1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Hypothesis Testing: Power Function

  1. Apr 15, 2009 #1
    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
    = 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]
  2. jcsd
  3. Apr 15, 2009 #2
    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].
  4. Apr 15, 2009 #3
    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!
    Last edited: Apr 15, 2009
  5. Apr 15, 2009 #4
    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]
  6. Apr 16, 2009 #5

    But in the end, I get a power function in terms of μ_a and n, not μ and n. What should I do?

  7. Apr 16, 2009 #6
    μ_a is right. When the question says μ and n, it really means μ_a and n, because μ_0=13.

    Good job.
  8. Apr 16, 2009 #7
    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...
  9. Apr 16, 2009 #8
    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.
  10. Apr 16, 2009 #9
    Yes, but you mean μ>13 as a stirct inequality, right?
  11. Apr 16, 2009 #10
    Put that if you'd rather. Maybe it's better. I'd probably forget even to mention the restriction at all.
  12. Apr 16, 2009 #11

    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?
  13. Apr 16, 2009 #12
  14. Apr 17, 2009 #13
    Thanks a lot!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook