# 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; 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 1.96sqrt(pi(1−pi)/n)=.01 where π is the sample estimated proportion of people

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; 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
$1.96\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=.01$
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 π 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:
$\pi ±1.96\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=.01$
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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vidovitogv5
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.
- ${\stackrel{^}{\pi }}_{n}$ 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 $Ber\left(\pi \right)$, which we know has a mean of $\mu =\pi$ and a standard deviation of $\sigma =\sqrt{\pi \left(1-\pi \right)}$.
The Central Limit Theorem implies that as $n\to \mathrm{\infty }$, our estimated proportion ${\stackrel{^}{\pi }}_{n}$ 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, ${\stackrel{^}{\pi }}_{n}$ 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:
$Pr\left(|{\stackrel{^}{\pi }}_{n}-\pi |\le 0.01\pi \right)>=0.95$
$|{\stackrel{^}{\pi }}_{n}-\pi |$ 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:
$\begin{array}{rl}Pr\left(|{\stackrel{^}{\pi }}_{n}-\pi |\le 0.01\pi \right)& =Pr\left(-0.01\pi \le {\stackrel{^}{\pi }}_{n}-\pi \le 0.01\pi \right)\\ & =Pr\left(\frac{-0.01\pi }{\sigma }\le \frac{{\stackrel{^}{\pi }}_{n}-\pi }{\sigma }\le \frac{0.01\pi }{\sigma }\right)\\ & =Pr\left(\frac{-0.01\pi }{\sigma }\le Z\le \frac{0.01\pi }{\sigma }\right)\end{array}$
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)
$1.96\sqrt{\frac{\pi \left(1-\pi \right)}{n}}=0.01\pi$
Since you know that $0.2\le \pi \le 0.4$, you can use the worst-case value of $\pi$, which is 0.2 in this case.
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