
#1
Apr1509, 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! [note: also discussing in other forum] 



#2
Apr1509, 09:27 AM

P: 394

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].




#3
Apr1509, 03:03 PM

P: 1,270

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! 



#4
Apr1509, 05:22 PM

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 [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] 



#5
Apr1609, 04:36 AM

P: 1,270

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! 



#6
Apr1609, 08:05 AM

P: 394

Good job. 



#7
Apr1609, 03:17 PM

P: 1,270

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



#8
Apr1609, 03:25 PM

P: 394

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.




#9
Apr1609, 03:27 PM

P: 1,270





#10
Apr1609, 03:29 PM

P: 394

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




#11
Apr1609, 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? 



#12
Apr1609, 05:47 PM

P: 394

Right!




#13
Apr1709, 02:33 AM

P: 1,270

Thanks a lot!!



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