In: Statistics and Probability
A research assistant working for a large company wanted to investigate attitudes towards a proposed management buy-out. Thus she asked a random sample of the company’s workforce whether or not they were in favour of the buy out. The research assistant was particularly interested in finding out whether the workers’ views on the buy out were related to the classification of their job (manual or non-manual).
Altogether 120 workers were interviewed by her, of these 70 were manual workers and 50 were non-manual workers. Of the manual workers 30 were in favour of the buy out, whereas 25 of the non-manual workers were in favour.
The 2 x 2 table showing the given information here is given as:
Manual Workers | Non Manual Workers | Total | |
Favour to buy | 30 | 25 | 55 |
Not in favour to buy | 40 | 25 | 65 |
Total | 70 | 50 | 120 |
The expected values for each of the 4 clls is computed here
as:
Ei = (Sum of row i)*(Sum of column i) / Grand Total
Manual Workers | Non Manual Workers | |
Favour to Buy | 55*70/120 = 32.08 | 55*50/120 = 22.92 |
Not in favour to buy | 65*70/120 = 37.92 | 65*50/120 = 27.08 |
Using the above expected and the observed frequencies, the chi square test statistic here is computed as:
Degrees of freedom here is obtained as:
Df = (num of col - 1) (Num of rows - 1) = 1
Therefore the p-value here is obtained from the chi square distribution tables as:
As the p-value here is very high, we have sufficient evidence here that there is no association between the two variables.