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I've become interested in the way various polls and surveys are conducted. I already know something about terms like sample size, margin of error and confidence level. I don't understand why such large sample sizes are required to get low margin of error with small population size.

Let's pretend that I'm conducting a poll about political preferences in the town with population of 1000. I want the poll to reflect preferences only in this town, not the whole country, etc. I made some calculations and it seems that in order to get a result with 2% margin of error and 95% confidence I'd need 707 people in my survey, which seems extremly high (70% of population).

To get 2% margin of error with 95% confidence in the country of 40 million people, I'd need just 2401 people (0,006% of population).

I'd like to ask for a mathematical expaination about why do I need such large sample sizes in small population and why do sample sizes seem to be so similar wheter I have a population of 20000 or 100000000 for example?

Piotr Koller
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  • Have a look at [Why do political polls have such large sample sizes?](https://stats.stackexchange.com/questions/87730/why-do-political-polls-have-such-large-sample-sizes/) - does that answer your question? You might also find relevant [Why should/does(?) statistical sampling work for politics (e.g. Gallup)?](https://stats.stackexchange.com/questions/244567/why-should-does-statistical-sampling-work-for-politics-e-g-gallup) – Silverfish Aug 09 '17 at 08:32
  • You might want to search for "finite population correction" which may be relevant to the issue here. – mdewey Aug 09 '17 at 09:15

1 Answers1

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The variance of a normally distributed population is smaller when the population size $ N $ is larger. This is because the population variance of a distribution is calculated using: $$ {\sigma}^2 = \frac {{\Sigma}(x_i - \bar x)^2}{N} $$

As you can tell, larger $ N $ implies smaller variance. Now, consider the formula for calculating your confidence interval:

$$ \mu \in \lbrack \bar x \pm z \frac {\sigma}{\sqrt n} \rbrack $$

You can see that a smaller population variance will make $z \frac {\sigma}{\sqrt n}$ smaller. This will make your margin of error smaller, so the amount of stray away from the sample mean becomes smaller. In conclusion, a higher level of confidence can be created with a smaller margin of error.

The missing piece of the puzzle is the fact that $ N $ can be treated independently of $ n $. It doesn't matter what proportion of the population you have. Large $ N $ will and large $ n $ will both contribute to a smaller margin of error, regardless of their ratio.

LloydTao
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  • This is incorrect. Variance is independent of population size. In your first formula, the sum expands as fast as $N$ grows. Variance of the mean depends on sample size, though, which is why we want larger samples that yield greater estimation precision of the mean. – Richard Hardy Jul 14 '21 at 06:40