Home > Statistics Every Writer Should Know > The Stats Board > Discusssion
margin of error
Message posted by alan m. seid on August 24, 2000 at 12:00 AM (ET)
given that the sample size is
1223 and the degree of
confidence is 95% what is the
margin of error for the
proportion?
READERS RESPOND:
(In chronological order. Most recent at the bottom.)
Re: margin of error
Message posted by nancy diehl on August 28, 2000 at 12:00 AM (ET)
You need to know what your proportion estimate, p, is. If you don't know, I suppose you could
let it equal .5. The formula for margin of error for a proportion is
Z*(SQRT(pq/n)) where Z is 1.96 for 95% confidence, q=(1-p) and your n is 1223.
Re: margin of error
Message posted by JG on September 6, 2000 at 12:00 AM (ET)
when p in not known you can assume .5 - worst case - or use a preliminary estimate or if a range is known use value closest to .5 such as if .1
Re: margin of error
Message posted by Richard Lowry on September 8, 2000 at 12:00 AM (ET)
Re: 'margin of error' and related matters, see
http://faculty.vassar.edu/~lowry/polls.html
This site also contains several calculators by which the visitor can
*calculate margins of error
*estimate the sample size of a poll when that information is not provided, and
*assess the statistical significance of the difference between the results for one candidate in two separate polls or between two candidates in a single poll.
Re: margin of error
Message posted by Austin Stevens on November 30, 2000 at 12:00 AM (ET)
Here's how I calculated it???E = z.025(p hat)*(q hat)/n
E = 1.96 SQRT [(.5)(.5)/1223]
E = 1.96 * .0143
E = .0280 = 2.8%
Rounded up to 3%
Confidence Interval
.500-.030.47047%
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