A dependent variable is regressed against three independent variables across 25 observations. The regression sum of squares is 119.25, and the total sum of squares is 294.45. The following are the estimated coefficient values and standard errors of the coefficients. Coefficient Value Standard error 1 2.43 1.4200 2 3.21 1.5500 3 0.18 0.0818 For which of the coefficients can the hypothesis that they are equal to zero be rejected at the 0.05 level of significance? A) 1 and 2 only. B) 3 only. C) 2 and 3 only. D) 1, 2, and 3. Your answer: C was incorrect. The correct answer was B) 3 only. The values of the t-statistics for the three coefficients are equal to the coefficients divided by the standard errors, which are 2.43 / 1.42 = 1.711, 3.21 / 1.55 = 2.070, and 0.18 / 0.0818 = 2.200. The statistic has 25 − 3 − 1 = 21 degrees of freedom. The critical value for a p-value of 0.025 (because this is a two-sided test) is 2.080, which means only coefficient 3 is significant. I don’t know why I am not getting this. I thought we use a t-test here, which would mean we use n-1 degrees of freedom right? Why are they using n-k-1 degrees of freedom? n-k-1 would be an F-test if I am not mistaken. And also, they didn’t give any critical values from the table, so how are they able to get the p-value of 2.080 out of nowhere? Do we need to memorize them or something? Thanks
They have 25-4=21 dof because of the 3 independent variables, and the intercept. Therefore, there are 4 coefficients. That is not a “p-value” that you are thinking of. The sentence clearly states “The critical value”. This p-value represents the area under the curve in the upper tail, which is half of .05, which is stated in the original question “at the 0.05 level of significance”.
But how do they get the 2.080, because no table was given in the question.
you will be given a t-table or a section of the table on the exam to pull the stat from.
For mult reg the t test has df = n - (k+1), where k is the number of indep var. It helps me remember if I look at the equation and see how many terms there are, if there are 4 b’s then the df = n - (# of b’s). In this q the equation was y = b0 + b1x + b2z + b3w + e, so 4 indep terms and then df = 25 - 4 = 21 An F test has df numerator = k = 3 and denominator = n - ( k+1) = 21. It helps me there if I draw the Anova table… first row k, second row n - ( k+1) and third row is the total and that should be n-1 = (k + n - ( k+1))… It also helps me remember that this requires a num and denom since to calc the test statistic you divide MSR/ MSE, again from the anova row one is msr w/ k and row 2 is mse w/ n - ( k+1). I do need to remember that this is a one-sided test. The p-value mention in the explanation has me confused. The p-value can be determined by looking at the t-table. Find the row w/ 21 df then move to the right until the test statistic falls between two given critical values. Here the test statistic for the second coefficient was 2.07 so the p-value is between 0.05 and 0.025 and since at 0.025 the critical value is 2.08, well I guess they can say that the p-value is essentially 0.025. You could potentially argue that it is a little higher than that… The p-value for the third coef would be different I think. I see that falling between .025 and .01 since the test stat is 2.20 and that falls between 2.08 and 2.518.