# Chi-square test on a sample from population having an unequal distribution

I got a sample with an unequal gender distribution ($30$ males to $200$ female). This distribution is quite "representative" for the population the sample is taken from.

Nevertheless, in a paper, does it make any sense to report chi-square for gender (with the purpose to show this inequality beyond descriptives)?

Intuitively, I would say no, because the assumption of equality is already not met in population. Hence, it would be some kind of circular reasoning.

Am I right?

• What's "$\chi^2$ for gender" mean? - what null hypothesis are you testing? Feb 23 '15 at 22:37
• I think this means you're comparing 30/200 with a hypothesis of equal frequencies. That chi-square test makes perfect sense, but doesn't sound interesting or useful. My calculations give $P$-value of the order of $10^{-29}$. There is no circularity in reporting that your sample data echo what is broadly known about the population; it's no more circular than reporting that most of the people attending a girls' school are female, there being some male staff. Feb 23 '15 at 22:40
• @Scortchi Its not a hypothesis itself, I wanted to use it for showing the inequality of male and female in sample description in an exporative manner. I wonder, if this is even a permissible strategy?
– Jens
Feb 23 '15 at 22:42
• @NickCox, yes, that's a good point. Reporting $Chi^2$ does not offer a plus of information in this case then. Feel free to write an answer, I will accept it.
– Jens
Feb 23 '15 at 22:44
• @Jens: Well there has to be a null hypothesis to carry out a test. It could be that the population frequencies of male and female are equal, or in any other proportion. As Nick implies, don't bother testing hypotheses that aren't interesting or useful - would a confidence interval for the proportion be? Feb 24 '15 at 0:29

I think this means you're comparing $30/200$ with a hypothesis of equal frequencies implying expected frequencies $115/115$.
My calculations give Pearson chi-square of $125.7$ and a $P$-value of about $4 \times 10^{-29}$: if that's not strong enough to convince the most sceptical of chance sceptics (someone asserting that you could have got this breakdown just by chance by sampling a population with equal frequencies), then your case is hopeless.
As @Scortchi underlines, a better way of quantifying uncertainty here is using a confidence interval. The proportion of females is $0.87$ and $95$% confidence intervals run from about $0.82$ to $0.91$, regardless of how you calculate them. That interval is clearly a long way from $0.5$.
These calculations are standard, but for the record I used chitesti and cii in Stata. The Stata manual entry on these confidence interval calculations is accessible to all and gives literature references on different methods for binomial confidence intervals. For probabilities very near $0$ or $1$, it matters what method you use, but that does not bite for this example.