I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

Any ideas?

I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.

In this case, the t-stat is effectively infinite. Can we
then state that the mean pay of men is higher than that
of women at our two standard deviation confidence level?
If so, it seems we cannot report a numeric t-stat - we can
only say "Yes, it's higher."

As always, my apologies if this is unclear. I wish I could
go back to school for a couple of years and study statistics,
so I could know how to handle these special cases.

Any ideas?

-- TRW
_______________________________________
t i m
a t
w i t o r t d o t c o m
_______________________________________

"Tim Witort" <tim_witort@hotmail.com> wrote in message
news:Xns98E696AF5A6B6timwitortwrotethis@216.196.97.131...
I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

Any ideas?

The first thing that I can think off is to use non-parametric statistical
methods. Big words, but actually easier to use than parametric statistics
(in many cases).

Check out:
http://mathworld.wolfram.com/WilcoxonSignedRankTest.html

or look up Wilcoxon Signed Rank Test for further details. The following is
my little shpeel on it:

I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.
[ snip, rest]

On Mar 2, 8:48 am, tim_wit...@hotmail.com (Tim Witort) wrote:
I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.

In this case, the t-stat is effectively infinite. Can we
then state that the mean pay of men is higher than that
of women at our two standard deviation confidence level?
If so, it seems we cannot report a numeric t-stat - we can
only say "Yes, it's higher."

As always, my apologies if this is unclear. I wish I could
go back to school for a couple of years and study statistics,
so I could know how to handle these special cases.

Any ideas?

-- TRW
_______________________________________
t i m
a t
w i t o r t d o t c o m
_______________________________________


Can you give us a bit more info about the procedure? It sounds like
your method has inadvertently taken a sample from a single class of
employee (e.g. executive) hence all the same pay. You may have to
consider a bigger sample, or a re-sample, or think about ways to make
your sample better represent the population.

I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.

In this case, the t-stat is effectively infinite. Can we
then state that the mean pay of men is higher than that
of women at our two standard deviation confidence level?
If so, it seems we cannot report a numeric t-stat - we can
only say "Yes, it's higher."

As always, my apologies if this is unclear. I wish I could
go back to school for a couple of years and study statistics,
so I could know how to handle these special cases.

Any ideas?

On Thu, 01 Mar 2007 16:48:14 -0600, tim_witort@hotmail.com (Tim
Witort) wrote:

I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.
[ snip, rest]

Speaking somewhat generally -
If the Universe of scores only contains two values, then
you have nothing more available than a sign test.

If these numbers are two groups within a larger universe,
such as a dozen sets of pairs of groups being compared for
salary, then it might be more sensible to do something like this -
Pool the data (at least, the ones that are reasonably in the same
range... if you don't want to argue about other differences in
standard deviations); get the common variance; and test each
pair as a contrast using the common variance.

Stratocaster wrote:
"Tim Witort" <tim_witort@hotmail.com> wrote in message
news:Xns98E696AF5A6B6timwitortwrotethis@216.196.97.131...
I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute the t-stat.

Any ideas?

The first thing that I can think off is to use non-parametric
statistical methods. Big words, but actually easier to use than
parametric statistics (in many cases).

Check out:
http://mathworld.wolfram.com/WilcoxonSignedRankTest.html

or look up Wilcoxon Signed Rank Test for further details. The
following is my little shpeel on it:

The Wilcoxon signed ranks test is for matched pairs. The OP's
description suggests two independent samples. So the usual
non-parametric test would be the Mann-Whitney U test, or the equivalent
Wilcoxon rank sum test for independent samples.

On the face of it there is another problem: Rank-based tests are not
well behaved when there are too many ties. However, for the
Mann-Whitney U test, it is between-groups ties that are problematic.
But in this case, all of the ties are within-groups. For more
discussion of this distinction, see John Whittington's post in
Medstats. You can search on the message ID given below by entering it
near the bottom of the Advanced (Groups) Search page.

Message-Id: 5.1.0.14.2.20051104132950.00abe780@pop3.powernet.co.uk

randovaro seemed to utter in news:1172802467.437879.251580@
31g2000cwt.googlegroups.com:

On Mar 2, 8:48 am, tim_wit...@hotmail.com (Tim Witort) wrote:
I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.

In this case, the t-stat is effectively infinite. Can we
then state that the mean pay of men is higher than that
of women at our two standard deviation confidence level?
If so, it seems we cannot report a numeric t-stat - we can
only say "Yes, it's higher."

As always, my apologies if this is unclear. I wish I could
go back to school for a couple of years and study statistics,
so I could know how to handle these special cases.

Any ideas?

-- TRW
_______________________________________
t i m
a t
w i t o r t d o t c o m
_______________________________________

Can you give us a bit more info about the procedure? It sounds like
your method has inadvertently taken a sample from a single class of
employee (e.g. executive) hence all the same pay. You may have to
consider a bigger sample, or a re-sample, or think about ways to make
your sample better represent the population.

You are correct, the samples are all of the current employees
with a given job title in a company's current workforce - but it
is in no way inadvertent.

So there is really no way to get a better/different sample. The
test requires that the workforce be examined at the end of the
company's fiscal year, so I have no leeway in how the sample
is taken. The sample is *all* of the employees with
a given title on the specified date.

-- TRW
_______________________________________
t i m
a t
w i t o r t d o t c o m
_______________________________________

On Thu, 01 Mar 2007 16:48:14 -0600, tim_witort@hotmail.com (Tim
Witort) wrote:

I have an application that compares the mean pay of men and
women using a T-test. The goal being to determine if the mean
pay of men is greater than the mean pay of women at a two standard
deviation confidence level. This has worked well for years, but
I came across an odd sample of pays where all of the men
were paid one salary and all of the women were paid another.
As a result, the variance in both samples was zero. This,
in turn gives me a divide by zero when trying to compute
the t-stat.

I've looked through a couple of statistics books for guidance
on how two samples can be compared when there is no variance,
but none of the books address this case.
[ snip, rest]

Speaking somewhat generally -
If the Universe of scores only contains two values, then
you have nothing more available than a sign test.

If these numbers are two groups within a larger universe,
such as a dozen sets of pairs of groups being compared for
salary, then it might be more sensible to do something like this -
Pool the data (at least, the ones that are reasonably in the same
range... if you don't want to argue about other differences in
standard deviations); get the common variance; and test each
pair as a contrast using the common variance.

As your data come from a census, not a sample, and because there is zero
variation, then in this case I don't see anything to test. You can
state with 100% confidence that the men's pay is greater than the
women's pay, for that year, in that job title cohort. You could perhaps
compare the difference this year to the difference last year, adjusted
for inflation.