Re: Power computation in Z test

Jack Tomsky wrote:
randovaro wrote:
quebecstat@xxxxxxx wrote:
On Aug 31, 12:47 am, Jack Tomsky
<jtom...@xxxxxxxxxxxxx> wrote:
A biologist is doing an experiment to measure
the
amount
of enzyme released by a process. This output is
known
to follow a
normal distribution with mean mu and std
deviation
sigma=5. The sample
size is n.
The test is
H0: mu=<10 versus
H1: mu>10
using a z-test
Question: What is the smallest sample size N
that the
biologist must
take in order to have at 83.7% power when the
true
value of the mean
is mu=12.95
I would be interested in getting some help with
this
problem.
My work so far:
P{Xbar > 10 | mu=12.95}>=.837
P{ sqrt(n)*(Xbar -12.95)/5 >
sqrt(n)*(10-12.95)/5
}>=.837
P{ Z > sqrt(n)*(-.59) }>=.837
1-Phi(sqrt(n)*(-.59)) =< .163
with Phi the cdf of the standard normal
sqrt(n)*(-.59)=<-.9822
sqrt(n) >=1.66
n>=2.77
So this means a sample size of N=3 minimum
Is my computation correct?
You would reject Ho not when Xbar > 10, but when
Xbar > 10 +
5*z(1-alpha)/sqrt(N), where alpha is the
significance level.
Jack
Dr. Tomsky, thanks so much for the help. I knew I
was wrong.
Is the answer wrong? I got the same result as you.
From:

http://www.itl.nist.gov/div898/handbook/prc/section2/p
rc222.htm
Sample size, N=(z_alpha/2 + z_beta)^2 *
(sigma/margin)^2
In order to minimize N, but keep power at 0.837,
set z_alpha=0 (i.e.
alpha=0.5)

So N=z_beta^2*(s/m)^2

= 0.9822*(5/2.95)^2

= 2.7714
Apologies. That should be 0.9822^2*(5/2.95)^2=2.7714



1. The alternative hypothesis H1 is one-sided, so that alpha/2 in your formula should be alpha.

Oops! Sorry, yes, you are quite correct.

2. While alpha can be chosen to be anything between zero and one, 0.05 is one of the conventional values, not 0.5. With alpha = 0.05, it turns out that N = 19.8, which can be rounded up to 20.

That's what I took to be the "trick" part of the question. Alpha wasn't stated. Certainly we could assume that it's the standard 0.05, but all we are given is that the biologist is after the "smallest sample size in order to have 83.7% power". Both of these conditions are met when alpha=0.5.
.

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