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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