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Hypothesis Testing: Power Function

by kingwinner
Tags: function, hypothesis, power, testing
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kingwinner
#1
Apr15-09, 04:46 AM
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!


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Billy Bob
#2
Apr15-09, 09:27 AM
P: 393
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].
kingwinner
#3
Apr15-09, 03:03 PM
P: 1,270
Quote Quote by Billy Bob View Post
Replace [tex]\overline{X}[/tex] with [tex]\mu_a+4Z/\sqrt{n}[/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!

Billy Bob
#4
Apr15-09, 05:22 PM
P: 393
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 [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]
kingwinner
#5
Apr16-09, 04:36 AM
P: 1,270
Quote Quote by Billy Bob View Post
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!
Billy Bob
#6
Apr16-09, 08:05 AM
P: 393
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.
kingwinner
#7
Apr16-09, 03:17 PM
P: 1,270
Quote Quote by Billy Bob View Post
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...
Billy Bob
#8
Apr16-09, 03:25 PM
P: 393
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.
kingwinner
#9
Apr16-09, 03:27 PM
P: 1,270
Quote Quote by Billy Bob View Post
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.
Yes, but you mean μ>13 as a stirct inequality, right?
Billy Bob
#10
Apr16-09, 03:29 PM
P: 393
Put that if you'd rather. Maybe it's better. I'd probably forget even to mention the restriction at all.
kingwinner
#11
Apr16-09, 03:32 PM
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?
Billy Bob
#12
Apr16-09, 05:47 PM
P: 393
Right!
kingwinner
#13
Apr17-09, 02:33 AM
P: 1,270
Thanks a lot!!


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