# Statistical test for a difference in ranked groups

I have an odd problem; I'm computing the axis lengths of ellipsoids, these are sorted based on their length (i.e. longest axis and shortest axis).

The problem is that I want to say the long axes are longer than the short ones, how can I properly do that? The 'long axis' group would be expected to have larger values purely due to the sorting. I imagine there must be a shuffle I can do to test if the difference I see between long and short axes are greater than chance but I can't work out what the shuffle should be...

Should I take random pairs of axis lengths from a shuffled population, rank them and compute the difference and then compare this shuffled difference to the one I observed?

Thanks for any help,

Rod.

• just for clarification, is the long axis/short axis group from sorting? That is, you have n axis length data points, you sort them and then use, for example, median split, to break your data into long axis/short axis group? So for example, if you have some length data of {1,4,6,7}, and you use a mean split, your short axis group is {1,4} and long axis group is {6,7}? – QmmmmLiu Jul 26 '18 at 13:41
• No. I measure the axes of a series of ellipses. Each ellipse can be described by two principal axes. If an ellipse is circular, or near circular, the axes will be essentially equal. If the ellipse deviates from a circle one axis will be larger than the other. So, I measure these for a series of ellipses. I then rank each pair to find the long axis and short axis for each ellipse. How can I say the long group is higher on average than the first? The two groups are dependent and the difference between them could be due to the sorting... – Metioche Jul 26 '18 at 15:27
• The question is, what would be your null hypothesis. It could be that short and long axes are always the same, i.e. your ellipses are actually circles. But then the question arises, why do you see differences in the first place? As far as I can see, that would have to be some kind of measurement error (the ellipses are truly circles, but you can't see it). That means, you have to have an idea how measurement errors arise. Without such information, your question has no answer. – A. Donda Jul 26 '18 at 15:48
• I guess what I'm asking is; if the ellipses are all circles then both axes will be the same (that's the null hypothesis). So, at what point can we say the sample of ellipses are not circles? If every ellipse has a long axis 10 x the length of its short axis it seems pretty clear that they are all elongated sausage shapes. However, I'm unclear how to show that statistically, given that the axes are ranked and sorted based on their length, someone might always say that the difference is just due to that... – Metioche Jul 26 '18 at 16:01

I am not sure if this would be something that works:

Denote the longest and shortest lengths of the axis of an ellipsoids as $s$ and $l$. For example, $s=1$ and $l=2$.

If say, for example, your null hypothesis is that the S and L share the same distribution assuming equal variance. For example, $S=\{1,.5\}$ and $L=\{2,1\}$. Here the first pair is $\{1,2\}$ and the second pair is $\{.5, 1\}$, Hence, 1 and .5 are respectively the short length and 1; 2 and 1 are respectively the long length.

we first create a pooled sample where $P=\{S,L\}$. For example, given $S$ and $L$ above, $P=\{1,.5,2,1\}$.

Then we employ a bootstrap procedure:

• create a bootstrapped sample of $P$, that is, draw $n$ numbers randomly with replacement from $P$ where $n$ is the size of $P$. Denote $P^*$ as the bootstrap sample.

• conduct a $t$-test between the first $\frac{n}{2}$ elements in $P^*$ and the second $\frac{n}{2}$ elements in $P^*$ and record the $t^*$ values (we use * to signify that the $t$ value is from the bootstrapped sample).

After that, we conduct a $t$-test between $S$ and $L$ and obtain the $t$-value. The $p$-value would be $\frac{1+\text{ number of }(|t^*| > |t|)}{m+1}$ where $m$ denotes the number of bootstrapped samples you created.

• I think this approach would work perfectly for two groups of data, as a non-parametric alternative to the t-test. However, I'm not sure it corrects for the fact that S and L are generated by sorting an initial pool of values. How about this; maybe we could take random pairs of numbers, without replacement, from P, sort them to get two groups: S* and L*, and then conduct a t-test on S* and L*, record the t* values and then continue as you describe above... – Metioche Jul 27 '18 at 16:10