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April 11th, 2017, 02:04 PM   #1
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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.
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April 12th, 2017, 01:18 AM   #2
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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
$p=0.75,~CI=[0.70,0.80]$

$\left(
\begin{array}{cc}
\text{Level}&\text{n}\\
0.5 & 7 \\
0.75 & 19 \\
0.9 & 39 \\
0.95 & 55 \\
0.99 & 94 \\
0.999 & 153 \\
\end{array}
\right)$
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