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st: RE: Statistics Question.


From   "FEIVESON, ALAN H. (AL) (JSC-SK) (NASA)" <alan.h.feiveson@nasa.gov>
To   "'statalist@hsphsun2.harvard.edu'" <statalist@hsphsun2.harvard.edu>
Subject   st: RE: Statistics Question.
Date   Thu, 7 Apr 2005 08:28:45 -0500

You could simply do a t-test on the ratios to see if their mean is one - or
do a t-test on the log ratios to see if their mean is zero. Which is
"better" depends on whether the ratios or log ratios are more likely to be
normally distributed. Doing a t-test on the log ratios is the same as doing
a paired t on the log of the original variable.

Al Feiveson

t-test on the ratios:
. ttest r=1

One-sample t test

----------------------------------------------------------------------------
--
Variable |     Obs        Mean    Std. Err.   Std. Dev.   [95% Conf.
Interval]
---------+------------------------------------------------------------------
--
       r |       6    .8761667    .0484282    .1186245    .7516779
1.000655
----------------------------------------------------------------------------
--
Degrees of freedom: 5

                               Ho: mean(r) = 1

     Ha: mean < 1               Ha: mean != 1              Ha: mean > 1
       t =  -2.5570                t =  -2.5570              t =  -2.5570
   P < t =   0.0254          P > |t| =   0.0508          P > t =   0.9746


t-test on the log ratios:
. ttest z=0

One-sample t test

----------------------------------------------------------------------------
--
Variable |     Obs        Mean    Std. Err.   Std. Dev.   [95% Conf.
Interval]
---------+------------------------------------------------------------------
--
       z |       6   -.1399694    .0559768    .1371145   -.2838622
.0039235
----------------------------------------------------------------------------
--
Degrees of freedom: 5

                               Ho: mean(z) = 0

     Ha: mean < 0               Ha: mean != 0              Ha: mean > 0
       t =  -2.5005                t =  -2.5005              t =  -2.5005
   P < t =   0.0272          P > |t| =   0.0545          P > t =   0.9728

. 





-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu]On Behalf Of Thomas J.
Flanagan
Sent: Wednesday, April 06, 2005 10:25 AM
To: statalist@hsphsun2.harvard.edu
Subject: st: Statistics Question.


Listhost:

is there any way to run a paired t-test for data values if one ONLY has the
ratios
of the values, and not the actual values.

The specific problem I am trying to work out is:

McElhoe and Conner (1986) use an instrument called VISIPLUME
to measure UV light. By comparing absorption in clear air
and absorption in polluted air, the concentration of SO2 in
the polluted air can be estimated. The EPA has a standard
method for measuring So2 in air, and we wish to compare the
two methods across a range of air samples. The recorded
response is the ratio of the Visiplume reading to the EPA
standard reading. The six observations on coal plant number
X are: .950, .978, .762, .733, .823, 1.011. If we make the
NULL HYPOTHESIS be that the Visiplume and standard
measurements are equivalent, then the ratios could, with
equal probability, have been observed as their reciprocals.
That is, the ratio of .950, could equally have been 1/.950=
1.053, since the labels are equivalent and assigned at
random. Supposed we take as our summary of the data the sum
of the ratios. We observer .95‚EUR¶+1.011= 5.257. Test (using
randomization methods) the null hypothesis of equivalent
measurement procedures AGAINST the alternative that
Visiplume reads higher than the standard. Report a P Value.

My current work thus far is this>.

I am using a paired t-test.

Ho is the mean of the difference between the Visiplume and
the Standard is 0.

To test this, I use the T statistic, with 5 degrees of
freedom,

d ‚EUR" U/ O‚EUR"(n)^1/2


D represents the mean of the differences between the two
treatments, and u is zero, per the null hypothesis.

N is 6, and the 0 is the pooled variance estimator.

This provides a p value nearby .30

Do you believe this method is accurate, and do you replicate
this p value? I am attempting next to work it out in STATA.


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