Hypothesis Testing: Power Function

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    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]
     
  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
    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!
     
  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
    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?
     
  13. Apr 16, 2009 #12
    Right!
     
  14. Apr 17, 2009 #13
    Thanks a lot!!
     
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