What sample size is needed to ensure a majority?

Question

The results of a sample of voters showed that $55\%$ voted for a given candidate. It was determined that at a confidence level of $0.95$ that candidate would be the winner (i.e. would receive the majority of the votes). What sample size is needed to ensure the accuracy of that statement?

Given the mean, standard deviation, and sample size, I can find a confidence interval using $Z_{1-\alpha/2}$ and without the standard deviation with $t_\alpha(n-1)$.

This question, however, doesn't seem to give the necessary information to use either approach.

Perhaps the possible $5\%$ difference between $55\%$ and "majority" should be used in the calculation, but I don't know how to approach making the correct formulas.

Answer 1

You are on the right track but not quite right. Unfortunately this question is a bit philosophically ambiguous.

If you decide to accept the claim that this candidate won with 95% confidence when the 95% confidence interval for $p$ lies entirely to the right of $0.5$, then you would want to have $0.55-1.96 \sqrt{0.55 \cdot 0.45/n}>0.5$. Here basically the idea is that for true/false data, we get the population standard deviation for free when we assume we know what the mean is.

So with some rounding you want $1.96 \cdot 0.497/\sqrt{n}<0.05$, so $n>\left ( \frac{0.05}{1.96 \cdot 0.497} \right )^{-2} \approx 380$. My rounding is crude so don't take all those numbers as being exactly right.

I claim that the above way of doing things has two philosophical flaws. One is that using confidence intervals is not really the right thing to do here in the first place, since the concept underlying confidence intervals is two-sided hypothesis tests, which aren't appropriate here (since you care about the claim $p>0.5$). The other is that you are estimating the standard deviation from the sample proportion, which is rather "optimistic". This can be seen by considering the extreme case of a sample with all votes for the same candidate. In this case the procedure above will conclude that this candidate won with any sample size and any level of significance.

I would argue that a way to work around these flaws is to use a null hypothesis of $p=0.5$ and run a one-sided test for rejecting this hypothesis when $\hat{p}=0.55$. Thus you reject the null hypothesis if $P(\hat{p} \geq 0.55)<0.05$. Now your test statistic is $\frac{0.55-0.5}{\sqrt{0.5^2/n}}=\frac{0.05}{0.5} \sqrt{n}=0.1 \sqrt{n}$ and you reject the null hypothesis if $P(Z \geq 0.1\sqrt{n})<0.05$, i.e. if $0.1\sqrt{n}>1.65$ so if $n \geq 273$. (Note that this is using the normal approximation but you could do it over again with an explicit binomial calculation, now that you know to look for the cutoff around $n=270$ or so.)

This number is made smaller than the previous number by the fact that we did a one-sided test whereas the confidence interval is effectively doing a two-sided test, but it is made larger by the fact that our estimate for the standard deviation is bigger since we computed it under the assumption $p=0.5$ instead of $p=0.55$.

Answer 2

I found a summary of confidence intervals for proportions and found the following formula:

$$\hat{p} \pm z_{1-\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$

I assume, using $0.55$ for $\hat{p}$ and $0.95$ for $1-\alpha$ is the approach to solving this problem.

Conditions for using the above formula:

$$n\hat{p}\geq5,\ n\left(1-\hat{p}\right)\geq5$$ $$n\ \geq\ 5/0.55$$ $$n\geq9.09$$ $$n\ \geq\ 5/0.45$$ $$n\geq 11.11$$

Plugging in values to the confidence interval formula:

$$0.55 - 1.960\sqrt{\frac{(0.55)(0.45)}{n}} = -1.960$$

$$n \geq 0.15$$

$$0.55 + 1.960\sqrt{\frac{(0.55)(0.45)}{n}} = 1.960$$

$$n \geq 0.48$$

Given all of the constraints on $n$, we can conclude $n$ can be any number larger than $11.11$ (in other words, a minimum sample size of $12$).