 # Quant - Hypothesis Testing, Correlations

With regards to hypothesis testing, is the reason why increasing sample size will reduce type 1 and 2 errors at the same time because you are more likely to get means and standard deviations closer to the population parameters? Also, i would really appreciate it if someone could explain to me how the correlation significance formula works. [r* sqrt(n - 2)/sqrt(1-r^2)]. So if I use my understanding of the z test using [mean - population mean/stdev], does this mean 1-r^2 is the variance, so square root of it is standard deviation? But how does the numerator come about? Thanks guys.

Check this out: http://janda.org/c10/Lectures/topic06/L24-significanceR.htm

Question: Suppose the covariance between Y and X is 10, the variance of Y is 25, and the variance of X is 64. The sample size is 30. Using a 5 percent level of significance, which of the following statements is TRUE? The null hypothesis of: A) no correlation cannot be rejected. B) no correlation is rejected. C)significant correlation is rejected. D)significant correlation cannot be rejected. My logic goes like this: the problem asks for a decision whether there is or not correlation, the null hypothesis is that there is not: 1. the calculus of the correlation as being Covariance/(square root of variances product) = 10/(25*64)^(1/2)=0.25, 2. than I determine the coeficient of determination as square of correlation coeficient, 0.25^2=0.0625, meaning that 6.25% of the variation in Y is determined by the variation of X. 3. For the level of significance required, 5%, and the N-2 = 28 degrees of freedom, the t test is 2.048 (one tail) 4. t calculated would be = coeficient of determination*square root of [(N-2)/(1-coef. of determination)] =0.25*(28/(1-0.0625))^(1/2)=1.366 5. Calculated t falls below test t, so the null hypothesis of no correlation cannot be rejected, correct answer is A