1. The problem statement, all variables and given/known data 2. Relevant equations Statistics: power function 3. The attempt at a solution a) Test statistic: Z = (X bar - 13) / (4/sqrt n) Let μ = μ_a > 13 be a particular value in H_a Power(μ_a) = 1- beta(μ_a) = P(reject Ho | H_a is true) = P(reject Ho | μ = μ_a) = P(Z>1.645 | μ = μ_a) = P(X bar > 13 + [(1.645 x 4)/(sqrt n)] | μ = μ_a) I also know that X bar ~ N(μ, 16/n) I am stuck here...How can I proceed from here and express the power in terms of μ and n? Thank you very much! [note: also discussing in other forum]
Replace [tex]\overline{X}[/tex] with [tex]\mu_a+4Z/\sqrt{n}[/tex], rearrange to get [tex]P(Z>\text{whatever})[/tex], then express in terms of [tex]\mu_a[/tex] and [tex]n[/tex] using the normal cdf [tex]\Phi(t)[/tex].
Um...why can we replace X bar by mu_a + 4Z/sqrt n ? What is Z here? I don't understand this step...can you please explain a little bit more? X bar has the distribution N(μ, 16/n). Here, we know the distribution of X bar, so I believe that we can express P(X bar > 13 + [(1.645 x 4)/(sqrt n)] when μ = μ_a) as an integral of the normal density and replace μ by μ_a, but how can I integrate it and get a compact answer? Thank you very much!
Z is N(0,1), a standard normal r.v. You can't integrate and get a compact answer. The way to get a compact answer is to express your answer using the cumulative density function [tex]\Phi(t)[/tex] where [tex]\Phi(t)=P(Z\le t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-x^2/2}\,dx[/tex]
OK! But in the end, I get a power function in terms of μ_a and n, not μ and n. What should I do? Thanks!
So the power function is in terms of μ_a, and μ_a > 13 should be a particular value in H_a, right? The quesiton says μ and n, which to me looks like μ can be anything...
Just replace your μ_a in your power function with simply μ, write [tex]\mu\ge13[/tex] as the restriction and you have the answer.
Put that if you'd rather. Maybe it's better. I'd probably forget even to mention the restriction at all.
Alright! For part b, Set Power(μ=15)=0.99 The left side should now be a function of n only, and using the standard normal table, we should be able to solve for n, right?