# "Invariance" of Pearson's Chi-squared test

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?

• I forgot to state the conclusion: your test is significant but not highly significant, being your pvalue in the middle between 5% and 1% significance level...As you are Italian, this is a good lecture for you quadernodiepidemiologia.it/epi/assoc/chi_qua.htm May 6, 2021 at 9:50
• Grazie very much for the reply and the lecture :) May 6, 2021 at 9:56

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
• Thanks very much for the quick reply! I guess I can't somehow grasp the difference between this kind of test and a Chi-squared goodness-of-fit one, in which I only compare the values from an observed distribution (in this case the bonus attribution between men and women?) with the expected, theoretical one! I mean, abusing your patience: why do I have to use both Bonus and Malus distribution info when they're clearly related? I expected that knowing one (hence, considering one in calculating $\chi^2$) was "enough" since it should contain all the information...sorry for making no sense" May 6, 2021 at 9:55