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I don't know the exact term for this, so googling didn't work. I will explain exactly what I need below.

I want to compare two sets of values such as blood sugar or blood pressure, where the values never start at 0. Please consider the example below:

Update

I compared the duration of motor block after spinal anesthesia. The results are not distributed normally:

Min. 1st Qu. Median Mean 3rd Qu. Max.

75.0 140.0 160.0 157.2 175.8 280.0

90.0 166.2 190.0 193.3 210.0 295.0

And Wilcoxon rank sum test with continuity correction resulted in W = 622.5, p-value = 1.475e-05.

So, is there any special concerns for comparing these groups aside from applying t-test or Mann-Whitney-U, regarding the fact that legal range does not start at 0?

@Peter Flom

Well, after reading your last comment, I tried it on R and saw that the two p values are the same:

tension1<-c(160,180,170,150,145,176,198,200)

deviation1<-c(20,40,30,10,5,36,58,60)

tension2<-tension1+12;deviation2<-deviation1+12

ks.test(tension1,tension2)

Two-sample Kolmogorov-Smirnov test

data: dev1 and dev2 D = 0.375, p-value = 0.6601 alternative hypothesis: two-sided

t.test(tension1,tension2)

Welch Two Sample t-test

data: tension1 and tension2

t = -1.1792, df = 14, p-value = 0.258

t.test(deviation1,deviation2)

Welch Two Sample t-test

data: deviation1 and deviation2

t = -1.1792, df = 14, p-value = 0.258

share|improve this question
    
I think for a t-test, the fact that the range starts at 0 or another value is not really important. What is important, however is that the mean of the two sequences are "similar" (how much is too much is somewhat open to interpretation) and that the data is normally distributed. For the Mann-Whitney U, because it is non parametric, you can use it on not normally distributed data. –  Antoine Vernet Dec 25 '12 at 15:19
2  
No, not that the means are similar, but that the variances are. Although, when means are very far apart, variances are likely to differ; but the you can use e.g. Satterthwaite to fix that. –  Peter Flom Dec 25 '12 at 17:53
    
@PeterFlom Thanks for correcting this. –  Antoine Vernet Dec 25 '12 at 17:57

2 Answers 2

up vote 2 down vote accepted

As far as I know, there is no test of difference in location (e.g. median, mean etc) that depends on either distribution having 0 as part of it. Certainly not t-test or Wilcoxon signed ranks or Mood's median test or bootstrapping.

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Can we say that it is better to compare the deviations from baseline in such cases to negate the effect of distance to 0? As an example, say the blood pressures deviated by 30% in one group, and only 15% in the other, instead of comparing the mean of measured blood pressure levels. –  barerd Dec 25 '12 at 18:42
    
It is not necessary to do this. There is no "effect of distance to 0" to negate. Blood pressure is not, as far as I know, measured in "deviation" but in some unit of pressure. You can compare the central tendencies in whatever units you've got. –  Peter Flom Dec 25 '12 at 20:05

The t-test is apt here. There is no dependency on the range involving zero. So long as $t = \frac{\delta \mu}{s}$ is within range (decided by the chosen p-value) for the given number of degrees of freedom you cannot reject the null hypothesis that the means are equal

share|improve this answer
    
I see the point. However, since you answered so specifically about parametric comparison, and my data is non parametric, does the case "There is no dependency on the range involving zero" apply to non parametric data, too? –  barerd Dec 25 '12 at 16:52

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