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April 11th, 2017, 02:04 PM  #1 
Newbie Joined: Apr 2017 From: London Posts: 1 Thanks: 0  Can anyone point me in the right direction with a stats problem please?
I have a problem remembering the right maths to use for a technical problem I am trying to solve. Is there anyone here that can point me in the right direction please? Here is the problem; Let's say there is a population of 1,000,000 people. Within this population, I want to know what percentage have a certain condition. The proportion of people who have this condition is likely to be in the range of circa 1 in 75. Let's assume that the people with the condition are distributed randomly distributed across the one million population. I would like to work out the number of people I would need to sample to know with a degree of certainty what proportion of people in the population had the particular condition. In terms of the level of accuracy I think would be necessary, if the actual answer was 1 in 75 people, I would be happy to know if the range was between 1 in 70 to 1 in 80. So I am not quite sure where to start to find how many people I need to sample in order to determine the proportion within the population. Any ideas or suggestions on the right approach very much appreciated! Last edited by skipjack; April 11th, 2017 at 09:18 PM. 
April 12th, 2017, 01:18 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 
Ok, you want to estimate the proportion of your population with this condition You have to estimate a range of values that the proportion will fall within with the confidence level you specify. Let the actual proportion be $p$ The sample mean mean will be distributed approximately as a normal random variable with mean $\mu = p$, and standard deviation $\sigma = \dfrac{p(1p)}{\sqrt{N}}$, where $N$ is the number of people you sample. Let's say you want the confidence level to be $\alpha$ Then the interval must contain $\alpha$ of the entire probability mass. So you look up that alpha in your standard normal table and find the associated zscore. Let's say $\alpha =0.95$ Then we must have $0.025$ below our interval and $0.025$ above it since the distribution is symmetric. The zscore of $0.025$ is $z_{low}=1.96$ The upper zscore will just be the negative of this. Our confidence interval is then given by $\begin {align*} CI_{0.95} &= \left[p+z_{low}\sigma,~p+z_{hi}\sigma\right] \\ \\ &= \left[p1.96\dfrac{p(1p)}{\sqrt{N}},~p+1.96 \dfrac{p(1p)}{\sqrt{N}}\right] \end{align*}$ Using your numbers of $p=0.75$ and you'd be happy with an interval of $[0.7,~0.8]$ and using $0.95$ confidence level we have $\sigma = (0.75)(0.25) = 0.1875$ $0.7 = 0.75  1.96 \dfrac{\sigma}{\sqrt{N}}$ $0.05 = (1.96)\dfrac{0.1875}{\sqrt{N}}$ $n\approx 54$ and we must round up to get $n=55$ Here's a table of $n$ for various confidence levels, using $p=0.75,~CI=[0.70,0.80]$ $\left( \begin{array}{cc} \text{Level}&\text{n}\\ 0.5 & 7 \\ 0.75 & 19 \\ 0.9 & 39 \\ 0.95 & 55 \\ 0.99 & 94 \\ 0.999 & 153 \\ \end{array} \right)$ 

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