I have a requirement where i want to perform testing of 1000 systems, but I want to limit or pick or sample size since the rest is assumed to have same configuration.

Whats criteria to pick sample size, Like I want to ensure high confidence rate e.g e.g if I pick 300 systems , with 5% or less margin of error the rest will have same error or issues.

How can math help me?

  • 1
    $\begingroup$ What does your performance scale look like? Is it binary, as in bad systems and good systems? $\endgroup$ Aug 3, 2018 at 13:10
  • $\begingroup$ Yes 1 and 0 as pass or fail $\endgroup$
    – asadz
    Aug 3, 2018 at 14:33
  • $\begingroup$ You want to look at the statistical notion of confidence intervals. To answer your question precisely one would need more information on what you mean by $5\%$ here. $\endgroup$ Aug 3, 2018 at 14:47
  • $\begingroup$ 5% means a deviation in results like 5% from remaining assets might fail on the same test. (show different result) $\endgroup$
    – asadz
    Aug 3, 2018 at 15:01

1 Answer 1


Your question is a little vague as it stands. But I hope I can show you how to think about this productively.

Assuming the process is stable so that the probability $p$ of success is constant over time, you could formulate this as a problem about confidence intervals.

First, you have to decide how close $d$ to the true value of $p$ you need to come. Suppose you will test $n$ items and estimate $p$ as $\hat p = X/n,$ where $X$ is the number of Successes among the $n.$ Then a 95% CI for the true $p$ is of the form $$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}}.$$

Then suppose you want $d = .05$ so that $1.96\sqrt{\frac{\hat p(1-\hat p)}{n}} = 0.05.$ Because you can't know in advance what $\hat p$ will be, it customary to take the 'worst-case-scenario' where $\hat p = 1/2.$ (It's the worst case because $\hat p(1-\hat p)$ is largest for $\hat p = 1/2.)$ Then you can find $n$ by solving $1.96\sqrt{1/4n} = .05.$ This is very nearly $1/\sqrt{n} = .05,$ so that $n \approx 400.$

In reporting results of public opinion polls, it is common to say that the 'margin of sampling error' is $1/\sqrt{n},$ where the number of respondents is $n.$

In your problem, you might stop after a hundred tests and re-compute $n$ based on your value of $\hat p_{100}.$ And again after two hundred tests. So if the true $p$ is far from $1/2,$ you might not need to do all of the $400$ projected tests. [If the true $p = .9,$ you may need $n < 200.$]

Notes: (a) A CI of the form $\check p \pm 1.96\sqrt{\frac{\check p(1-\check p)}{\check n}},$ where $\check n = n+4$ and $\check p = (X+2)/\check n,$ is known to be better in terms of achieving a true 95% coverage than CI shown above. But for purposes of planning sample size $n,$ it is simpler and usually OK to use the CI shown above. [Perhaps see this Q&A and its references.] (b) The method (in my last paragraph) of revising $n$ based on intermediate results is formalized as 'sequential analysis', which you can google if you like.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .