# Working out minimum sample size

I have just started a course in statistics and have some general questions that have arisen trying to solve the following question:

A survey organisation wants to take a simple random sample in order to estimate the percentage of people who have seen a certain programme. The sample is to be as small as possible. The estimate is specified to be within 1 percentage point of the true value; $\textit{i.e.}$, the width of the interval centered on the sample proportion who watched the programme should be 1%. The population from which the sample is to be taken is very large. Past experience suggests the population percentage to be in the range 20% to 40%. What size sample should be taken?

I think I have to use this and solve for n $$1.96\sqrt{\frac{\pi (1-\pi)}{n}} = .01$$ where $\pi$ is the sample estimated proportion of people who watch the programme.

Now does this mean that I am 95% sure that I am within 1% accuracy? I also am not aware as to how I can find $\pi$ though I have read I could use the population standard deviation instead and suspect I would have to use that as I am given some information- that the pop proportion is 20%-40%.

Finally in general what is being said here:

$$\pi \pm 1.96\sqrt{\frac{\pi (1-\pi)}{n}} = .01$$

My notes at the moment just say it contains the population mean 95% of the time.... why? I think if I had some graphical understanding of what was going on everything would be much simpler for me.

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Let's get some terminology down here:

• $\pi$ will denote the true percentage of people that have seen a certain programme. We do not know this value, and are trying to estimate it by sampling people and asking them.
• $\hat{\pi}_n$ will denote the estimated value that we will get by asking $n$ people and calculating the proportion. This is a random variable.

If we ask one person at random, we know that our response will follow a Bernoulli Distribution $Ber(\pi)$, which we know has a mean of $\mu = \pi$ and a standard deviation of $\sigma = \sqrt{\pi(1-\pi)}$.

The Central Limit Theorem implies that as $n \to \infty$, our estimated proportion $\hat{\pi}_n$ will approach a Normal Distribution with mean $\pi$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

From here on, we replace "approach" with "is" in the previous sentence, and are basically saying that for a large enough $n$, $\hat{\pi}_n$ is a normal distribution.

Now we can formalize the specific questions you are asking:

What does it mean that I am 95% sure that I am within 1% accuracy? This translates to:

$$Pr(|\hat{\pi}_n-\pi| \leq 0.01\pi) >= 0.95$$

$|\hat{\pi}_n-\pi|$ represents how off you are from the truth, and $0.01\pi$ is the most you want to be off. So you want the probability of this happening to be at least 95%.

Now, to solve this we should normalize the left-hand side:

\begin{align} Pr(|\hat{\pi}_n-\pi| \leq 0.01\pi) &= Pr(-0.01\pi \leq \hat{\pi}_n-\pi \leq 0.01\pi) \\ &= Pr\left(\frac{-0.01\pi}{\sigma} \leq \frac{\hat{\pi}_n-\pi}{\sigma} \leq \frac{0.01\pi}{\sigma}\right) \\ &= Pr\left(\frac{-0.01\pi}{\sigma} \leq Z \leq \frac{0.01\pi}{\sigma}\right) \end{align}

where $Z$ is the standard normal distribution. We know that this probability is equal to $0.95$ when $$\frac{0.01\pi}{\sigma} \approx 1.96$$

So we get to essentially your initial formula (where you interpreted 1% as absolute)

$$1.96\sqrt{\frac{\pi(1-\pi)}{n}} = 0.01\pi$$

Since you know that $0.2 \leq \pi \leq 0.4$, you can use the worst-case value of $\pi$, which is 0.2 in this case.

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Assuming that $n$ is large enough to justify a normal approximation to the binomial, and that other assumptions such as independent and unbiased nature of the sample, in frequentist terms, able to say something like:
Your second expression $\pi \pm 1.96\sqrt{\dfrac{\pi (1-\pi)}{n}} = .01$ does not really mean anything. If your estimate of the proportion is $\hat\pi$ and you have solved your first expression for $n$ then you might be able to say $$\Pr\left(\hat\pi \in \left(\pi - 1.96\sqrt{\frac{\pi (1-\pi)}{n}} , \pi + 1.96\sqrt{\frac{\pi (1-\pi)}{n}} \right) \right) = 0.95$$ but what you may want to say is the similar but reversed $$\Pr\left(\pi \in \left(\hat\pi - 1.96\sqrt{\frac{\hat\pi (1-\hat\pi)}{n}} , \hat\pi + 1.96\sqrt{\frac{\hat\pi (1-\hat\pi)}{n}} \right) \right) = 0.95.$$
Since you do not know $\pi$ or before the sample know $\hat\pi$, you might take a worst case for $\pi$ to work out your sample size. This happens when $\pi=\frac12$ in which case you get $n= 1.96^2 \times {\frac14} / 0.01^2 =9604$.