Hypothesis testing- how to find a p-value? I cannot get my head around a statistics issue. 
So I have run a test with 10 participants as a basic short memory test. It was found that the mean difference in correct answers to the test between before and after treatment was 2.4 with a standard deviation of 0.8. My null hypothesis is that the difference in the wider population would be <2.4 and my alternative is that the difference would be greater or equal to 2.4.
How would I go about getting a p-value in this case?
Any help would be much appreciated
 A: It is more usual in a case like this to let the null hypothesis be that there is no effect of the treatment. The alternative is then that the effect is positive (one-sided test) or that there is an effect at all (two-sided test). 
Now let's assume the difference is zero and normally distributed. The p-value is then how surprised you are to see 2.4 as the actual mean, and 0.8 as the actual standard deviation of your sample. 
Much simplified, we estimate the standard deviation of the mean of such a sample to be 0.8 (over the whole population), we find a mean of 2.4, which is 3 times the standard deviation, so the p value is roughly 0.01 (using the one-sided test).  
That is the amount of samples from a normal distribution with standard deviation 0.8 that exceed the value 2.4, or in the graph below, the surface under the density (the probability), right of the red line. 

It is useful to understand this procedure, but you should be aware that this is just a model, where you assume that everything is normally distributed, there are no selection issues in your data gathering and so on. All of these things are never actually true, and it takes common sense, an understanding of the underlying real-world problem, and statistical understanding to define a level of trust in these findings. Even with all of that expertise, many errors are made. 
A more technical problem is that estimating the standard deviation of your population, and also the mean, actually introduces more uncertainty than is estimated here. 
A: The statistic you have used is $$t = \frac{d - d_0}{sd(d)}$$ which theoretically under the null hypothesis follows the t distribution with $n-1$ degrees of freedom. We suppose that all the basic assumptions are valid for that paired t-test. Then, p-value is the probability $$P[t_{n-1} > t^{obs}] = P[t_{9} > \frac{2.4 - 2.4}{0.8}] = P[t_{9} > 0] $$ which is of course 0.5 because the $t$ distribution is symmetric. In case of not having the value $t^{obs} = 0$, you should compute p-value with a statistical software. In R for example we take
tobs = 0
1-pt(tobs,9)
[1] 0.5

I hope I helped you.
