Thanks - just wondering how do I account for scores of zero i.e. (0%)
using this formula:

lower bound formula would be
= A1-1.96*sqrt(((100%-A1)*A1)/(B1-1))
upper bound formula would be
= A1+1.96*sqrt(((100%-A1)*A1)/(B1-1))

Regardless of the sample size I always get lower bound and upper bound
scores of zero?

Thanks again,

Heather


"Tom Ogilvy" wrote in message ...
the confidence worksheet function assumes a confidence for a mean. It is
unclear from the description given by the OP whether her % is a mean of a
bunch of percentages or if it represents the sample percentage (I assumed
the latter). If so, the sample percentage is actually modelled by a
binomial distribution which can be approximated by a normal distribution for
large samples. Her example really isn't a large sample based on the
"rules", but this isn't the forum to teach statistics.

So back to the confidence worksheet function, one of the inputs is the
standard deviation of the source population. If this is a sample
percentage, then there is no distribution for the source population - just
the population percentage. An estimate of the standard error is calculate
from this sample percentage, but this is not what the confidence worksheet
function is looking for. The standard error is dependent on the sample
size, 1.96 is a constant for 95% confidence interval.

If I take the standard error and multiply it by the squareroot of the sample
size and feed that as the second argument to the confidence function, then
it returns .25303 or 25.3% as I calculated in my post. So I suppose you
could use it with that adjustment.

--
Regards,
Tom Ogilvy



"Kevin Beckham" wrote in message
...
Is it valid to use
=CONFIDENCE(Alpha, Standard_dev, Size),
(returns the confidence interval for a population mean)
i.e.
=CONFIDENCE(0.05, 0.5, 16)
(= 24.5%) ?
Is the 1.96 factor also sample size dependent ?
(My stats are rusty)

Merry Christmas to all

Kevin Beckham

-----Original Message-----
the standard error for your sample percentage is =
sqrt(((100-percentage)*percentage)/n-1)

assume 50% is in A1, 16 in B1

= sqrt(((100%-A1)*A1)/(B1-1))

This comes out to 12.91%

assuming your percentage is normally distributed, then a
95% confidence
interval says you should go +/- 1.96 standard errors from
the mean

50% - (1.96 * 12.91%) as the lower bound and

50% + (1.96 * 12.91%) as the upper bound


(1.96 * 12.91%) = 25.303%

so you lower bound formula would be

= A1-1.96*sqrt(((100%-A1)*A1)/(B1-1))
your upper bound formula would be
= A1+1.96*sqrt(((100%-A1)*A1)/(B1-1))

--
Regards,
Tom Ogilvy


Heather Rabbitt wrote in message
. com...
Hi,

I'm looking for a formula in excel to give me the
maximum and minimum
margin of error at the 95% confidence interval for a
given percentage
and sample size.

For example the percentage may be 50% I have a sample
size of 16 and
using a stat testing program (STATCHCK) I know the
margin of error is
+/- 25% so my maximum would be 75% and my minimum would
be 25%.

My problem is I have over 10,000 numbers to check and I
want to
automate this in excel. I know there is a data analysis
add in excel
but not sure if it can be used to solve my problem.

Any help with my problem would be greatly appreciated.

If you think this should be posted somewhere else
please let me know.

Thanks in advance,

Heather


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