At 04:49 PM 9/26/01 +0000, John Jackson wrote:
>re: the formula:
>
> n = (Z?/e)2
>
>
>could you express E as a % of a standard deviation .
>
>In other words does a .02 error translate into .02/1 standard deviations,
>assuming you are dealing w/a normal distribution?
well, let's see ... e is the margin of error ... using the formula for a CI
for a population mean ..
X bar +/- z * stan error of the mean
so, the margin of error or e ... is z * standard error of the mean
now, let's assume that we stick to 95% CIs ... so the z will be about 2 ...
that leaves us with the standard error of the mean ... or, sigma / sqrt n
let's say that we were estimating SAT M scores and assumed a sigma of about
100 and were taking a sample size of n=100 (to make my figuring simple) ...
this would give us a standard error of 100/10 = 10 so, the margin of error
would be:
e = 2 * 10 or about 20
so, 20/100 = .2 ... that is, the e or margin of error is about .2 of the
population sd
if we had used a sample size of 400 ... then the standard error would have
been: 100/20 = 5
and our e or margin of error would be 2 * 5 = 10
so, the margin of error is now 10/100 or .1 of a sigma unit OR 1/2 the size
it was before
but, i don't see what you have accomplished by doing this ... rather than
just reporting the margin of error ... 10 versus 20 ... which is also 1/2
the size
since z * stan error is really score UNITS ... and, the way you done it ...
.2 or .1 would represent fractions of sigma ... which still amounts to
score UNITS ... i don't think anything new has been done ... certainly, no
new information has been created
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_________________________________________________________
dennis roberts, educational psychology, penn state university
208 cedar, AC 8148632401, mailto:[EMAIL PROTECTED]
http://roberts.ed.psu.edu/users/droberts/drober~1.htm
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