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
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