**Date:** Thu, 16 Jun 2011 09:19:06 -0400
**Reply-To:** Arthur Kramer <akramer@NJCU.edu>
**Sender:** "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
**From:** Arthur Kramer <akramer@NJCU.edu>
**Subject:** Re: Margin of error redux
**In-Reply-To:** <1308217813784-4494317.post@n5.nabble.com>
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I understand what you said and agree that it is better to report the margin
of error for each question rather than for the survey, as a whole. But when
reporting to most of the members of the upper administration of my
institution, who do not have a statistical background and are accustomed to
seeing the margin of error reported in newspapers, they want to know the
margin of error of my survey. Giving them an explanation similar to the one
you provided leaves them "unsatisfied" with the "statistical jargon."

Arthur Kramer

"Believe half of what you see and none of what you hear."

N. Whitfield
B. Strong

-----Original Message-----
From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of
jmdpulido
Sent: Thursday, June 16, 2011 5:50 AM
To: SPSSX-L@LISTSERV.UGA.EDU
Subject: Re: Margin of error redux

Dear Arthur,

The formula you are talking about 1,96*sqrt[(p*(1-p)/n] comes from the
following:

+- 1,96 is the point in an Standard Normal Distribution (0,1) that
correspond to 95% probability. Thus, 1,96 implies you are approximating a
binomial distribution by a normal distribution, which is right for large
samples (you can check Anderson, Sweeney and Williams: Statistics for
Business Administration and Economics for a prove of this result).

sqrt[p*(1-p)/n] is the Standard Error of a point estimate of a proportion.

The usual margin error reported in samples asume p=50%=(1-p) and N for the
total sample. As you can easily check with an excel file sqrt[p*(1-p)/n] is
a parabol, with a maximum at p=50%. Thus, the SE is maximum at p=50% for any
given N. That's why in many samples they say that the reported margin of
error is calculated under the hypothesis of maximum uncertainty.

However, this is not always the case. On the first hand, some times the
sample is not a pure random sample, so not all the individuals have the same
probability of being sample. If this is the case, this formula will
underestimate the "true" standard error of your estimation of the
proportion.

Secondly, sometimes not all the sample have a "valid" answer. So, your "n"
is not the total sample size "N", but some smaller "n", not counting the
"missing data". Because if you either delete the missing data or impute them
by any method (e.g. regression), you do not have N independent observations,
but an smaller one.
E.g., if you want to calculate the percentage of women that are unemployed,
your "n" is not the total number of people (men & women), but only the women
with valid data for their labour status (which is an n much smaller than the
total N).

Thridly, there is the "finite population" correction that Art Kendall wisely
talks to you about.

Summing up, I prefer to calculate the margin of error for each question,
rather than using the whole sample. This some times gives me a smaller
confidence interval that the reported margin of error for the whole sample,
and some times a bigger one. I also tend to prefer a 99% confidence
interval, specially for large samples, as in large samples everything tends
to be statistically significant.

Don't hesitate to contact me if you have any more queries

Jmdpulido@yahoo.es
PhD student in Applied Economics.

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