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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 $\sigma = \dfrac{p(1-p)}{\sqrt{N}}$, 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 z-score. 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 z-score of $0.025$ is $z_{low}=-1.96$ The upper z-score 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[p-1.96\dfrac{p(1-p)}{\sqrt{N}},~p+1.96 \dfrac{p(1-p)}{\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 $p=0.75,~CI=[0.70,0.80]$ $\left( \begin{array}{cc} \text{Level}&\text{n}\\ 0.5 & 7 \\ 0.75 & 19 \\ 0.9 & 39 \\ 0.95 & 55 \\ 0.99 & 94 \\ 0.999 & 153 \\ \end{array} \right)$ Tags direction, point, problem, stats Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post james777 Calculus 1 February 27th, 2015 02:15 PM mophiejoe Calculus 1 February 18th, 2015 03:49 AM ROSJ61 Advanced Statistics 1 September 25th, 2013 06:01 PM tsl182forever8 Algebra 3 February 12th, 2012 07:45 PM cmmcnamara Trigonometry 2 December 4th, 2009 03:58 PM

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