Way of measuring students' performance Using their previous results and the results form an aptitude test, students are given a target score. This score is a decimal number between $0$ and $120$. After their final exams they are graded on the same scale (although they receive a letter corresponding to intervals).
I would like to devise a statistical test to compare their target score with their actual score.
My idea is to use the $\chi^2$-test of significance, where the target scores are the expected frequencies $E_i$, and their actual scores are the observed frequencies $O_i$. Say that I have $20$ students, then
$$X^2 = \sum_{i=1}^{20} \frac{(O_i-E_i)^2}{E_i}$$
My null hypothesis, $H_0$, would be that there is no difference in teaching and learning in my college than at the average college. (The target scores come from a team of government statisticians performing statistical analysis on all past outcomes, for all institutions.)
Clearly, if $O_i=E_i$ for all $1 \le i \le 20$ then $X^2=0$, and we can have $0\%$ confidence in rejecting the null hypothesis $H_0$. Let's say that we wanted to be $95\%$ confident that we could reject the null hypothesis, then we would find $\chi_{20}^2(5\%) = 31.41$ and require that $X^2 \ge 31.41$.
I get the degrees of freedom to be the same as the number of students because there are no constraints. The total final scores of the candidates, i.e. $\sum O_i$ isn't fixed.
Does this model seem like it would work, or is it a miss-application of the test?
Can you recommend any other models or any refinements to the current one?
 A: 1) The problem is that the chi-square arises because it's a sum of squares of standardized deviations of (approximately) normally distributed variables. 
The numerator you propose is fine - under the null hypothesis it will be small. The problem arises with the denominator. In the case of sets of Poisson (or multinomial) counts, a sum of squares of standardized deviations will be (or will simplify to) dividing by the expected values.
The $E_i$ in the denominator of the chi-square doesn't seem to apply to your situation. To make it a chi-square test in your problem, you'd need to specify the variance of $O_i-E_i$. 
You seem to be doing this on a per-student basis, so you'd need to have a variance-per-student. You might assume they have equal variance (which I doubt can be true, since the variability in scores getting near the limits of 0 and 120 will be smaller than the variability in scores when they're near the middle.
2) I am also concerned that your choice of statistic might not correspond to a question of interest. What is the underlying question you're trying to answer? Or, more directly, what are the alternatives you're most interested in being able to identify?
A: You can't use $\chi^2$ test here, because it is for counts (frequency) data. $E_i$ in this test is the frequency of observing value $i$. In your case it is a single score of a student, i.e. the outcome of exactly one observation. The motivation for $\chi^2$ test is that you know the probability $P_i$ of an outcome $i$, then you conduct N experiments and observe $O_i$ number of outcome $i$, where $\sum_iO_i=N$, so you compare it to expected frequencies $E_i=N\times P_i$.
UPDATE: If this was USA, then the measurement errors of the test scores would have been available from College Board, they have statistical tables available, such as these. They claim that the measurement error is ~30 points. So, you can use this sort of information to see whether an individual student's score is different from the target score. 


*

*You could also test whether the entire group of students scored differently than the target. In this case the standard deviation of the mean score of N students is $\sigma_N=\sigma/\sqrt{N}$. So, you can get the t-statistics by $t=\frac{\bar{T}-\bar{S}}{\sigma_N}$, where the numerator is a difference between an averages of target scores and the test results. Based on the t-stat you can say whether your test scores are significantly different from the target or not.

*In your case, you don't have the measurement error $\sigma$. You can try to estimate it under reasonable assumptions. The mechanics are simple: $\hat\sigma^2=Var[T_i-S_i]$, where $T_i,S_i$ - target and test scores of individual students. Basically, get the variance of the deviations from the target scores. This will give you an estimate of the measurement errors, which you can plug into the $\hat\sigma_N$ equation to get the estimate of the measurement error of the average class score similar to the first case.
Now, how would you interpret this result? Let's say that you got the class average lower than the target. Does it mean that you are teaching worse than the other schools? It would depend on how the scores are computed. For instance, if it is possible that all colleges had lower scores than the target in entire UK, then it would be possible that your college faired as well as others. On the other hand, if they somehow rescale the test scores so they match the target somehow in average UK, then it's a different story.
A: I wonder if a simple rank sum test for stochastic dominance (or, if the assumptions of same shape and distributions differing only with respect to central location, test for median difference) would work. You have paired observations, and two measures that are not strictly normal (i.e. possible scores do not range from $-\infty$ to $\infty$). Seems a straightforward application. Added advantage that it is implemented in all the major software packages.
A: Well, I'm not sure, but you could wonder if the target score can predict the actual score. I think that a positive correlation between target and actual scores is a reasonable assumption, so you could try $O_i=\alpha + \beta E_i + \varepsilon$.
A toy example in R:
> set.seed(123)
> e <- rnorm(20, 80, 20)
> range(e)
[1]  40.67 115.74
> o <- e - rnorm(20, 20, 10)
> range(o)
[1]  21.29 111.17
> fit <- lm(o ~ e)
> summary(fit)
[...]
Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   -22.73       8.51   -2.67    0.016 *  
e               1.04       0.10   10.38    5e-09 ***

In this example, you get:
$$O_i = -\underset{(8.51)}{22.73}+\underset{(0.10)}{1.04}\;E_i+\varepsilon$$
(standard errors under the estimates.)
This would mean that:


*

*actual scores are lesser than target scores by 22.7 on average;

*high actual scores are slightly more likely when the target score is high.


If a regression doesn't look absurd to you, and if you get a reasonable explication (i.e., a reasonable $R^2$), you could add some predictors, e.g. gender.
