[prev in list] [next in list] [prev in thread] [next in thread] 

List:       sas-l
Subject:    Re: FPC Correction on Sample Size
From:       Warren Schlechte <Warren.Schlechte () TPWD ! TEXAS ! GOV>
Date:       2014-01-31 22:34:50
Message-ID: 8f37264b48e147c0a90e32d4ad27af3a () BL2PR09MB017 ! namprd09 ! prod ! outlook ! com
[Download RAW message or body]

I think the issue you're running into is that you need more subjects than are \
available to get the precision you desire.

Let's keep this simple.  Say I have 10 subjects, and I want to split them into two \
groups and test whether Trt_A varies from Trt_B by <=1%.  I cannot do this, as with \
10 people, my error will always be larger than I want it to be.  For example, if \
Trt_A has a mean response of 50%, even if I put 9 of the 10 people into Trt_A, my SE \
is still 17%.  I have no ability to differentiate small differences because I have \
too few samples.

Further, I have to question whether the fpc is appropriate under such a situation.  \
Is 10 truly the entire population, or is it really a sample from a much larger \
population?  Recall, the fpc is supposed to shrink the variance down to zero to \
accommodate the idea that as you sample all the population, you arrive at a census; \
you no longer have a sample.  However, in the instance where you are administering an \
experimental treatment, does the fpc make sense?  I would suggest not since the \
response is a random variable, even if applied to all units.  As such, you never get \
0 variance; that is, you never get a census.  If you repeated this experiment several \
times, it would seem unlikely that you would get exactly the same proportional \
response to Trt_A each time.

What I would do is to go back to Proc Power, create a sample of 400, with the split \
that I wanted (say 2:1), and see what power I had to detect different proportional \
differences.

Alternatively, you could create a simulation with some underlying idea of the \
variance around the response (see Agresti and Caffo: Simple and Effective Confidence \
Intervals for Proportions and Differences of Proportions Result from Adding Two \
Successes and Two Failures), and use simulation to estimate power.


HTH,

Warren Schlechte
Learn how you can help Texas State Parks


-----Original Message-----
From: Brian [mailto:b_miner@live.com]
Sent: Friday, January 31, 2014 3:48 PM
To: Warren Schlechte; SAS-L@LISTSERV.UGA.EDU
Subject: Re: FPC Correction on Sample Size

Warren,

The background is conducting a power analysis for an experimental design, where one \
group gets treatment 1 and the other gets treatment 2. The population is actually \
finite as it pertains to a database where there is only a fixed number of applicable \
subjects. The response rates are really small and as such, assuming an infinite \
population (or equivalently, thinking of this as a data generating process) suggests \
a sample size that is many times exhaustive of the "population".

I am not sure I follow your statement about ratio dependence. Can you illustrate?

I don't think #1 works as described above, there is only 400 subjects available (in \
my example). #2  cant work as these are two treatments that are mutually exclusive.


On 1/31/2014 4:24 PM, Warren Schlechte wrote:
> Brian,
> 
> Gotcha.  Missed that the sum of the two>400.
> 
> Seems then that what you're saying is that n1 comes from a population of size \
> Pop_N1, and n2 comes from a population of size Pop_N2, and that Pop_N1+Pop_N2=400. 
> Maybe I need a little more background to this problem.  I would suggest the \
> approach you have taken is not entirely appropriate, for if you keep the same ratio \
> of n1:n2, but change the sample sizes, and then use the formula for FPC to get the \
> corrected n1 and n2, you will see that the ratio of the corrected n1:n2 changes \
> dependent upon n1 and n2. 
> Could you use the formula with the separate Pop_Ns?
> 
> Or, could you do either of these:
> 
> 1. If they are independent could you simply think of sampling twice from the same \
> population?  In such a case, your population for N1=400, and your population for \
> N2=400; your total population = 800. 
> 2. Could you get both pieces of information from a single draw.  Then you could use \
> the larger of the two sample sizes (600), make the correction (241), but from that \
> sample of 241 objects, get both p1 and p2? 
> Warren Schlechte
> 
> 
> -----Original Message-----
> From: Brian [mailto:b_miner@live.com]
> Sent: Friday, January 31, 2014 2:55 PM
> To: Warren Schlechte
> Cc: SAS-L@LISTSERV.UGA.EDU
> Subject: Re: FPC Correction on Sample Size
> 
> Hi Warren,
> 
> That is why I adjusted the total and then reallocated between the two population, \
> versus adjusting each separately --- your two values are actually > 400. (241+172). \
>  I am not sure if this is correct however.
> 
> On 1/31/2014 3:26 PM, Warren Schlechte wrote:
> > According to a formula I found:
> > 
> > N_Adj= N_NotAdj*Pop_N/( N_NotAdj + (Pop_N-1)
> > 
> > So, assuming you got N_NotAdj=300, with Pop_N=400, the N_Adj should be 172; and \
> > for N_NotAdj=600, N_Adj = 241. 
> > HTH.
> > 
> > Warren Schlechte
> > Texas Parks and Wildlife
> > IF Research
> > 
> > -----Original Message-----
> > From: Brian Miner [mailto:b_miner@LIVE.COM]
> > Sent: Thursday, January 30, 2014 12:13 PM
> > Subject: FPC Correction on Sample Size
> > 
> > Hi All,
> > 
> > Since Proc Power does not allow incorporation of a finite population correction \
> > to its output, is the adjustment after the fact simply a matter of adjusting the \
> > total 'n' from proc power? 
> > Specifically, my problem is a power analysis for a test of two
> > independent proportions (n1 and n2). I want to allow for the ratio of
> > n2/n1 to not be 1 (i.e. equal sample sizes). This is easy with proc power \
> > assuming infinite populations. 
> > If the required sample size for population 1 and population 2 end up being, from \
> > proc power, for example: 
> > n1=300
> > n2=600
> > 
> > (as n2/n1  was set to 2)
> > 
> > and the population size is 400 can the adjustment be made after the fact:
> > 
> > N=300+600=900
> > 
> > adjusted_N = 900/(1+(900-1)/400)) = 278
> > 
> > so that adjusted_n2 = 278(2/3) = 186
> > and adjusted_n1 = 92
> > 
> > 
> > Is this correct? Thanks!
> > 
> > 
> 


[prev in list] [next in list] [prev in thread] [next in thread]