Mathematical Methods for Physics and Engineering - Matematica.NET

STATISTICS◮Suppose that two classes of students take the same mathematics examination and thefollowing percentage marks are obtained:Class 1: 66 62 34 55 77 80 55 60 69 47 50Class 2: 64 90 76 56 81 72 70Assuming that the two sets of examinations marks are drawn from Gaussian distributions,test the hypothesis H 0 : σ1 2 = σ2 2at the 5% significance level.The variances of the two samples are s 2 1 =(12.8)2 and s 2 2 =(10.3)2 and the sample sizesare N 1 =11andN 2 = 7. Thus, we haveu 2 = N 1s 2 1N 1 − 1 = 180.2 and v2 = N 2s 2 2N 2 − 1 = 123.8,where we have taken u 2 to be the larger value. Thus, F = u 2 /v 2 =1.46 to two decimalplaces. Since the first sample contains eleven values and the second contains seven values,we take n 1 =10andn 2 = 6. Consulting table 31.4, we see that, at the 5% significancelevel, F crit =4.06. Since our value lies comfortably below this, we conclude that there isno statistical evidence for rejecting the hypothesis that the two samples were drawn fromGaussian distributions with a common variance. ◭It is also common to define the variable z = 1 2ln F, the distribution of whichcan be found straightfowardly from (31.126). This is a useful change of variablesince it can be shown that, for large values of n 1 and n 2 , the variable z isdistributed approximately as a Gaussian with mean 1 2 (n−1 2 − n −11 ) and variance12 (n−1 2 + n −11 ). 31.7.7 Goodness of fit in least-squares problemsWe conclude our discussion of hypothesis testing with an example of a goodnessof-fittest. In section 31.6, we discussed the use of the method of least squares inestimating the best-fit values of a set of parameters a in a given model y = f(x; a)for a data set (x i ,y i ), i =1, 2,...,N. We have not addressed, however, the questionof whether the best-fit model y = f(x; â) does, in fact, provide a good fit to thedata. In other words, we have not considered thus far how to verify that thefunctional form f of our assumed model is indeed correct. In the language ofhypothesis testing, we wish to distinguish between the two hypothesesH 0 : model is correct and H 1 : model is incorrect.Given the vague nature of the alternative hypothesis H 1 , we clearly cannot usethe generalised likelihood-ratio test. Nevertheless, it is still possible to test thenull hypothesis H 0 at a given significance level α.The least-squares estimates of the parameters â 1 , â 2 ,...,â M , as discussed insection 31.6, are those values that minimise the quantityN∑χ 2 (a) = [y i − f(x i ; a)](N −1 ) ij [y j − f(x j ; a)] = (y − f) T N −1 (y − f).i,j=11296