6
$\begingroup$

Unfortunately, it's a long while since I did statistics and despite reading & research I'm not 'confident' as to how to calculate this correctly.

I would like to know the smallest sample size required in order to have a given confidence level that the sample mean would be with a given % of the population mean.

Whilst arbitrary, the following would be known (can be calculated):

  • population size
  • population mean
  • population standard deviation

Sampling would be without replacement.

Distribution is normal.

For example, for a population of 1,000,000 with a mean of 0.90 and a population standard deviation of 1.32 I would need a sample n to be 99% confident that the sample mean is within 1% of the population mean.

I'm interested in understanding the formula as I have to solve this many times for different populations, different confidence levels, and different margins of error. Thank you.

$\endgroup$
6
$\begingroup$

For example, for a population of 1,000,000 with a mean of 0.90 and a population standard deviation of 1.32 I would need a sample n to be 99% confident that the sample mean is within 1% of the population mean.

Okay.

Sampling would be without replacement.

With a million in the population?

To a first approximation, it doesn't matter enough to be worth worrying about

Actually, turns out in this case it does. I'll do it both without replacement and with. With replacement is simpler, and I do it first.

Distribution is normal.

Don't need it. The sample size will be large enough that with the other assumptions, only really strongly non-normal distributions will have any impact.

Can we assume independence (apart from the effect of sampling without replacement)? e.g. sampling completely at random? I'll take it that we can.

$\mu = 0.90$

$\sigma = 1.32$

Want 'to be 99% confident that the sample mean is within 1% of the population mean'.

i.e. Find $n$ such that $P(|\bar{x}-\mu| < .01\mu) = 0.99$

$\bar{x}-\mu \sim N(0, \frac{\sigma^2}{n})$

99% of a normal distribution is within 2.576 s.d.'s of the population mean (this figure is gettable from normal tables, or using a function in a program. I used R) ` Thus I need $2.576 \times \sigma/\sqrt{n} < 0.01 \mu = 0.009$

Hence $2.576^2 \sigma^2/n < 0.009^2$

Hence $2.576^2 \sigma^2 < n \times 0.009^2$

Or $n > (2.576 \times 1.32/0.009)^2 = 142742.9$

So if $n$ is about 142700, (the means and sd's and normal table values were only accurate to about the same number of figures - only the first 3-4 digits will be meaningful) then the required probability statement should hold.

If we allow for the 'without replacement' the sample size would reduce about 14% percent (google for finite population correction to the variance); other factors are likely to affect you by more than a couple of percent (like not having perfectly random sampling, for one example)


Let's look at the without replacement case using the finite population correction now.

The finite population correction multiplies the variance by a factor $f = \frac{N-n}{N-1} = 1-\frac{n-1}{N-1}$.

Some people approximate this by $1 -\, n/N$, which is easily accurate enough with the large numbers for $n$ and $N$ involved here. However, I'll try to do the first version there.

$2.576^2 \sigma^2 (N-n)/(N-1) < n \times 0.009^2$

$(2.576\sigma/0.009)^2 /(N-1) < n/(N-n) $

$(2.576\sigma/0.009)^2 /(N-1) < 1/[N/n\,\,\, -1] $

$142743 \times 1000000/1142742 < n$

So (if I did that right), $n > 124912.7$

Or to the accuracy in the normal value, $n$ should be about $124900$.

(assuming the mean and s.d. are actually accurate to at least 4 figures, too)

Calculation check:

Interval half-width =

$(2.576\times 1.32/\sqrt{124900})\sqrt{(1000000-124900)/999999}$

$= 0.00900$

$\endgroup$
  • $\begingroup$ Thank you so much Glen. I think I follow most of that and will practise with some examples. Given that I told you I was very rusty, how do I represent this in the form 'n = ...." The equation n = ... will help me test my understanding in Excel. Thank you again. $\endgroup$ – Clair.Gibbon Mar 26 '13 at 9:22
  • $\begingroup$ Clair - I screwed up the calculation, sorry, leaving out the 2.576 in later calculation. Please look again. Because of that the finite population correction matters, too. $\endgroup$ – Glen_b -Reinstate Monica Mar 26 '13 at 9:37
  • $\begingroup$ Thanks very much for the update & explanation Glen. That is much closer to the answer I was expecting. Just to confirm, this example allows the sample mean to be plus or minus 1% of the population mean? $\endgroup$ – Clair.Gibbon Mar 26 '13 at 10:56
  • $\begingroup$ Just saw your most recent edits. The detail you have gone to is wonderful. Thank you. I am impressed by how far the sample size falls for finite populations. Terrific. Thanks, Clair. $\endgroup$ – Clair.Gibbon Mar 26 '13 at 11:02
  • $\begingroup$ How far it falls depends on how big a fraction of the population you'd be sampling. If you're only sampling 1% of the population, it makes about 1% difference to the sample size. If you're sampling about 10% of the population, the FPC makes roughly 10% difference to the answer. $\endgroup$ – Glen_b -Reinstate Monica Mar 26 '13 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.