# Statistics power function problem

• Apr 9th 2010, 07:36 PM
bryandn
Statistics power function problem
Here's the problem, I'm stuck:

Consider the distributions N(mu1, 400) and N(mu2, 225). Let theta = mu1-mu2 and x and y be the observed means of two independent random samples, each of size n, from these two disbtibutions. We reject H(0) : theta = 0 and accept H(a): theta >0 if and only if x-y >=C. If pi(theta) is the power function of this test, find n and C so that pi(theta=10) = 0.95 at significance level alpha = 0.05.

Thank for anyone's help.
• Apr 9th 2010, 09:59 PM
matheagle
you reject the null hypothesis and accept the alternative if........

$\displaystyle {X-Y-0\over\sqrt{{400\over n}+{225\over n}}}>1.645$

or

$\displaystyle X-Y>1.645\sqrt{{400\over n}+{225\over n}}$

Now beta=.05 when theta=10

$\displaystyle {X-Y-10\over\sqrt{{400\over n}+{225\over n}}}<-1.645$
• Apr 12th 2010, 12:54 PM
bryandn
Wait, so what are n and C? Don't you need two equations with two unknowns to find them? Where are those two equations?
• Apr 12th 2010, 04:40 PM
matheagle
FROM .....

$\displaystyle X-Y>1.645\sqrt{{400\over n}+{225\over n}}$

$\displaystyle 1.645\sqrt{{400\over n}+{225\over n}}$
• Apr 12th 2010, 08:18 PM
bryandn
I feel kinda dumb, but where do you get the value of n from? Do you get it by solving the following system of equations?

$\displaystyle {X-Y-10\over\sqrt{{400\over n}+{225\over n}}}=-1.645$

and:

$\displaystyle X-Y=1.645\sqrt{{400\over n}+{225\over n}}$

So if I subtract the first equation from the 2nd, I get: $\displaystyle 10=3.29\sqrt{{400\over n}+{225\over n}}$

and n = 67.65, but since it should be a whole number, we round up to 68?