Hypothesis Testing: Power Function

by kingwinner
Tags: function, hypothesis, power, testing
 P: 1,270 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]
 P: 394 Replace $$\overline{X}$$ with $$\mu_a+4Z/\sqrt{n}$$, rearrange to get $$P(Z>\text{whatever})$$, then express in terms of $$\mu_a$$ and $$n$$ using the normal cdf $$\Phi(t)$$.
P: 1,270
 Quote by Billy Bob Replace $$\overline{X}$$ with $$\mu_a+4Z/\sqrt{n}$$
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!

P: 394

Hypothesis Testing: Power Function

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 $$\Phi(t)$$ where $$\Phi(t)=P(Z\le t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-x^2/2}\,dx$$
P: 1,270
 Quote by Billy Bob 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 $$\Phi(t)$$ where $$\Phi(t)=P(Z\le t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-x^2/2}\,dx$$
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!
P: 394
 But in the end, I get a power function in terms of μ_a and n, not μ and n. What should I do?
μ_a is right. When the question says μ and n, it really means μ_a and n, because μ_0=13.

Good job.
P: 1,270
 Quote by Billy Bob μ_a is right. When the question says μ and n, it really means μ_a and n, because μ_0=13. Good job.
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...
 P: 394 Just replace your μ_a in your power function with simply μ, write $$\mu\ge13$$ as the restriction and you have the answer.
P: 1,270
 Quote by Billy Bob Just replace your μ_a in your power function with simply μ, write $$\mu\ge13$$ as the restriction and you have the answer.
Yes, but you mean μ>13 as a stirct inequality, right?
 P: 394 Put that if you'd rather. Maybe it's better. I'd probably forget even to mention the restriction at all.
 P: 1,270 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?
 P: 394 Right!
 P: 1,270 Thanks a lot!!

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