# T-test for percentage change

Suppose I have some paired data (for example weight before and after), and I wish to find a confidence interval for the percentage increase in weight, can I simply apply the t-test to the percentage difference (A-B)/A or should I use another test?

• are you intetested in confidence for mean or confidence intrrval of %age change? Moreover, it is not clear for what purpose t-test is desired ? May 17 '16 at 16:15
• @subhashc.davar, I'm interested in the confidence interval for the %age change. May 21 '16 at 7:58
• In addition to the answer below from @FrankHarrell, consider his answer to this related question; you might be better off modeling the weight after as a function of weight before (with both weights perhaps in a log scale) along with your other covariates, and then expressing the final result in percentage terms if still you think that is appropriate. That will allow you to test whether weight behaves additively, proportionately, or as some mixture, as he notes.
– EdM
May 21 '16 at 16:20

Percent change is an improper measure because of asymmetry. If you really think that weight operates proportionally (it usually operates as a mixture of additively and proportionately) then analyze log ratio, which is a symmetric measure. Get a confidence interval for that using standard methods, then anti-log to get fold change and its asymmetric confidence interval. This fold change is a ratio of medians (also a ratio of means in this case, I think).

• I agree that using logarithms is natural and convenient. The "fold change" won't be a ratio of means--but it could be a close approximation. When it is a close approximation, the asymmetry you refer to is likely mild to nonexistent. Regardless, even if percent change is an "improper" measure, if it's the measure needed for the application, then perhaps finding a CI for it should not be dismissed out of hand.
– whuber
May 22 '16 at 20:36
• What I was trying to say is that when a measure is improper on the scale for which it is being used, it should not be used . The correct average of a 100% change and a 50% decrease is 0% change. The math doesn't work. For the lognormal case with equal $\sigma$, the fold change (ratio) is both the ratio of means and the ratio of medians. May 22 '16 at 21:44
• If you're really interested in the percentage increase in something, it's hard to see how the math could not work. There might be concerns about the distribution of observed variations from a (hypothetical, theoretical) true percentage increase, but that increase itself is well defined. It's certainly good to warn naive users of the t-test about the likely failure of related confidence interval procedures when the observations cover a range as wide as -50% to + 100%. Nevertheless, there's nothing inherently wrong either with percentage increases or even with taking their arithmetic means.
– whuber
May 22 '16 at 22:18
• @whuber I respectfully disagree. Taking the arithmetic mean of percent change is not correct. Think about one subject who increases from 1.0 to 2.0 and another starts at 2.0 and decreases to 1.0. The mean % change is +25% which is misleading. May 23 '16 at 3:00

Using normally distributed dataset one can run a t-test for both, the absolute and the percentage change. Have a look here

To keep things simple, if you apply a paired sample t-test for the previously computed %difference, i.e. $\frac{A-B}{A}*100$, then all you can do is a one-sample t-test which will test your new variable against a single value which I suspect would be 0 (0% change) using a one-sample t-test.

You could instead run a paired-samples t-test (assuming that the assumptions of a t-test hold in your dataset) and then express the absolute mean difference which will be computed by any softare running a t-test as a %

• Correct me if I am wrong, but surely percentage change breaks normality, for example if my original value was 5, this can go to 0 which is -100% or to 10 which is +100% Firstly -100% is the lowest possible drop, a normal distribution should not have a maximum or a minimum and secondly due to the nature of percentage change I have found in my datasets the difference between the mean and the upper quartile to be larger than the difference between the mean and the lower quartile.
– par
Apr 2 '16 at 19:55
• Wouldn't one use the Delta method for such a CI ? Jul 8 '19 at 21:50