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FRM Part I · FRM Exam Part I · Regression with Multiple Explanatory Variables

A model of 100 observations with five explanatory variables has an unrestricted R-squared of 0.55. A researcher tests the joint null that the coefficients on two of the variables are both zero. The restricted model has an R-squared of 0.52. The 5% critical value of F(2, 94) is approximately 3.09. Which is the correct computation and conclusion?

F is about 3.13, which exceeds the 5% critical value of 3.09, so the joint null is rejected. The calculation is 0.015 divided by 0.004787, using two restrictions and 94 residual degrees of freedom. The two variables are jointly significant.

  1. AF is about 3.13; fail to reject the null
  2. BF is about 1.57; fail to reject the null
  3. CF is about 6.27; reject the null
  4. DF is about 3.13; reject the null at the 5% levelCorrect

Explanation

Residual df = 100 - 5 - 1 = 94. F = [(0.55 - 0.52)/2] / [0.45/94] = 0.015 / 0.004787 = 3.13. This exceeds 3.09, so the null is rejected at 5%. Forgetting to divide by q gives 6.27, and dividing by q twice gives 1.57. Keeping 3.13 but not rejecting misreads the decision rule.

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