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).


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!

  • $\begingroup$ 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 $\endgroup$
    – tommik
    May 6, 2021 at 9:50
  • $\begingroup$ Grazie very much for the reply and the lecture :) $\endgroup$ May 6, 2021 at 9:56

1 Answer 1


No, your test is wrong. Here is the correct one

enter image description here

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

  • $\begingroup$ 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" $\endgroup$ May 6, 2021 at 9:55
  • 1
    $\begingroup$ @luciadefinetti : I think that a quick lecture of the above link can explain you how this important test works. It is a very suitable test for various situations: checking independence, checking omogeneity and goodness of fit...finally in your data the cell with 4 observation is very borderline to apply this test...usually it is requested a minimum of 5...but 4 is good too... $\endgroup$
    – tommik
    May 6, 2021 at 10:01
  • 1
    $\begingroup$ @luciadefinetti : here calvino.polito.it/~riganti/Prob_Stat_teoria_esercizi.pdf is a very complete lecture that can be very useful for you: Cap 10 on non parametric tests. In particulare, Example 10.10 on page 301 shows you how suitable is chi squared test $\endgroup$
    – tommik
    May 6, 2021 at 10:06
  • $\begingroup$ Thanks again for the reply and all this material! $\endgroup$ May 6, 2021 at 10:10

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