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I have two random samples drawn without replacement from the same population. Each sample received a different treatment (in fact both were drawn for different studies). I am inclined to believe these can be treated as if they were randomly assigned because each member had an equal probability of being selected into the first sample (or this case assigned to the first study). Thoughts?

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  • $\begingroup$ If the samples were drawn and then one sample was assigned to one study and the other sample to the other study...this would be a random sample. $\endgroup$ – Gregg H Apr 3 '18 at 15:03
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A solid way to analyze situations involving complex sampling programs (without replacement from a finite population) is to look at samples as subsets of the population. Two sampling programs are equivalent when they draw each possible sample with the same probabilities. (Arguably, it's enough that they would have the same chance of drawing your particular sample.) In this case, a "sample" is the ordered pair consisting of the subjects selected for the first study and those selected for the second study.

A quick elegant solution is to note that both sampling procedures are uniform distributions of all possible ordered pairs of samples: no particular set of possible subjects is favored over any other set in either procedure. But if you find that unconvincing (or opaque), you can carry out an explicit calculation. It's fairly easy to do.

As it happened, the first sample (say of size $k_1$) was one of the $\binom{n}{k_1}$ possible $k_1$-subsets of the population, a set of size $n.$ All of them were equally likely to be drawn. Conditional on that, the second sample (say of size $k_2$) was one of the $\binom{n-k_1}{k_2}$ possible $k_2$-subsets of the elements that remained. That gives

$$\binom{n}{k_1}\binom{n-k_1}{k_2} = \frac{n!}{k_1!(n-k_1)!} \frac{(n-k_1)!}{k_2!(n-k_1-k_2)!} = \frac{n!}{k_1! k_2! (n-k_1-k_2)!}$$

equiprobable ordered pairs of samples to choose from under this procedure.

The alternative consists of selecting all $k_1+k_2$ subjects at random and then choosing $k_1$ among them to serve as the first sample (leaving the remaining $k_2$ for the second sample). The number of possible ways to do this (all equally probable) is therefore the product of the appropriate Binomial coefficients, equal to

$$\binom{n}{k_1+k_2} \binom{k_1+k_2}{k_1} = \frac{n!}{(k_1+k_2)!(n-k_1-k_2)!} \frac{(k_1+k_2)!}{k_1!k_2!} = \frac{n!}{k_1! k_2! (n-k_1-k_2)!},$$

the same as before. Since both sampling procedures give the same probability of obtaining any given ordered pair of samples, they are equivalent.

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