Inadequate information is given to permit a sensible reply (at least,
without making some rather arbitrary assumptions). Comments below.
On 21 Mar 2003, Ken Mintz wrote:
> Suppose we measure a process and determine that the average is
> 100 +/- 8% with 99% confidence; that is, the average is between
> 92 and 108.
1. Does "%" apply only the the uncertainty measure(s),
or are we to read it as also applying to the "100"
(and also, presumably, the "92" and the "108"?
> We improve the process, and when we measure again, the average
> is 120 +/- 5% with 99% confidence; that is, the average is
> beteen 114 and 126. Assume that a larger number is an improvement.
2. This suggests that "%" does not apply to the average reported:
that "100" and "120" are values, in some unspecified units, on an
unspecified variable. (If it were "120% +/- 5%", the interval in
question would be 115 to 125, not 114 to 126, presumably.)
> What is the percentage of the improvement?
This question cannot be answered without knowing what the *whole* is
with respect to which one could calculate a percentage. So far, all
we see is that the change in the interval between measurements has been
an increase of 20 scale points. Nothing so far indicates whether it be
reasonable to consider this increase a per centum of anything.
(Counterexample: if "100" and "120" represent the temperature, in
degrees Fahrenheit, of a volume of air after passing through a furnace
at a specified flow rate, the difference is 20 degrees Fahrenheit. It
is specifically and emphatically NOT an increase of 20%. If it were
appropriate to refer this result to an absolute temperature scale,
either Kelvin or Rankine, it would arguably be an increase of 4.3%.
But this is, so far as one can tell, irrelevant to your question.)
> And what is the confidence interval, if any, of that percentage
> improvement?
Again, not enough information. We apparently know something about the
uncertainty of measurement: the 8 scale points (from the first
measurement) or the 6 scale points (from the second) may each represent
2.6 standard errors of their respective means (this might be consistent
with the "99% confidence" reported), but the improved precision at the
second measurement might as readily derive from having made more
measurements (i.e., larger sample size) the second time as from
improvements in the process between times.
It would be straightforward to put a confidence interval around the
observed difference of 20 units. Whether any of the numbers involved in
that process could reasonably be called "percentages" of anything would
then have to be dealt with. Of course, there is no (obvious) reason why
you need to report the results as percentages, and in nearly all cases
my recommendation would be to report in terms of the measured variable.
People often know (or think they know) what "100 degrees Fahrenheit" or
"25 miles per gallon" mean, and they can relate to an increase of 20
degrees Fahrenheit or 5 miles per gallon much more readily than to an
increase of 20%.
Again, an example: suppose an automobile, rather carelessly
maintained, is getting 20 miles to the gallon of gasoline. A careful
tune-up and other maintenance is carried out: the automobile now gets
25 miles to the gallon. What is the percentage of "improvement"?
Well, THAT depends on how you want to specify the performance.
From 20 mpg to 25 mpg is, possibly reasonably, an increase of 25%.
But, expressed as fuel consumption in gallons per hundred miles,
the change is from 5 gallons per 100 miles to 4 gallons per hundred
miles: a decrease of 20%.
What then is the "improvement", expressed as a percentage? 20% or 25%?
Percentages, you see, are rather slippery things: often difficult to
interpret (although many people seem to think them easy to understand),
and remarkably difficult to justify in computation, without a lot of
rather hard thinking.
> Do we simply state the improvement of the average (20%)?
We can state the *change* in the average (20 score units). Can't
convert that to a percentage without having some [defensible!] basis for
defining 100%.
> That does not seem right to me. The improvement might be as low
> as 5.6% and as high as 37%.
Not with standard statistical procedures, it mightn't -- not at any rate
for a 99% confidence interval on the difference.
And you haven't said whether the two measurements are at least
conceptually "independent" or unrelated, or whether the measurements,
being taken on the same entities at both times, are in a one-to-one
correspondence that would induce a correlation between measurements.
If this correlation were positive, as is often the case, the uncertainty
in the difference would be less, perhaps considerably less, than would
be the case for uncorrelated measures.
If your measurements are uncorrelated, and with about equal sample sizes
on both occasions, we can use the fact that the variance of a difference
(between two uncorrelated measures) is the sum of the variances of the
two separate measures, and estimate the uncertainty in the difference
(with 99% confidence) as 8^2 (at the first time) + 6^2 (at the second
time) = 10^2. You could then report an increase of 20 scale points,
+/- 10 scale points, with 99% confidence.
> Or do we state the improvement as an average of the deltas of the
> 9 pairs of lows, highs, and averages of the two ranges (20.5%)?
I recommend that you find an elementary statistics text (in your local
library; or, better, in the library of your local college or
university) and consult it on the topic of confidence intervals around
differences in means. If you can find a local statistician to consult,
so much the better.
<snip, results of assorted arithmetic exercises>
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Donald F. Burrill [EMAIL PROTECTED]
56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816
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