R help - May 2009 - Power function for ratio of lognormal means: two equally valid results? [SEC=Unclassified]

Hi All

This is a general stats problem that I am dealing with using R, so any help is
greater appreciated.

I have two lognormal distributions with means M1 and M2.

If we have:

H0: log(M1/M2)=0

H1: log(M1/M2) !=0 equivalent to log(M1/M2)=log(1+P) where P<0 or P>0.

If we calculate the power for a value of P=0.1 or P=-0.1 (i.e. a 10% difference)
and say assume SE{log(M1/M2)}=0.05, and confidence level= 100(1-alpha),
alpha=0.05, then how is the power function calculated?

As far as I can see we can calculate the power in the two ways given below and
if there is no assumed direction to difference between M1 and M2 are not both
calculations valid?
> # P=-0.1
> qn <- qnorm(p=0.05, mean=0, sd=0.05, lower.tail=T)
> Power.1 <- pnorm(q=qn, mean=log(0.9), sd=0.05, lower.tail=T)
> # P=0.1
> qn <- qnorm(p=0.95, mean=0, sd=0.05, lower.tail=T)
> Power.2 <- pnorm(q=qn, mean=log(1.1), sd=0.05, lower.tail=F)
>
> print(c(Power.1,Power.2))
[1] 0.6780872 0.6030887
>
 So which value should I use? Or would the average of the two values be
appropriate to use? Or is there a fault in my logic? After a quick lit search I
contacted a colleague who has written two stats text books and taught at
University level for many years and he has not come across this problem and
suggested averaging the values. This post is to ask if anyone has come across
this pretty basic problem and has a suggested approach.

thanks

Steve



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On 12-May-09 12:00:50, Steve Candy wrote:
> Hi All
> This is a general stats problem that I am dealing with using R,
> so any help is greater appreciated.
> 
> I have two lognormal distributions with means M1 and M2.
> If we have:
> 
> H0: log(M1/M2)=0
> 
> H1: log(M1/M2) !=0 equivalent to log(M1/M2)=log(1+P) where P<0 or
P>0.
> 
> If we calculate the power for a value of P=0.1 or P=-0.1 (i.e. a 10%
> difference) and say assume SE{log(M1/M2)}=0.05, and confidence level>
100(1-alpha), alpha=0.05, then how is the power function calculated?
> 
> As far as I can see we can calculate the power in the two ways given
> below and if there is no assumed direction to difference between M1 and
> M2 are not both calculations valid?
> 
>> # P=-0.1
>> qn <- qnorm(p=0.05, mean=0, sd=0.05, lower.tail=T)
>> Power.1 <- pnorm(q=qn, mean=log(0.9), sd=0.05, lower.tail=T)
>> # P=0.1
>> qn <- qnorm(p=0.95, mean=0, sd=0.05, lower.tail=T)
>> Power.2 <- pnorm(q=qn, mean=log(1.1), sd=0.05, lower.tail=F)
>>
>> print(c(Power.1,Power.2))
> [1] 0.6780872 0.6030887
>>
>  So which value should I use? Or would the average of the two values be
> appropriate to use? Or is there a fault in my logic? After a quick lit
> search I contacted a colleague who has written two stats text books and
> taught at University level for many years and he has not come across
> this problem and suggested averaging the values. This post is to ask if
> anyone has come across this pretty basic problem and has a suggested
> approach.
> 
> thanks
> Steve
The Power Function is the probability of rejecting H0 for a particular
instance of H1 (defined by a specific non-null value of a parameter),
considered as a function of that parameter.

So the power for H1: +10% is not the same as the power for H1: -10%.

So then you face the problem of what to do about that when you want
to consider "What is the power when H1 is +/-10%?". The answer, in
general terms, is whatever best suits the context in which the problem
arises.

One "general" solution which makes a certain kind of sense is: Adopt
the smaller value of the two results (0.6780872 or 0.6030887). Then
you know that the power is at least 0.6030887 if H1 (either +10% or -10%)
is true. If your objective is to know "at least" what power can
be achieved in the circumstances, then that answers the question. If
0.6030887 is adequate for your needs, then your ivestigation is in
a satisfactory state.

On the other hand, if in the context of the problem, you are concerned
that there may not be enough power, then you may want to know "at
most" what power can be achieved. In that case, you can achieve at most
0.6780872. If that is inadequate for your needs, then you need to
re-consider the investigation as a whole, with a view to increasing
the power..

You might also be considering a situation in which you are prepared
to think as though +/-10% may arise "at random" with probabilities
(P+) and (P-), and are interested in the "average power" as being
representative of what you may expect to achieve overall (even though
this will never be true for any particular case).

In that case, you may reasonably adopt 0.6780872*(P-) + 0.6030887*(P+).
Your colleague's suggestion is equivalent to (P-) = (P+) = 0.5.

Additionally, in such a context, you may have different utilities
for successfully rejecting H0 when H1: -10% is true, versus successfully
rejecting H0 when H1: +10% is true. In that case, you would be
thinking of computing the expected utility (the preceding case is
tantamount to having equal utilities).

Summary: Apply the results in the way that best fits into what you
know about the situation you are investigating, and best meets the
objectives you have in that aituation. There is not a universal
answer to your question!

Hoping this helps,
Ted.

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Date: 12-May-09                                       Time: 14:19:41
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