0
$\begingroup$

I have the following data

X 1 2 3 4 5…

Y 10 12 13 14 15…

X/Y 10% 16% 23% etc.

How do I find the standard deviation of percentage (last line)? Can I treat the ratio as a normal distribution and apply regular SD formula?

$\endgroup$
  • $\begingroup$ This looks like a standard textbook problem. Please review the tag wiki info of the self-study tag and add it if at all applicable. $\endgroup$ – Glen_b Nov 6 '13 at 23:06
2
$\begingroup$

You should clarify your question to be clear whether you want the sample standard deviation of the collection of percentages or the estimated standard deviation of each percentage.

I'll discuss the second case. You should also clarify the question enough that I can remove some of the 'if's below.

If the Y's are total counts of objects of which the X's are counts of some particular subset (such as X='number of people with red hair in a classroom', and Y='number of people in the classroom'), and if you can assume independence of occurrence of the characteristic being counted in X and if you can assume constant probability of occurrence of that characteristic, ... then conditional on Y, you're in a binomial sampling situation and the estimated s.d. of the fraction X/Y is $\sqrt{\frac{1}{Y} \frac{X}{Y}(1-\frac{X}{Y})}$, which you can convert to percentage terms by multiplying by 100%.

$\endgroup$
-1
$\begingroup$

The classical definition of the standard deviation estimate is independent from the theoretical distribution of the data, so you can perfectly apply it to a set of percentages $$ s = \sqrt{\frac{1}{n-1}\sum_{i=1}^n \left( z_i - \overline z\right)^2 } .$$

Depending on the distribution, you can have other estimates, though, with different properties.

$\endgroup$
  • 1
    $\begingroup$ You should divide by $N$ of $N-1$ within the square root $\endgroup$ – Luis Mendo Nov 6 '13 at 23:09
  • $\begingroup$ @Luis Mendo means $n$ or $n - 1$. (I'm matching the $n$ notation.) $\endgroup$ – Nick Cox Nov 7 '13 at 0:28
  • $\begingroup$ Indeed! Correction made. $\endgroup$ – Vincent Guillemot Nov 7 '13 at 1:14
  • $\begingroup$ It is incorrect that the standard deviation is independent of the distribution of the data. These data are binomial proportions. The binomial distribution has only a single parameter, $p$ which is used to calculate both the mean and the variance. The standard deviation is simple the square root of the variance. There are several other distributions where the mean and standard deviation are related by the same set of parameters. $\endgroup$ – Dalton Hance Feb 26 '16 at 17:44
  • $\begingroup$ My answer, as was the question, was about the estimate, not the distribution parameter. I modified my answer to make it clearer. $\endgroup$ – Vincent Guillemot Feb 29 '16 at 10:28

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.