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

Nathanael Perkins 2022-09-26 Answered
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 π ( 1 π ) n = .01
where π 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:
π ± 1.96 π ( 1 π ) 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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Answers (1)

vidovitogv5
Answered 2022-09-27 Author has 10 answers
Let's get some terminology down here:
π 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.
- π ^ 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 B e r ( π ), which we know has a mean of μ = π and a standard deviation of σ = π ( 1 π ) .
The Central Limit Theorem implies that as n , our estimated proportion π ^ n will approach a Normal Distribution with mean π and standard deviation σ n .
From here on, we replace "approach" with "is" in the previous sentence, and are basically saying that for a large enough n, π ^ 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:
P r ( | π ^ n π | 0.01 π ) >= 0.95
| π ^ n π | represents how off you are from the truth, and 0.01 π 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:
P r ( | π ^ n π | 0.01 π ) = P r ( 0.01 π π ^ n π 0.01 π ) = P r ( 0.01 π σ π ^ n π σ 0.01 π σ ) = P r ( 0.01 π σ Z 0.01 π σ )
where Z is the standard normal distribution. We know that this probability is equal to 0.95 when
0.01 π σ 1.96
So we get to essentially your initial formula (where you interpreted 1% as absolute)
1.96 π ( 1 π ) n = 0.01 π
Since you know that 0.2 π 0.4, you can use the worst-case value of π, which is 0.2 in this case.
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