Test to see what population an observation came from I have a population with a mean (mean_1) (which was calculated through many trials and so have a very very small uncertainty in the mean). The standard deviation of the population is 'S'.
I have a second population with mean_2 (which again has a very very small uncertainty) and a standard deviation which is again 'S'.
Say I make a single further observation and the value falls somewhere in between mean_1 and mean_2, is there a test that I can perform to say with what certainty the observation came from population 1 and what certainty it came from population 2?
Thank you.
 A: Suppose that the populations are distributed as $N(\mu_1, \sigma^2)$ and $N(\mu_2, \sigma^2)$ , and the value is $x$, then 
$$P(\text{X is from population 1}|X=x) = {P(X=x|\text{X is from population 1})P(\text{X is from population 1}) \over P(X=x)} $$
But 
$$P(X=x) = \sum_{i=1}^2 P(X=x|\text{X is from population i})$$
Therefore
$$P(\text{X is from population 1}|X=x) = {P(X=x|\text{X is from population 1})P(\text{X is from population 1}) \over \sum_{i=1}^2 P(X=x|\text{X is from population i})} = {(1/\sqrt{2\pi}\sigma) e^-{\left(x-\mu_1\right)^2 \over 2\sigma^2} P(\text{X is from population 1}) \over \sum_{i=1}^2 (1/\sqrt{2\pi}\sigma) e^-{\left(x-\mu_i\right)^2 \over 2\sigma^2} P(\text{X is from population i})} = {e^-{\left(x-\mu_1\right)^2 \over 2\sigma^2} P(\text{X is from population 1}) \over \sum_{i=1}^2 e^-{\left(x-\mu_i\right)^2 \over 2\sigma^2} P(\text{X is from population i})}$$
If $P(\text{X is from population 1}) = P(\text{X is from population 2})$ (we have no prior knowledge what population is X from), then the probability is equal to:
$$P(\text{X is from population 1}|X=x) = {e^-{\left(x-\mu_1\right)^2 \over 2\sigma^2}  \over \sum_{i=1}^2 e^-{\left(x-\mu_i\right)^2 \over 2\sigma^2} }$$
A: I think whuber was asking about the proportion of further observations that may come from population 1 and 2. For example, if population 1 contains 100,000 individuals and population 2 contains 200,000, and you randomly select one individual from both populations combined, you will have 1/3 chance of getting an individual from pop1 and 2/3 from pop2.
If we assume that the chances are equal, simply convert the new observation to z-scores for both populations by subtracting the mean and dividing by the corresponding S. The population with the less extreme z-score is more likely to be the one that the observation came from. This assumes that the distributions of both populations are normal.
Not sure how I would calculate exact chances. A crude way would be to select a small range around the z-scores (e.g., if your z is 1.22, select 1.21 to 1.23), find the chance that you would get that z-score for both populations, then divide the proportion for pop1 by pop2. You can get a relative chance this way, but I think there may be better methods.
