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A charity runs a pilot survey of 200 households, chosen at random, and 60 of them say they would…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A charity runs a pilot survey of $200$ households, chosen at random, and $60$ of them say they would give to a new appeal. Find the margin of error of the pilot's 95% interval, to four decimal places. The charity wants a margin of at most $0.03$ in the full survey. Using the pilot's share as the guess, how many households must the full survey ask? How many times as large as the pilot is that?

Need a reminder? The key ideaShow

The interval for a proportion is $\hat p \pm 1.96\sqrt{\dfrac{\hat p(1 - \hat p)}{n}}$ . It needs a random sample with at least $10$ successes and $10$ failures. For a margin $m$ , $n \ge \left(\dfrac{1.96}{m}\right)^2 \hat p(1 - \hat p)$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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