Hi, In Q5 of the September 2020 CS2 paper A, in the mark scheme for ASET it says we can't use a normal approximation for the binomial as the number of age groups is not sufficiently large. The binomial distribution we're trying to approximate in the Q is Bin(11,0.5), and I thought from CS1 that the test of when we can use a normal approximation for bin(n,p) is as long as n*p is greater than or equal to 5 and n*(1-p) is also greater than or equal to 5. Here n*p = 5.5 and n*(1-p) =5.5 also , so I wasn't sure why this method was not given any marks in the solutions? Also, in this question the signs test is two-sided, is that always/generally the case for signs tests? Thanks in advance!
Hi f12378 That is indeed the rule of thumb we give in CS1. That's a fair point. Looking at this, the approximation seems pretty close for this particular probability: > pnorm(3.5, 5.5, sqrt(11 * 0.5^2)) [1] 0.1139 > pbinom(3, 11, 0.5) [1] 0.1132812 The whole CDF seems pretty well approximated in my view: norm.values = pnorm(0:11 + 0.5, 5.5, sqrt(11 * 0.5^2)) binom.values = pbinom(0:11, 11, 0.5) plot(0:11, norm.values, type = "b") lines(0:11, binom.values, col = "red") points(0:11, binom.values, col = "red", pch = 20) The comment in ASET is just taken from the Examiners' report. I'm afraid I'm not sure why marks weren't awarded in this case. Perhaps assume that if you can interpolate using the tables then they don't want you to use the approximation. The signs test is always two-sided (unless for some reason they say otherwise but that would be quite unusual). We are interested in too many or too few positive deviations, both of which would indicate an issue with the implied distribution of the zx's under H0. Hope this helps Andy
