0
$\begingroup$

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?

$\endgroup$
4
  • 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

0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

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