your formula is right on the money, but suppose your problem supplies no
SD - see my recent message in this thread.

"Dennis Roberts" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> this is the typical margin of error formula for building a confidence
> interval were the sample mean is desired to be within a certain distance
of
> the population mean
>
> n = sample size
> z = z score from nd that will produce desired confidence level (usually
> 1.96 for 95% CI)
> e = margin of error
>
> so, typical CI for mu would be:
>
> samp mean +/- z times standard error of mean
>
>   e or the margin of error here is z * stan error of the mean (let me
> symbolize se)
>
>      e = z * se
>
> for 95% CI .. e = 1.96 * se
>
>                 e = 1.96 * (sigma / sqrt n)
>
> now, what n might it take to produce some e? we can rearrange the formula
...
>
>                 sqrt n = (1.96 * sigma) / e
>
> but, we don't want sqrt n ... we WANT n!
>
>                     n = ((1.96 * sigma)/ e) ^2
>
> so, what if we wanted to be within 3 points of mu with our sample mean the
> population standard deviation or sigma were 15?
>
>                    n = ((1.96 * 5) / 3)^2 = about 11 ...
>
> only would take a SRS of about 11 to be within 3 points of the true mu
> value in your 95% confidence interval
>
> unless i made a mistake someplace
>
>
> At 09:54 AM 9/28/01 -0400, Randy Poe wrote:
> >John Jackson wrote:
> >
> > > the forumla I was using was n = (Z?/e)^2  and attempting to express
.05
> > as a
> > > fraction of a std dev.
> >
> >I think you posted that before, and it's still getting
> >garbled. We see a Z followed by a question mark, and
> >have no idea what was actually intended.
> >
> >      - Randy
> >
> >
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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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