Hi, I am currently looking at CP2 Paper 1 April 2022 and have carried out the check on the data to determine if it is safe to say the data provided is indeed from the N(0,1) distribution. My question is about how we determine the theoretical test statistic that we should compare with our observed test statistic. Keeping in line with this exam, lets assume that we wish to carry out the test at a 85% confidence level i.e. 15% significance level (not sure why the exam seems to do this when usually we use a 95% confidence level i.e 5% significance level). Also, that the degrees of freedom is 15. Also, lets assume that my observed test statistic is S and that the theoretical test statistic I wish to compare it to is denoted T. It's been a while since I studied the material about goodness of fit tests, but my understanding was that, given the assumptions above, I would calculate T like this: T = CHISQ.INV.RT(15%, 15) and then if S < T we would say that there is no evidence to reject the null hypothesis. Below is the picture I have in my head when I do this (in a situation where we would NOT reject the null hypothesis that the data is N(0,1)) https://app.gemoo.com/share/image-a...codeId=vz228mRzJdRz0&origin=imageurlgenerator However, the solutions use the following formula: T = CHIINV(85%, 15) I am aware that the CHIINV and the CHISQ.INV.RT functions produce the same results, I'm just confused on why they seem to use the right tail probability as 85% whereas I would've though they would use the right tail probability being 15% (for an 85% confidence level). I am confused about this mainly as solutions seem to be inconsistent across past papers. For example, in Paper 1 April 2020 they seem to use the right tail probability of 5% (which is what I would think to do) but in this paper they seem to use the right tail probability of 85% (instead of 15% as I would think to use) Hopefully I haven't waffled too much can anyone help me with understanding this as I feel like I just cant recall how to do this test properly..
Hi David You're right to be confused about this. The examiners have applied the chi-squared test in a very unusual manner. Normally we would think of a test as being powerful if we make it hard to reject the null hypothesis (eg a significance level of 1% say). Here however we want to accept H0, so a high significance level makes it harder to pass the test. The examiners have used a significance of 85%, which is a very high bar to jump to show consistency with a given distribution. However this inevitably raises the question of whether they really intended the test to be at the 15% significance level. It's possible this was worked out backwards. The p-value of this data is 88%, ie it shows very strong consistency with the expected distribution. This test passes at a significance of 85% but would fail at 90%. This is possibly where the decision to use 85% came from. In short I think you're understanding is correct, and the method as you have stated it would have scored full marks. Hope that helps. Dave
