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I have a data set of 10,000 patients (roughly). Around 6300 are women and 3700 are men. Is there anyway for me to say that within this patient group, a randomly selected patient is significantly more likely to be a women than a man? I considered trying boot-strapping, but was wondering if this would be the right approach or if there is anything simpler to try first.

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    $\begingroup$ A randomly selected patient, by construction, has a 6300/10000 chance of being female and 3700/10000 chance of being male. Please explain, then, what you mean by "significantly more likely:" in what sense are you using "significant"? $\endgroup$ – whuber Sep 6 '18 at 17:04
  • $\begingroup$ In order to say anything a control group is needed. For example, suppose we are looking at people who are 80+ y/o. Then we are looking at 80+ y/o patients with hyperthyoridism. There are more women than men alive who are 80+ y/o. Without accounting for the prevalence of sex in the age group (for which tables are available), we could not state that women have a higher likelihood than men of becoming hyperthyroid. $\endgroup$ – Carl Sep 7 '18 at 3:45
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Your question is framed in a rather ambiguous fashion since "a randomly selected patient is significantly more likely to be a women than a man" is not a crystal clear statement. However, one way to approach this question is the following.

Since you want to test whether it's more likely to draw a woman than a man, you can define a Bernoulli random variable capturing this attribute. Suppose that this random variable takes value equal to $1$ if the subject is a female, and that this event occurs with probability $p$. In other words,

$$ \begin{equation} Z= \begin{cases} 1 & \text{with probability }\ p \\ 0 & \text{with probability }\ 1-p \end{cases} \end{equation} $$

Now, we can formulate your question as a hypothesis testing problem. Let's say that you happen to know what would define the cutoff "equally likely" given the example at hand. For instance, this can be a probability can be pinned down based on an idealized or actual control population that includes factors such as male/female predominance for a control population is a specific age group (thanks Carl for the suggestion, see comment section). Let's denote that threshold $p_0$, and you're conjecturing that it's more likely to draw a woman than a man. Then

$$ H_0: p=p_0\:\:\text{ vs } H_a: p>p_0 $$

For the sake of simplicity and since you have a sample size $n=10,000$, we will use the normal approximation to make inference. Let $\hat{p}$ be the proportion of female in your sample (6300/10000). Recall that

$$ t=\frac{\sqrt{n}(\hat{p}-p_0)}{\sqrt{p_0(1-p_0)}}\overset{d}{\rightarrow}\mathcal{N}(0,1) $$

Let $\alpha=0.05$, then since this is a one-sided, upper-tail test, then if $t>=1.645$ we reject the null hypothesis, and we conclude that we have enough evidence in favor of the alternative hypothesis i.e. it's more likely to draw a woman than a man.

I hope this helps.

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  • $\begingroup$ You should not assume a 0.5 probability of choosing a particular sex. $\endgroup$ – Carl Sep 7 '18 at 3:36
  • $\begingroup$ I am not assuming is 0.5, the probability of choosing a particular sex is hypothesized to be 0.5 and that's the one we want to reject (to break the even). The test statistic is calculated under the null, and the limiting distribution is determined under the null too. $\endgroup$ – MauOlivares Sep 7 '18 at 15:05
  • $\begingroup$ You should not null-hypothesize that a breakeven occurs at a probability of 0.5 because in general that does not occur for frequency of sex predominance in any particular sub-population of ages. $\endgroup$ – Carl Sep 7 '18 at 15:24
  • $\begingroup$ I think you're making this problem unnecessarily more complicated. If they were equally likely then it would be reasonable to work under the understanding that the the probability is 0.5 (like in all textbook examples when you flip a coin). In the link you provide, by the way, the author does subgroup analysis based on observables we don't even have in this question, making your point too stringent and you down-vote unfair. $\endgroup$ – MauOlivares Sep 7 '18 at 15:34
  • $\begingroup$ Genetics is not a perfect coin. More males than females are conceived (approx. 106%) and they live shorter lives. That is why at least a theoretical control group would be needed to solve this problem. If you would generalize your response to include a control or baseline probability, then it would be improved. $\endgroup$ – Carl Sep 7 '18 at 15:42

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