One-sided chi^2-test? Someone can conduct a t-test and test unidirectional instead of bidirectional. As far as I know, unidirectional testing is not allowed when doing a chi^2-test, correct? If I am right, is it because there is no "directionality" in chi^2-distribution?
Can there be done something similar like chi^2-test that allows unidirectional testing? I am asking because I want to test a concrete hypothesis assuming imbalanced distribution of frequencies in 2x2-design (and I have an expectation which cells are overrepresented).
 A: If you consider the form of the classical Pearson chi-squared statistic, as the sum of (observed frequency $-$ expected frequency)$^2$ / expected frequency, you can see that the sign of the discrepancy between observed and expected is washed out by squaring, making the procedure one-tailed only. However, it is good practice to examine signed residuals, such as the so-called Pearson residuals, to look more closely at the structure underlying lack of fit.
Worth mentioning too is that extremely high P-values, close to 1, can be troubling in some cases, on grounds often informally but well summarized as "too good to be true". A classic case, giving rise to much discussion and repeated analyses over several decades, is whether some of Gregor Mendel's data agree with null hypotheses too closely for comfort. Interpretations here and in other cases range from the null hypothesis genuinely capturing the structure of the data to practices verging on fudging, faking or fraud, or (more politely) a stopping rule for measurement ceasing if results are satisfactory.
A: One of the problems here is that there is no such thing as "the" chi squared test.  There are lots of classical hypothesis tests where the null distribution is a chi-squared distribution.  If you are referring to the classical Pearson test then that does not have this property --- higher values of the test statistic always constitute greater evidence for the alternative hypothesis.
In order to get a test with a rejection region on both sides of the distribution, you would need to have a test where very small and very large values of the test statistic constitute greater evidence for the alternative than values in the "middle" of the distribution.  In theory that is possible, but in all of the tests I've seen that have a chi-squared null distribution, it is not the case.
