Alternative Hypotheses with equal sign I have been having some trouble formalizing the hypotheses for an experiment. My claim, it's not this one, but it basically poses the same problem I am having: 

Studying math in the morning or in the evening has no effect on a
  student's performance.

I know this claim might be a little unconventional. Most examples I have read in books and online are that something either increases or decreases. However, for the research I am currently doing, what's relevant it's to see if there's no change.
For what I have read, the null hypotheses should always have an equality and the alternative hypotheses an inequality. However, in this case, I want precisely the opposite.
So, let's consider M as the test scores of people studying in the morning, and E the tests scores of people studying in the evening.
$$H_0: \overline{M} \neq \overline{E}$$
$$H_1: \overline{M} = \overline{E}$$
This formalization is what I though should be correct, since $H_{1}$ is my claim (what I am trying to demonstrate) and $H_{0}$ is the complement of that. However, it goes against the hypothesis formulation definition.
I think, in the end, what I am asking is how can one formalize hypotheses for an experiment that is designed to show no correlation between two levels of a factor?
Thanks.
 A: Why you can't really do it
The difficulty is in creating the sampling distribution for $M\neq E$. In other words, what would be the probability of seeing the data you observed if $M \neq E$. More specifically: what is the probability of seeing your estimate of the parameter if the hypothesis is true.
The usual example is a normal distribution, and an estimate of the mean. In which case, given the mean and variance under the null you can work out how likely estimating various values of the mean are. When working with means, if $\mu_0$ is the mean under the null hypothesis, and $\hat{\mu}$ is the estimation from your data, you can get a probability distribution $D$, so
$$\hat{\mu} \sim D(\mu_0)$$
this is called a sampling distribution. Your $p$-value, or what have you, will be related the probability of your estimate according to this.
If you cannot work out this distribution, then you can't do this kind of test. Basically, one can only do it when the parameters of your hypothetical distribution have exact values, not a range of values. This can only really be done in a few cases, for example a coin toss where $H_1: \text{not heads} = H_1: \text{tails}$.
What you could do instead
There are a couple of solutions:


*

*Just go with the $H_0: E=M$ and accept that it is weak thing to do. Remember to report the $\beta$ value (assuming that the null isn't rejected). (usually the standard thing to do)

*Use a Bayesian approach.

*Give up on testing completely, and just quantify the difference in an intuitive way. (and use this to supplement 1)

*Formulate a specific an alternative hypothesis $H_1: E=\text{somthing specific}$, and use a different form of testing - a log likelihood test for example.

*There's probably something else...
A: Maybe I did not understand your question correct but I think you should look towards simple/composite hypothesis testing. When you fully specify the population distribution you have a simple hypothesis (for example $x\sim \mathcal{N}(\mu,\sigma)$ with known $\sigma$ so $\mathrm{H_0}:\mu=\mu_0$ and $\mathrm{H_1}:\mu=\mu_1$ where $\mu_0,\mu_1$ are known). Otherwise, when you do not know exactly some parameters it will be a composite hypothesis (for the same example, when $\mathrm{H_0}:\mu=\mu_0$ and $\mathrm{H_1}:\mu>\mu_1$ or $\mathrm{H_1}:\mu<\mu_1$ or $\mathrm{H_1}:\mu\neq\mu_1$).
A: One approach you could try is the "Two One-Sided Tests" (TOST) procedure.  This involves looking at the difference between the mean values for the two populations then testing a joint null hypothesis: that this difference is less than or equal to $-\delta$ OR that this difference is greater than or equal to $\delta$ (for some sufficiently small and positive $\delta$).  If you can reject this joint null hypothesis, then you can conclude that the absolute value of the difference between the means is less than $\delta$.
