Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.
R version 3.1.1 (2014-07-10) -- "Sock it to Me"
> bl <- c(140, 138, 150, 148, 135)
> fu <- c(138, 136, 148, 146, 133)
> t.test(fu, bl, alternative = "two.sided", paired = TRUE)
Error in t.test.default(fu, bl, alternative = "two.sided", paired = TRUE) : 
data are essentially constant

Then I change just a single character in my fu dataset:

> fu <- c(138, 136, 148, 146, 132)

and it runs...

> t.test(fu, bl, alternative = "two.sided", paired = TRUE)

    Paired t-test

What am I missing here?

share|improve this question
3  
Type bl-fu. Now sd(bl-fu). If it's not obvious, yet, do these: dif=bl-fu then n=length(dif) then mean(dif)/(sd(dif)/sqrt(n))... do you see now? –  Glen_b Aug 23 at 5:35
    
whoops, thanks :) agree with me that the error message could have been more newbie-friendly. So this means that as far as statistics go, there's no need for fancy t.test and its a certainty that for each subject there would be a -2 reduction in the fu compared to the bl? –  ihadanny Aug 23 at 5:46

1 Answer 1

up vote 3 down vote accepted

As covered in comments, the issue was that the differences were all 2 (or -2, depending on which way around you write the pairs).


Responding to the question in comments:

So this means that as far as statistics go, there's no need for fancy t.test and its a certainty that for each subject there would be a -2 reduction in the fu compared to the bl?

Well, that depends.

If the distribution of differences really was normal, that would be the conclusion, but it might be that the normality assumption is wrong and the distribution of differences in measurements is actually discrete (maybe in the population you wish to make inference about it's usually -2 but occasionally different from -2).

In fact, seeing that all the numbers are integers, it seems like discreteness is probably the case.

... in which case there's no such certainty that all differences will be -2 in the population -- it's more that there's a lack of evidence in the sample of a difference in the population means any different from -2.

(For example, if 87% of the population differences were -2, there's only a 50-50 chance that any of the 5 sample differences would be anything other than -2. So the sample is quite consistent with there being variation from -2 in the population)

But you would also be led to question the suitability of the assumptions for the t-test -- especially in such a small sample.

share|improve this answer
    
they are blood pressures in mmHg in a baseline and followup checks, so I'm pretty relaxed about assuming normality and of course non-discreteness. It was just an exercise that showed me how much more powerful is paired-t-test (when available) over non-paired. –  ihadanny Aug 23 at 6:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.