1
$\begingroup$

Is it correct to calculate the standard deviation of percentages/proportions as you would for four numbers (i.e. non-percentages)?

Thanks for your help @user2974951 @whuber and @asdf. Yes, you're right I was alluding to using the common SD formula. To explain further, my four percentages are from an experiment where we measured the uptake of a chemical by 4 plants. Each plant took up between 40% and 55% of what was applied, therefore the percentage can't ever be >100%. So, if I understand correctly I will need to use the proportion SD which is different to the common SD formula? Thanks for the link to the thread @user2974951

Improve this question
$\endgroup$
9
  • $\begingroup$ Yes. 55% is just 0.55. You can also think in absolute numbers: 1st plant ate 55 g, 2nd 52 g, etc., out of 100 g applied. On average, they ate 53.7 g. Each had the same amount to begin with (100 g), so the average proportion is 53.7/100 = 0.537 = 53.7%. $\endgroup$
    – corey979
    Commented Feb 20, 2019 at 9:03
  • $\begingroup$ Thanks corey979. I guess the question I was wondering is, can you calculate the standard deviation of these four percentages? Or would that be statistically incorrect? $\endgroup$
    – Andrew
    Commented Feb 21, 2019 at 5:43
  • $\begingroup$ @Andrew You can do that, as for whether it is sensible is another matter. SD is really only sensible when you have a normally distributed variable. In your case that seems to be the case. $\endgroup$
    – user2974951
    Commented Feb 21, 2019 at 8:38
  • $\begingroup$ Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. $\endgroup$
    – gung - Reinstate Monica
    Commented Feb 21, 2019 at 12:26
  • 1
    $\begingroup$ @user2974951 SD makes sense, period. It's just a statistic and has nothing inherently to do with normality. $\endgroup$
    – whuber ♦
    Commented Feb 21, 2019 at 13:28

1 Answer 1

Reset to default
1
$\begingroup$

It depends. If you are just expressing a number that can take any value as a percent (for example, relative growth of number of sales), then go ahead! -15% is just -0.15, 133% is just 1.33 and so on.

However, if what you're dealing with is a proportion (as in "73% of students passed the test, ie: a value between 0% and 100%), then you should calculate standard deviation from the binomial distribution, in other words, for a population of n and a proportion of p (measured from 0 to 1), your variance is np(1-p), and your standard deviation is the square root of that number

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ In the proportion case there might be overdispersion, in which case the binomial standard deviation would be to small. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Jun 23, 2020 at 16:10

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.