A company has 100 employees, 40 women and 60 men.
At the end of the year, the company decides to give bonuses to 20 people: 16 men and 4 women.
I'm asked to find a way to test the presence of a connection between the gender and the bonus attribution.
I thought to use a standard Pearson's Chi-squared test, hence building the following table:
| Received bonus | $O_i$ | $E_i$ |
|---|---|---|
| Men | 16 | 12 |
| Women | 4 | 8 |
$O_i$ is the observed frequency for the two categories and $E_i$ is the expected one, assuming the bonus attribution does not depend on gender (and all employees have similar features if not for the gender).
I proceed to calculate the $\chi^2$ statistics as follows: $$\chi^2 = \frac{(16-12)^2}{12} + \frac{(4-8)^2}{8} = 3.\overline{3}$$
And then (since $\chi^2_{(0,05, 1)} \approx 3,841$) I can conclude that I cannot refuse the null hypothesis of independence between gender and bonus attribution at a level of significance of $0.05$ (I can refuse the null hypothesis at a level of significance of $0.1$, though).
[MY QUESTION]
Assuming all I did is correct (please tell me if it's not!), I wondered if testing for the independence between gender and NOT receiving a bonus should yield to the same conclusion and I concluded that it must do so! (right?)
So, I built the following table for the character bonus NOT received:
| NOT received bonus | $O_i$ | $E_i$ |
|---|---|---|
| Men | 44 | 48 |
| Women | 36 | 32 |
Now, calculating the Pearson's chi-squared statistics from this I get $\chi^2=0.8\overline{3}$. This statistics brings me to different conclusions from the previous ones: for example, now I cannot refuse the null hypothesis with a level of significance $0.1$!
What am I doing wrong? Am I testing for two different things without realizing it?
Thanks very much in advance for your help!
No, your test is wrong. Here is the correct one
In the first two-way table (tabella a doppia entrata) you have your observed distribution. In the second one you have the theoretical one, assuming independence between the two characters, sex and bonus; in the third one there is your $\chi^2$ test, where the total value has to be compared with the critical value from a $(k-1)\times(h-1)=(2-1)\times(2-1)=1$ d.o.f. chi square distribution