Are my data values too small for chi-squared for trend? If so what trend test can I do instead? I'm doing a project about dilated kidneys, these can be arranged into mild, moderate and severe. There are 53 patients in total.
I want to see if there is a trend in the severity of dilatation between those that have an 'adverse outcome' (like surgery, scarring etc) and those who do not. 
Here is my data:
              mild(1)       mod(2)     severe(3)       Total

Adverse outcome   ---0---------------- 4-----------------5--------------------9
No adverse outcome-28--------------10---------------- 6------------------ 44
Total----------------------28--------------14---------------11----------------- 53
Now, i realise these numbers are fairly small. I am new to statistics and initially tried a chi-square test for linear trend using software which came out as P=0.0003. However on reflection, the data values in some of the boxes are very low (and one is zero!) - do the same minimum values apply for chi-square for linear trend as with pearsons chi-square? ie 80% expected values should be above or equal to 5. Nothing online seems to mention any particular threshold for chi-squared for trend. 
If i cannot use this test, please could you let me know of an alternative ASAP? 
thanks a lot! 
 A: For your data, I would use the IOT (Interocular Trauma Test – the results are obviously statistically significant):

But in general, if you want a proper statistical test, I recommend the Wilcoxon–Mann–Whitney test. It’s based on the following idea: If you choose a random ‘adverse outcome patient’ and a random ‘non-adverse outcome’ patient, what’s the probability that that the ‘adverse’ patient will have the most severe dilatation. The null hypothesis is that this probability is 50%.
Here’s some R code to perform the (two-sided version of the) test:
> adv = rep(1:3, times=c(0,4,5))
> no_adv = rep(1:3, times=c(28,10,6))
> 
> library(exactRankTests)
> wilcox.exact(adv, no_adv)

    Exact Wilcoxon rank sum test

data:  adv and no_adv
W = 337, p-value = 0.0001798
alternative hypothesis: true mu is not equal to 0

Since you have lots of ‘ties’, I used the exact version of the test.
But really, you have so much data, at least in one of the groups, that even an ordinary Welch’s t-test would work well. (Note that it’s important that it’s the ‘Welch’ version, which doesn’t assume equal variance in the two groups.) For comparison, it gives a p-value of 0.00013.
For summarising the data, you might consider the concordance measure, the c-index, sometimes known as the AUC. (See my explanation on how to how to interpret and calculate it, which includes a couple of graphical representations.)
