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)
Step 3: Compute the statistic χ ²=Σ(O-E)² / E
Where O is the observed frequency, E is the expected
frequency, and the sum Σ is over all cells.
Step 4: Find the critical value χ²α 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.
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 χ²α.
Step5: Compare the sample statistic χ² of Step 3 with the critical value of χ²α of Step 4. If the sample statistic is larger, reject the null hypothesis of independence. Otherwise, do not reject the null hypothesis.
The following table shows the Myers-Briggs personality preference and professions for a random sample of 2408 people in the listed professions.
Personality Preference Type
Occupation Extrovert Introvert Row Total
_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.
The solution provides step by step method for the calculation of chi square test for association. Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.