# Marginal Probabilities in Re-Sampling

The background for this problem is as follows: There is a population of $$10,000$$ females and $$12,000$$ males. Researchers separately sample $$n_1=100$$ females and $$n_2=200$$ males, asking them if they are currently living with their parents. The resulting table is the following :

$$\begin{array}{|c|c|c|c|} \hline & Female & Male & Total \\ \hline \text{Living with parents}& 8& 22& 30\\ \hline \text{Not living with parents} & 92 & 178 &270\\ \hline Total & 100&200 &300\\ \hline \end{array}$$ The problem is as follows:

Here is a shuffling simulation (a resampling procedure) that is a good representation of how the data were actually collected (if male and female millenials from this high school actually do live with their parents in equal proportions.) Use two stacks of cards: one with $$10,000$$ cards and one with $$12,000$$ cards to represent the $$10,000$$ females and $$12,000$$ males, respectively. Split the pile of $$10,000$$ cards into $$1,000$$ red cards and $$9,000$$ black cards.

I am confused about the bolded question. The cards are being split into $$10$$ percent red cards, and $$90$$ percent black cards, which exactly follows the marginal probability of a person living with his/her parents. Yet the $$10,000$$ cards represents only the female population, so why should we use the marginal probability, and not the $$8$$ percent of females who are living with their parents?

## 1 Answer

Your null hypothesis is that, in the population, the proportion of females living with their parents is equal to the proportion of males living with their parents

The cards analogy supposes that these proportions are $$0.1\%$$, to match the overall proportion from the combination of the two samples

It could instead have supposed this proportion was something else, such as supposing these proportions were each $$\dfrac{\frac{8}{100}\times 10000 + \frac{22}{200}\times 12000}{10000+12000} \approx 0.0986$$ (the population-weighted mean) to make a very small difference to the later calculations, but the key point is to assume they are equal to each other and then test the null hypothesis by calculating how unlikely the results actually seen were in terms of the probability of seeing these results or more extreme results