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We randomly choose 1000 people and run a survey on whether they prefer Apples or Berries. The results are $71\%$ for Apples and $29\%$ for Berries.

We want to calculate the Margin of error of the poll, with confidence level $90\%$.

My attempt, as I have just started to study probability and statistics (out of personal interest - I am not a student):

For confidence level of $90\%$, we get the value of $z = 1.645$.

So the MOE is $z*\sqrt{\frac{p(1-p)}{n}} = 1.645*\sqrt{\frac{0.71(0.29)}{1000}} = 2.36\%$?

Thank you!

tommik
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Samuel
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1 Answers1

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Yes, it is correct. The Moe is $\pm 2.36\%$ in the sense that the confidence interval at $90\%$ is

$$(71 \pm 2.36) \%$$ for apples and

$$(29 \pm 2.36)\%$$ for berries

If you want to type % you have to type $\text{\%}$ otherwise the symbol will not appear.

tommik
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