Calculating the size of Type 1 error, Type 2 error and power of the test

Let \(X\) have a binomial distribution with parameter \(n=5\) and \(P\in [p:p=\frac{1}{4},\frac{1}{2}]\).

The null hypothesis \(H_{0}:p=\frac{1}{4}\) is rejected, and The alternative hypothesis \(H_a:p=\frac{1}{2}\) is accepted.

If the observed value of \(X_1\), a random sample of size one, is less than or equal to \(3\).

Find the size of Type 1 error, Type 2 error and power of the test.

I have no idea to solve the question. I only know that

###Size of a Type 1 error = Pr[rejecting\(H_0|H_0 \)is true]
###Size of a Type 2 error = Pr[not rejecting[/math]H_0|H_0 [/math]is False]

The sign \(|\) denotes "given that".


TS Contributor
this looks like homework and you need to show a credible effort to get help.

You are given a decision rule ( lines 2 and 3 in your post) so you can define the event of rejecting/accepting H0 in terms of your decision rule. Once you have that, calculating the probabilities is easy.

BTW the rule you gave does not make sense to me. Are you sure it is " less then" not greater then?



Ambassador to the humans
I think what rogojel is saying is that if it's true that

Ho: p = 1/4
Ha: p = 1/2

then it would make more sense to *fail* to reject the null if X <= 3 (since under the null we expect X<=3 to happen more than under the alternative). But this could just be a silly exam question and you just need to do the calculations.