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I have some paired data, obtained by measuring the response of subjects to the same stimulus given twice. I would like to show that the response does not change the second time, which means that the ratio between the two responses is 1.

Is it correct to use a one sample t-test to do compare the ratio of the two responses with a theoretical mean of 1?

Some R code to give you an example:

response.1 <- c(100, 2000, 4000, 50000, 3500)
response.2 <- c(120, 1980, 4030, 40900, 3300)
ratio <- response.2 / response.1
t.test(ratio, mu = 1)

EDIT : just a few more details about the measurements I am performing. These are experimental data ($n \sim 30$) coming from cells treated with a certain drug. I am measuring the level of an intracellular response to the drug, and the values I get are the area under the curve of the response, so they are strictly positive. The cells are exposed to the same drug twice and I want to show that the response does not decay over time. The absolute values are extremely variable for both biological and technical reasons.

My rational for choosing to use the ratio was to express the second response as percentage of the first response, hence removing much of the dispersion of the values.

It's important to point out at this point that the absolute values don't have much biological relevance, and that what is interesting in my context is the relative change of intensity in the response.

Nick Cox
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nico
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  • Why not just do a paired test for 0 on the difference? Given your data it looks like the issue might be that they are on different scales and are heteroskedastic? – Josh Magarick Oct 11 '16 at 19:00
  • @JoshMagarick yes, that's exactly the issue. Any particular reason to use difference instead of ratio? – nico Oct 11 '16 at 20:05
  • @nico the sample mean of the ratios of two normally distributed variables is not t-student distributed, even if the two normal variables had the same variance (which is quite clearly not the case here). Thus there's no way to justify the use of the t-statistics with your approach. – DeltaIV Oct 11 '16 at 20:28
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    @DeltaIV on the other hand I bet the original values aren't really normal either, so the premise (of your correct statement) likely doesn't hold – Glen_b Oct 12 '16 at 00:00
  • @Glen_b I wasn't suggesting the user should use the differences, I was explaining why he couldn't use the ratios. But yes, even difference of original data would't work because data don't look normal. If there were more data, then maybe the t-statistics could tend to be t-distributed, even if the original population wasn't normal. But with a sample of size 5...nope. Anyway, I second your point that unless the OP gives more info on the data generating process, it's hard to answer. – DeltaIV Oct 12 '16 at 07:08
  • @DeltaIV Your statement about the mean of the ratios relies on normality, which you won't have. – Glen_b Oct 12 '16 at 08:05
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    @Glen_b ok, poorly worded. "Even if your two responses were normally distributed and had the same variance, which doesn't seem to be the case, the sample mean of the ratios of two normally distributed variables with common variance would not be t-distributed, thus there's no ground on which to use the t-test". Better? – DeltaIV Oct 12 '16 at 09:18

1 Answers1

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(I take it these measures are going to be strictly positive)

I'd suggest that a more likely useful approach would be to consider working on the log scale.

That is, if your interest is on whether the ratio $Y_{2i}/Y_{1i}$ typically differs from $1$, you may be better to look at whether $\log(Y_{2i}/Y_{1i})=\log(Y_{2i})-\log(Y_{1i})$ differs from $0$.

However, it's hard to say much because you explain so little about your measurements and your hypotheses (and the underlying theoretical ideas) in relation to them.

Glen_b
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  • Thanks for the answer Glen, I've updated the question with more details. – nico Oct 12 '16 at 07:54
  • +1. Exactly what I was going to suggest. Note that a merit of this approach is that the interval $Y_2 < Y_1$ maps to negative logarithms and $Y_2 > Y_1$ maps to positive logarithms. In contrast the first maps to a ratio $>1$ but otherwise unbounded while the second maps to a ratio between $0$ and $1$. In all this we're assuming that responses are positive. So skewness of ratios is in broad arm-waving terms much more likely (and much more likely to be problematic) than skewness of their logarithms. But your data are the only check here. So ratios make sense, but look at their logarithms. – Nick Cox Oct 12 '16 at 08:23