Power difference between t-test and chi-squared test on the same data Say I have two groups of patients and like to test the association between this grouping and another pathological feature on these patients. Given that the pathological features is nearly continous. I can do a t-test / rank-sum test. Alternatively, I can set a cutoff on the pathlogical feature and make a 2*2 contingency table and apply chi-squared test. 
My question the difference between these to method. My initial guess is that I would loss power due to the lost of continous value in the 2*2 contigency table. Is t-test / rank-sum test always prefered due to the reason above? Or there is some deeper discussion on this matter?
 A: 
I can set a cutoff on the pathlogical feature and make a 2*2 contingency table and apply chi-squared test.

This is perhaps the worst thing you could do.  Binarizing a continuous outcome destroys information (quite literally, there are more bits of information in the continuous variable than there are in a binary variable).  You also leave yourself open to residual confounding if the outcome in your (arbitrarily selected groups) are not completely flat.
I would list out all 10+ issues with this approach (as Frank Harrell does in his excellent course notes) but instead I will just point you to section 2.4.1 of the linked PDF.  As for power, if I recall correctly you are best off if you split at the population median (which is unknown to you anyway) and even then the test has reduced power as compared to a more appropriate test.
Ok, that's enough harping on categorization.
A: You are correct that discretizing your continuous data and using a chi-squared test is probably a bad idea because of the loss of information and ensuing decrease in the power of the test. If you have two groups and independent observations with a more-or-less continuous outcome, then the t test or rank-sum test are two good options, depending what you want to test and what assumptions you're willing to make.
You should read up on them more to decide what you need. If you have a reasonably large sample and simply want to test for a difference in means, the t test is often the best choice. Rank-sum tests the null hypothesis that $P(X>Y) = P(Y>X)$ where $X$ and $Y$ are the random variables representing the two groups. This is a bit of a bizarre hypothesis, but if you have reason to believe the means are similar yet the distributions are differently shaped, it could be a good choice. Kolmogorov-Smirnov or tests for dominance are also options, though they are more difficult to understand.
A: The power of test: t-test > rank-sum test > $2\times2$ table test
Requirements: t-test---strong (Normal, at least near symmetric); rank-sum test --- middle (random variable is continuous); $2\times2$ table test --- weak (following binomial distribution after cutoff)
So according to your confidence on data properties, select the most powerful test.
This principle is very useful for small sample size. When sample size is large, the power is high enough for 3 tests to detect the meaningful difference. When sample size is large, using t-test is always reasonable because of central limit theorem. For small sample size, power of test is important, also the validation of test heavily depends on data meeting the requirements. 
