Schweser Book 1 page 144. (Hypothesis test for the Correlation Coefficient). The example on this page shows how to compute the test statistic using the sample correlation. t=r*(n-2)^1/2 / (1-r^2)^1/2 Schweser Book 1 page 153. (Hypothesis test for the Correlation Coefficient). This example shows how to calculate the test statistic similiar to what I saw in Level I. t= estimated value - population paramater divided by standard error term Aren’t we doing the same thing in both these examples? (Hypothesis test on a correlation coefficient) Why are the test statistics being calculated differently? What am I missing? Thanks.
testing different things. the first is testing the significance of the correlation coeff as a measure of the degree to which 2 variables move together. second is testing significance of a particular slope or intercept coeff in explaining variations in the dependent variable, in isolation from the other slope or intercept coefficient(s) that may be present in the relationship. The denominator is the std error of that particular coefficient, whereas the denominator in the correlation test relates to the whole relationship.
I like slouiscar’s answer better except if he wasn’t so lazy he would do the al juh bruh
By definition the standard error of the sample correlation coefficient is given by sqrt[(1-r2)/(n-2)]. The null hypothesis is rho=0, alternative hypothesis is rho =/= 0 Thus the t stat is given by (r-0)/standard error of the sample correlation coefficient =[r*sqrt(n-2)]/[sqrt(1-r2)]
lol - By definition? A statistic is a random quantity and as a random quantity it has a distribution called its sampling distribution. In this case, a correlation coefficient is a statistic calculated from two random quantities and you are looking for the standard deviation of its sampling distribution. You can’t just “define” that standard deviation as sqrt((1-r2)/(n-2). Even if you get this, you still have some work to show that r/se® has a t-distribution. That doesn’t quite require that t is normal and that se® is chi-square but is close to that (and is the case here).
Hmmm, then I’m wondering, what about the formula for calculating the variance of a sample, that thing about [1/(n-1)]*sigma(x-x bar)^2? The impression I got was that it was “by definition”? Hmm, and I am not sure how that it has a t distribution comes about too. I am very curious about this too…Can a kind soul give the detailed working pls??
JoeyDVivre Wrote: ------------------------------------------------------- > I like slouiscar’s answer better except if he > wasn’t so lazy he would do the al juh bruh true dat. the sad part is that I would probably enjoy the exercise. oh well back to the LOS
For any distribution, there are “central moments”, i.e., E(X-u)^k. The population variance is a central moment with k =2 so the natural question is how to estimate it. The sample standard variance is a really natural estimator of it. Then you go about proving nice properties of it. Unlike using the sample mean for estimating the population mean, the sample variance has some problems for estimating the population variance and reasonable people can disagree. In particular it is not “admissable” and not maximum likelihood for normality. The sample standard deviation calculated from it is biased. The t-distribution comes about by taking the ratio of a standard normal r.v. and the sqrt of an independent chi-square r.v… Which of course leads to this odd thing with the usual one-sample test of the mean where you have X-bar upstairs and the sum(x[i] - X-bar)^2 downstairs. For normality, they are independent and for nothing else. I know that, can prove it, but it blows my mind once in awhile.
Joey, I did not quite get this: That doesn’t quite require that t is normal and that se® is chi-square but is close to that (and is the case here). Did you mean …that r is normal and that se® is chi-squared? So is that to mean that it might not always be the case that r is normally distributed and that s.e. ® is chi-squared,i.e. these 2 variables can take any other distribution and it is by assumption that r is normally distributed and that s.e.® is chi-square?
In fact, r is certainly not normally distributed or even symmetric unless rho = 0. There is something called a “Fisher z transform” that makes r approximately normally distributed but is not part of CFA curriculum.
woah. ok. i think i have digressed way ttoo much. -__-"