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I was given the following question:

A survey found that 89% of a random sample of 1024 American adults approved of cloning endangered animals. Find the margin of error for this survey if we want 90% confidence in our estimate of the percent of American adults who approve of cloning endangered animals.

I know that for 90% Confidence, $\text{ME}\sim 0.82/\sqrt{n}$.

I attempted using this formula with n equal to both 1024 and (.89)1024. I got 0.025625 and 0.02716, respectively. The answer given for the problem is 1.61%. I do not understand where I went wrong. Perhaps I am using the Margin of Error formula incorrectly?

Thanks. :)

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  • $\begingroup$ the ME should be $1.645$ and not $0.82$ $\endgroup$
    – probabilityislogic
    Commented Apr 19, 2011 at 11:45

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Because you are dealing with proportions, the variance is given by:

$$\frac{p(1-p)}{n}$$

And so the 90% CI ME is equal to $1.645\times \sqrt{\frac{p(1-p)}{n}}=1.645\times \sqrt{\frac{0.89(1-0.89)}{1024}}=0.016$

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