The ratio paired t test

Paired data: Differences vs. ratios

The paired t test analyzes the differences between pairs. For each pair, you calculate the difference. Then you calculate the average difference, the 95% CI of that difference, and a P value testing the null hypothesis that the set of differences is sampled from a population where the mean difference is zero.

The paired t test makes sense when the difference is consistent. The control values might bounce around, but the difference between treated and control is a consistent measure of the effect you are looking for.

With some kinds of data, the difference between control and treated is not a consistent measure of effect. Instead, the differences are larger when the control values are larger. In this case, the ratio (treated/control) may be a much more consistent way to quantify the effect of the treatment.

Analyzing ratios can lead to problems because ratios are intrinsically asymmetric – all decreases are expressed as ratios between zero and one, while all increases are expressed as ratios greater than 1.0. A doubling is 2.0. But if you look at it the other way (swap numerator and denominator), the ratio is 0.5. These two values are not symmetrical around 1.0 (no effect). 

Instead it makes more sense to look at the logarithm of ratios. Then no change is zero (the logarithm of 1.0), increases are positive and decreases are negative. The log of 2.0 is 0.301. The log of 0.5 is -0.301. Those two values are symmetrical around 0.0.

The ratio paired t test

A ratio t test averages the logarithm of the ratio of treated/control and then tests the null hypothesis that the mean of those logarithms is really zero.  The P value answers this question:

 

If there really were no differences between control and treated values, what is the chance of obtaining a ratio as far from 1.0 as was observed? If the P value is small, you have evidence that the ratio between the paired values is not 1.0.

The ratio t test also reports the geometric mean of the ratios, and the confidence interval of the geometric mean.

Prism 6 and later can do the ratio t test directly. Just choose that choice in the t test dialog.

With Prism 5 or earlier (or other programs) you need to follow these steps:

1. Starting from your data table, click Analyze and choose Transform. On the Transform dialog, choose the transform Y=log(Y).
2. From the results of this transform, click Analyze and choose to do a t test. On the t test dialog, choose a paired t test. Notice that you are chaining the transform analysis with the t test analysis.
3. Manually compute the antilog of the means of the differences. This is the geometric mean of the ratios.
4. Manually compute the antilog of each end of the confidence interval of the difference between the means of the logarithms. The result is the 95% confidence interval of the geometric mean.

Example

You measure the Km of a kidney enzyme (in nM) before and after a treatment. Each experiment was done with renal tissue from a different animal.

 

Control	Treated	Difference	Ratio
4.2	8.7	4.3	0.483
2.5	4.9	2.4	0.510
6.5	13.1	6.6	0.496

The ratio paired t test reports a P value (two-tailed) of 0.0005. The treatment has an effect that is highly statistically significant.

The geometric mean of the ratio of control/treated is 0.496. In other words, the control values are about half the treated values.

The 95% confidence interval of the ratio extends from 0.463 to 0.531. We are 95% sure that, on average, the control values are between 46% and 53% of the treated values.

You might want to look at the ratios the other way around. If you entered the treated values into column A and the control in column B, the results would have been the reciprocal of what we just computed. It is easy enough to compute the reciprocal of the ratio at both ends of its confidence interval. On average, the treated values are 2.02 times larger than the control values, and the 95% confidence interval extends from 1.88 to 2.16.

If you did a conventional paired t test instead, the P value would have been 0.07. The difference between control and treated is not consistent enough to be statistically significant. This makes sense because the paired t test looks at differences, and the differences are not very consistent. The 95% confidence interval for the difference between control and treated Km value is -0.72 to 9.72, which includes zero.

With these data, the results of a paired t test were unconvincing, or at least somewhat ambiguous. In contrast, a ratio t test produced persuasive results – the treatment doubled the Km of the enzyme, which is unlikely to be due to chance. This makes sense, since the differences between control and treated values were not very consistent, while the ratio is very consistent.