In the following scenario, what is the table t value to test whether a regressio

Question

In the following scenario, what is the table t value to test whether a regression coefficient is statistically significant at the 05 level (one tailed) for this problem? The personnel department of a large manufacturing firm selected a random sample of 23 workers. The workers were interviewed and given several tests. On the basis of the test results, the following variables were investigated: X2 manual dexterity score, x3 mental aptitude score, and X4 personnel assessment score. Subsequently, the workers were observed in order to determine the average number of units of work completed (Y) in a given time period for each worker. Regression analysis yielded these results Y =-212 + 1.90x2 + 2.00x3 + 0.25x4, R2 = .75. (.050) (.060) (.20) The quantities in parentheses are the standard errors of the regression coeficients. The standard error of the regression is 25, and the standard deviation of the dependent variable is 50. O 1) 1.729 o 2) 2.093 3) 1.725 4) 2.086 O 5) 1.328

Explanation / Answer

To test the significance of a regression coeffecient with a one-sided alternative, we must calculate the t-statistic given by, T = b-est / sqrt(MSE) where MSE is the MSE is the mean square error = SSE / n-p, where SSE is the sum of sqaures of error and follows a chi-square distribution with n-p d.f. where n is the total number of observations and p is the total number of predictors.

b-est is the estimate of beta coeffecient of the respective predictor which is needed to be tested.

Now under H0: the predictor is not significant, T ~ t(n-p) distn. i.e. a t-distn with n-p d.f.

Since the alternative hypotheses is a one-sided hypotheses hence at 0.05 level of significance, we must compare it against t(n-p).0.05 which is the upper 1-0.05 = 0.95 percentile of the t(n-p) distribution. Here n=23 and p=3, so n-p=20. From Biometrika table we must find the 95th percentile of the t(20) distribution whih is equal to 1.725.

From the Biometrika table we have 95th percentile of the t(20) distribution is 1.725.

So answer is 1.725.

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