Probability of two normally distributed random variables to be equal
Answers
Answer:
The test
if (a) X1∼N(μ1,σ1) and (b) X1∼N(μ2,σ2) and (c) X1 and X2 are independent then if you draw a sample of size N1 from the distribution of X1 and a sample of size N2 from the distribution of X2, then the arithemtic average of the sample from Xi is x¯i and is distributed X¯i∼N(μi,σiNi√). If the variables are independent then X¯1−X¯2∼N(μ1−μ2,σ21N1+σ22N2−−−−−−−√).
You seem to have some a priori knowledge about the 'truth' namely that μ1≤μ2 so if you want to find evidence that μ1<μ2 then (see What follows if we fail to reject the null hypothesis?) your H1 should be H1:μ1<μ2 and in order to 'demonstrate' this, one assumes the opposite, but as you say that μ1≤μ2 the opposite is H0:μ1=μ2.
So you have a one-sided test H0:μ1−μ2=0 against H1:μ1−μ2<0.
By the above, if H0 is true, then μ1−μ2=0 and therefore X¯1−X¯2∼N(0,σ21N1+σ22N2−−−−−−−√). With this you can define a left tail, one sided test and using the Neyman-Pearson lemma it can be shown tbe be the unformly most powerfull test. If you known σi then you can use the normal distribution to define the critical region in the left tail, if you don't know them then you estimate them from the samples and then you have to define the critical region using the t-distribution.
To define the test you have to (1) define a significance level α (2) draw a sample of size N1 from X1 and of size N2 from X2, then (3) compute x¯i for each of these samples and then compute the p-value of x¯1−x¯2 knowing that X¯1−X¯2∼N(0,σ21N1+σ22N2−−−−−−−√). If the obtained p-value is below α then the test concludes that there is evidence in favour of H1. One can also compute the α-quantile of the distribution under H0, : q0α and H0 will be rejected when x¯1−x¯2≤q0α.
The probability of a false positive is equal to the significance level α that you choose, the probability of a false negative can only be computed for a given value of μ1−μ2.
The sample sizes
If you have such a value for μ1−μ2 e.g. the difference is -0.05, then you can compute the type II error for this value:
The type II error is the probability that H0 is accepted when it is false, to compute it, let us assume that it is false and that μ1−μ2=−0.05. In that case X¯1−X¯2∼N(−0.05,σ21N1+σ22N2−−−−−−−√) (the mean has changed).
Now we have to compute the probability that H0 is accepted when the latter is true. This is the probability of observing q0α under the above distribution with mean -0.05.
it will be a function of N1 and N2. You can then fix a value for N2 (the one that is difficult to sample) and find N1 by fixing a type II error that is acceptable for you.
Step-by-step explanation: