Question

4. [8]A company is considering an organizational change by adopting the use of self-managed work teams. To assess the attitudes of employees of the company toward this change, a sample of 400 employees is selected and asked whether they favor the institution of self-managed work teams in the organization. Three responses were permitted: favor, neutral, or oppose. The results of the survey, cross-classified by type of job and attitude toward self-managed work teams, are summarized as follows: Attitude Toward Self-managed Work Teams Favor Neutral Oppose Total Type of Job 108 46 71 225 18 12 30 60 35 14 26 75 Hourly worker Supervisor Middle Management Upper Management Total 24 7 9 40 185 79 136 400 At the 0.05 level of significance, is there evidence of a relationship between attitude toward self-managed work teams and type of job?

Answer 1

For the above 16 cells, we obtain the expected frequencies for each cell as:
E_i = (Sum of column i)*(Sum of row i) / Grand Total

Also the chi square test statistic contribution for each cell here is computed as:

$\chi^2_i = \frac{(O_i- E_i)^2}{E_i}$

These computations are given in the following table here:

Hourly worker Supervisor Middle Management Upper Management Favor Neutral 108 (104.06) [0.15] 46 (44.44) [0.05] 18 (27.75) [3.43] 12 (11.85) [0.00] 35 (34.69) [0.00] 14 (14.81) [0.04] 24 (18.50) (1.64] 7 (7.90) [0.10] Oppose 71 (76.50) [0.40] 30 (20.40) [4.52] 26 (25.50) [0.01] 9 (13.60) (1.56]

The circular bracket here contains the expected values while the square bracket contains the chi square test statistic contribution for each cell.

Therefore they are added together to get the chi square test statistic value here as:

$\chi^2 = \sum \chi^2_i = \sum \frac{(O_i- E_i)^2}{E_i} = 11.8953$

The degrees of freedom here is computed as:
Df = (num of columns - 1)(num of rows - 1) = 2*3 = 6

Therefore for 6 degrees of freedom, the p-value here is obtained from chi square distribution tables as:

$p =P(\chi^2_6 > 11.8953) = 0.0643$

As the p-value here is 0.0643 > 0.05 which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here that the 2 variables are associated.