How to check hypothesis about estimation of random variable with unknown distribution? I have one data sample of non-negative random variable $X$ with unknown distribution and predefined expected value $y$. Is there any test able to check null hypothesis $\mathbb{E}[X]\geq y$ or $\mathbb{E}[X]\leq y$?
Actual data samples are gathered in realtime. More specifically, it's an intervals between HTTP-requests coming to web-server from one client. Pearson's test shown that it is not normally distributed variable.
 A: The problem with allowing any distribution is that it could have a tiny chance of yielding a huge value.  That eliminates any possibility of testing the mean with satisfactory confidence.  
Here are the details.  Choose a unit of measurement in which $y$ is hugely greater than $1$. Let $\alpha$ be the desired significance for the hypothesis test ($0 \lt \alpha \lt 1$) and $n$ be the sample size.   Choose any $p$ for which $0 \lt p \lt 1 - \alpha^{1/n}$ and define $\mu = 1 + y/p$.  Consider the two-point distribution for which $1$ has probability $1-p$ and $\mu$ has probability $p$.  The chance that a sample of size $n$ from this distribution consists entirely of $1$s is 
$$(1-p)^n \gt (\alpha^{1/n})^n = \alpha,$$
yet its expectation is
$$1(1-p) + \mu (p) = (1-p) + (1 + y/p)p = y+1 \gt y.$$
Because $y$ can be made arbitrarily large compared to $1$, no hypothesis test of any positive power will conclude that the true mean exceeds $y$ when $n$ $1$s are observed.  Therefore the test will fail to detect that the mean exceeds $y$ with probability greater than $\alpha$ when this two-point distribution is the true distribution.  Because this analysis places no restrictions on $\alpha$ or $n$, this proves that no test with positive power, with any amount of sampling, can achieve any positive level of significance.
A: Intervals between things like requests are often modeled well with exponential, gamma, and Weibull distributions. These can have pretty fat tails, so @whuber's concern is already accounted for, to some extent, when you calculate your confidence intervals. 
A: Let's take an extreme example where you have just one value in your sample.  Then you have no information about the dispersion of $X$ either from your knowledge of the distribution or from your sample and so no way of testing your hypotheses.  
It does not take much to change this: for example if you know that $X$ is always non-negative then, taking the null hypothesis $\mathbb{E}[X] \le y$, by Markov's inequality you have 
$$\Pr(X \geq x) \leq \frac{y}{x}$$
