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In a voting intention polling, we use a a random sample of 400 people and the two candidates, party A and party B get 32% and 28% respectively. Check whether party A will win against party B with probability 99%.

I assume that, by "probability" they mean "confidence". So the question is, if with the given sample size, we can secure a 99% confidence interval that A will win.

All I managed to find is critical value *z for 99% confidence, $z = 2.58$.

Then we calculate $β = \frac {32}{100}$.

How do we continue?

Thank you very much.

Juan Manuel Prada
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    What does "win" mean here? $.32+.28=.6$ so "Other" got $.4$, beating both $A$ and $B$. Does winning just require $A$ to get more votes than $B$? – lulu Oct 05 '20 at 18:05
  • @lulu: My understanding is that the remaining 40% corresponds to undecided voters, so yes, winning means A gets more votes than B. – Juan Manuel Prada Oct 05 '20 at 18:17
  • Well, then you can quick a quick answer by assuming that exactly $60%$ of the pool will vote for $A,B$. Then, within that group, a random vote will be for $A$ with probability $p=\frac {.32}{.32+.38}$ which gives a simple binomial process (which you can further approximate with a normal if you want to). You can then compute (or estimate) the probability that $A$ gets $>50%$ of the vote from that group. A more careful analysis would require you to consider the possibility that you get more or less than $60%$ of the group to vote...but I'm not sure how important that is. – lulu Oct 05 '20 at 18:28
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    My point is that the information is quite vague. You have to assume, for instance, that the three totals are independent, which is probably not realistic. It could be the case that, say, $B's$ voters are solidly for $B$ while the others are split between $A$ voters and "undecided". Thus a bigger than expected Undecided block is bad news for $A$. Or one could come up with all sorts of possibilities. Given that degree of uncertainty, a few numerical shortcuts might seem negligible. – lulu Oct 05 '20 at 18:41
  • @lulu: thank you for your comments. I am more towards the assumption that the remaining 40% is (proportionally? maybe not) split between A and B. But, like you say, it is quite vague. – Juan Manuel Prada Oct 05 '20 at 18:55
  • Well, the method I sketched assumes that. That is, it assumes that of those who vote, $\frac {.32}{.32+.28}$ vote for $A$. My point wasn't that this assumption is necessarily wrong, but rather that there was no way to either defend it or attack it. It's just a plausible sounding assumption that we make to get a result. – lulu Oct 05 '20 at 19:01

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