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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 >>> >>> >>> >>> >>> >>> * >>> * For searches and help try: >>> * http://www.stata.com/help.cgi?search >>> * http://www.stata.com/support/statalist/faq >>> * http://www.ats.ucla.edu/stat/stata/ >> _________________________________________________________________ >> Stay in the loop and chat with friends, right from your inbox! >> http://go.microsoft.com/?linkid=9671354 >> * >> * For searches and help try: >> * http://www.stata.com/help.cgi?search >> * http://www.stata.com/support/statalist/faq >> * http://www.ats.ucla.edu/stat/stata/ >> > > > > -- > Steven Samuels > sjsamuels@gmail.com > 18 Cantine's Island > Saugerties NY 12477 > USA > 845-246-0774 > -- Steven Samuels sjsamuels@gmail.com 18 Cantine's Island Saugerties NY 12477 USA 845-246-0774 * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**Re: st: Uninstalling Stata***From:*Sergiy Radyakin <serjradyakin@gmail.com>

**st: sampsi and power***From:*Ricardo Ovaldia <ovaldia@yahoo.com>

**RE: st: sampsi and power***From:*Cameron McIntosh <cnm100@hotmail.com>

**Re: st: sampsi and power***From:*sjsamuels@gmail.com

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