Hypothesis Testing: Power Function

[kingwinner]

Homework Statement

http://www.geocities.com/asdfasdf23135/stat14.JPG

Homework Equations

Statistics: power function

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]

 

[Billy Bob]

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 $$\Phi(t)$$.

 

[kingwinner]

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!

 

[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 $$\Phi(t)=P(Z\le t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-x^2/2}\,dx$$

 

[kingwinner]

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!

 

[Billy Bob]

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]

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 question says μ and n, which to me looks like μ can be anything...

 

[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.

 

[kingwinner]

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?

 

[Billy Bob]

Put that if you'd rather. Maybe it's better. I'd probably forget even to mention the restriction at all.

 

[kingwinner]

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]

Right!

 

[kingwinner]

Thanks a lot!