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Who pays for this blog? I receive no fees, considerations, etc. etc. for the posts on this blog nor do I have any plans to accept any. My salary is paid by Canwest Global Communications Corp. I work for that company as the Ottawa-based National Affairs Correspondent for Canwest News Service. The blog publishing platform used here is called Blogware and it's developed by Tucows Inc. of Toronto, Ontario, Canada. My use of Blogware should not be taken as an endorsement of that company. Like all Blogware users, I do not pay any fees for the use of this service. I participate in program. Google pays me some money and, for that, I give Google some space on this site to display ads. Google sells those ads and Google, not me, decides what advertising content you are seeing. I do not filter these ads and take no responsibility for them. Readers should not assume I endorse any of the products or services advertised here. If you think other disclosures are appropriate in this space, I'd like to hear from you. All of my contact details are always at www.davidakin.com You can read more about this section |
Re: The KLR VU poll in Guelph: Dirty tricks or business development?
by
Stephen Taylor
To state a confidence level of 95% (pollsters will say "accurate 19 times out of 20"), the margin of error is calculated from a simple equation if we assume a normal population distribution (we do). Margin of error at the 95% confidence interval is calculated like so 1.96((0.5(1-0.5))/n)^0.5 or 0.98 divided by the square-root of n where n is the number in the sample which is to be representative of the population.
so, with a sample of 3,400 homes, that's a direct margin of error of (0.98 / (3400)^0.5) = 1.68%.
This formula gives margins of error of 3.1% on polls involving 1000 people which is usually the standard sample size for political polls.
When we wish a sample to be representative of a normally distributed population we state the margin of error. If we want a 19 times out of 20 confidence level we use the equation above. If we say 99 times out of 100 the margin of error equation becomes 1.29/n^0.5 and the margin of error goes up.
The point is, just by stating the sample size and the confidence interval, the margin of error is implicitly stated.
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