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Before a vote on a new park, a newspaper runs an online poll and gets 50 000 replies, with 70% in…

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

Before a vote on a new park, a newspaper runs an online poll and gets $50\,000$ replies, with $70\%$ in favour. A survey company asks a random sample of 1000 adults, with a high response rate, and finds $52\%$ in favour. For each poll, find the usual luck in percentage points, $2\sqrt{\dfrac{q(100 - q)}{n}}$ , where $q$ is the poll's percentage and $n$ its number of replies. Give both to two decimal places. Then decide which poll to trust and explain why the giant poll's tiny luck does not help it.

Need a reminder? The key ideaShow

When two groups reply at different rates $r_y$ and $r_n$ , the share of the first group among the replies is $\dfrac{p\,r_y}{p\,r_y + (100 - p)\,r_n} \times 100\%$ . A larger sample shrinks the luck, about $\pm 2\sqrt{\dfrac{q(100 - q)}{n}}$ percentage points, but never the bias.

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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