the expected
frequency, and the sum &#931; is over all cells.

Step 4: Find the critical value &#967;²&#945; in Table A. 10 page 651. Use the level of significance of 0.01 and the number of degrees of d.f. to find the critical value.
d.f.= (R-1)(C-1)

Page 2

where R is the number of rows and C is the number of columns of cells in the contingency table. The critical region consists of all values of &#967;²&#945;.
Step5: Compare the sample statistic &#967;² of Step 3 with the critical value of &#967;²&#945; of Step 4. If the sample statistic is larger, reject the null hypothesis of independence. Otherwise, do not reject the null hypothesis.

Problem

The following table shows the Myers-Briggs personality preference and professions for a random sample of 2408 people in the listed professions.

Page 3

____________________________________________________
Personality Preference Type

Occupation Extrovert Introvert Row Total
_Clergy_____________________________________308_____________226________534__
_M.D._______________________667________936______ _1603_ Lawyer_______________________112________159________271____Column total___________________ 1087_____- 1321 2408 _

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.01 level of significance.