# Statistical comparison of two means with a range not starting at 0

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

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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
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

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