1
$\begingroup$

Sorry for the incorrect formatting, my first time posting here.

So I've got the following question - the probability of A is 0.2. Given 5000 trials, what will be the variation from the expected outcome (which is 1000 as far as I understand) with the probability of 0.9128?

I had to ideas - if we assume this falls under the Gaussian distribution then the probability of 0.9128 would mean we have to check the normal distribution tables to find t for Ψ(t) = 0.4564 (since the curve is symmetrical), get the corresponding outcome from t.

But then again shouldn't this scenario fall under the binomial distribution? Then I guess I should find the correct k via the Bernoulli formula, but then I get an equation with k factorial and k in the exponent, which I cant solve.

Given that I'm not simply asking to do everything in my place, and have provided the options I came down to, can anyone advise me what is the correct solution to this?

UPDATE

I really should learn Latex, but in the mean time from the link in the comments I assume what I need is:

z = 1.71 (since error in my case is 0.0872, 1 - 0.0872 / 2 = 0,9564 and from the table the corresponding z value is 1.71)

1.71 * sqrt(0.16/5000) = 0.00967

and the variation then is (0.2 +/- 0.00967) * 5000. Did I get this correctly?

Am I correct?

Improve this question
$\endgroup$
5
  • $\begingroup$ en.wikipedia.org/wiki/Binomial_proportion_confidence_interval $\endgroup$
    – Trisoloriansunscreen
    Commented Mar 8, 2016 at 21:44
  • $\begingroup$ @Tal - there was an answer posted here a little ago using this (which I had not heard of), but after there was a comment that it is not the appropriate method from another person, the answer was deleted. $\endgroup$
    – Rudolfs Bundulis
    Commented Mar 8, 2016 at 22:15
  • $\begingroup$ IMHO your problem is almost the definition of a binomial proportion confidence interval, I wonder what was the criticism about. Two comments: (1) The distribution is not symmetrical (2) you indeed can't compute the limits of an exact confidence interval using a closed form solution, but you can find them by searching (varying k systematically). $\endgroup$
    – Trisoloriansunscreen
    Commented Mar 8, 2016 at 22:32
  • $\begingroup$ @Tal, yes, the comment mentioned that the distribution is not symmetrical. $\endgroup$
    – Rudolfs Bundulis
    Commented Mar 8, 2016 at 22:39
  • $\begingroup$ That's a good reason for avoiding the Gaussian approximation and using instead an exact binomial confidence interval (see the 'Clopper-Pearson interval' on that wikipedia page). Note that there's some issue with the discontinuity of the binomial distribution that will probably preclude you from defining an interval with a probability of exactly 0.9128 (but you can find the smallest interval that holds p>0.9128). $\endgroup$
    – Trisoloriansunscreen
    Commented Mar 8, 2016 at 22:51

1 Answer 1

Reset to default
1
$\begingroup$

Your normal approximation of the binomial confidence interval is correctly calculated. The exact interval (Clopper-Pearson), which is based on the binomial distribution instead of the normal (see the Wikipedia link) is slightly asymmetric and also a bit wider: 0.2-0.0096351 to 0.2+0.0099127 (I calculated it using Matlab's binofit). This discrepancy doesn't look so bad for your example, but it will be worse for more extreme proportions or for smaller samples. For example, zero empirical success rate will lead to a zero width interval, which is clearly wrong for any finite sample of trials. For the empirical proportion of 0.2 observed in a very small sample of trials (e.g. 10), the normal interval will include 0 and extend beyond, which is again a clear contradiction.

Improve this answer
$\endgroup$
1
  • $\begingroup$ ok thanks for the explanations, I will look at the Clopper-Pearson method. $\endgroup$
    – Rudolfs Bundulis
    Commented Mar 9, 2016 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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