# Re: st: sampsi and power

 From sjsamuels@gmail.com To statalist@hsphsun2.harvard.edu Subject Re: st: sampsi and power Date Thu, 13 Aug 2009 07:58:23 -0400

```Ricardo.  I didn't really answer your question about "why".  Let me
illustrated the reason using the .05 p-value case.

Think of one-sided testing for a normal mean H0: m =0, with H1: m=
m1>0, with known SD =1.  The critical value is 1.645.  This is because
P(N(0,1)>1.645) = 0.05.    Suppose you observe z =1.645.  Then you get
p= 0.05 exactly.

If the true mean is m1>0, the the power of the test is
P(N(m1,1)>1.645).    if m1 = 1.645 (the observed value of Z), this is
P(N(1.645,1)>1.645) or the probability that a normal variate is
greater than its mean.  This is exactly 50%.

Similar reasoning applies to other situations,

-Steve

On Tue, Aug 11, 2009 at 12:12 PM, <sjsamuels@gmail.com> wrote:
> -
> There is no surprise here. In fact, if you had seen p exactly equal to
> 0.05, the post-hoc power would have been 50%.  You can work this out
> for yourself from the derivations for power calculations contained in
> most introductory texts.
>
> -Steve
>
>
> On Mon, Aug 10, 2009 at 3:41 PM, Cameron McIntosh<cnm100@hotmail.com> wrote:
>> What does it tell you about sample size if you try and estimate it for 80% power?
>> http://www.ats.ucla.edu/stat/stata/dae/proportionpow.htm
>> Cam
>>
>> ----------------------------------------
>>> Date: Mon, 10 Aug 2009 10:50:39 -0700
>>> From: ovaldia@yahoo.com
>>> Subject: st: sampsi and power
>>> To: statalist@hsphsun2.harvard.edu
>>>
>>> Dear all,
>>>
>>> With this question I will demonstrate my basic misunderstanding of power analyses.
>>>
>>> A journal reviewer has asked for a power calculation of a chi-square 2x2 table from a submitted manuscript. Although I am aware of the futility of doing power analyses after the fact, and will most likely not include it in the manuscript, I was intrigued by the results. The odds ratio from the table is significant 3.284 (95%CI=1.145 to 9.419). Plugging the observed percentages and sample sizes in -sampsi- I get power=0.624. I was expecting the power to be higher. Although I suspect that this is correct, I am having a hard time explaining why I do not get 80% or better?
>>>
>>> Thank you,
>>> Ricardo.
>>>
>>> Ricardo Ovaldia, MS
>>> Statistician
>>> Oklahoma City, OK
>>>
>>>
>>>
>>>
>>>
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>
>
> --
> Steven Samuels
> sjsamuels@gmail.com
> 18 Cantine's Island
> Saugerties NY 12477
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> 845-246-0774
>

--
Steven Samuels
sjsamuels@gmail.com
18 Cantine's Island
Saugerties NY 12477
USA
845-246-0774

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```