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Test several regression coefficients together with joint hypothesis tests, build confidence intervals for them, and interpret the findings.

FRM® Part I · Quantitative Analysis · Regression with Multiple Explanatory Variables · FRM1.QA.8.d

The idea

A joint hypothesis on several coefficients is tested with an F-test. Estimate the unrestricted model and the model with the restrictions imposed. F = [(SSRr − SSRu) ÷ q] ÷ [SSRu ÷ (n − k − 1)]. Here SSRr and SSRu are the residual sums of squares of the restricted and unrestricted models, q is the number of restrictions and k is the number of regressors in the unrestricted model. The same test can be written with R² when the dependent variable is identical. In the overall regression F-test, the number of restrictions q equals the number of regressors k. Reject when F exceeds the critical value with (q, n − k − 1) degrees of freedom. Do not run separate t-tests for a joint hypothesis. Individual t-statistics can all be insignificant while the joint F-test rejects, a typical symptom of multicollinearity. A confidence interval is the estimate plus or minus the critical t value times the standard error, using n − k − 1 degrees of freedom.

Formulas

F-statistic for q joint restrictions

F = [(SSRr − SSRu) ÷ q] ÷ [SSRu ÷ (n − k − 1)]

SSRr
the residual sum of squares of the restricted model
SSRu
the residual sum of squares of the unrestricted model
q
the number of restrictions
n
the number of observations
k
the number of regressors in the unrestricted model

Compare with F(q, n − k − 1). It can be written with R² when the dependent variable is the same; the trap is putting the restricted SSR in the denominator.

A practice question

From the question bank, with the reason each wrong option is wrong.

A risk manager wants to test H0: beta2 = beta3 = beta4 = 0 in a multiple regression. Which procedure is appropriate?

  1. Three individual t-tests at 5 percent, rejecting if any is significantWhy not: Multiple separate tests inflate the overall size and ignore the correlation among the coefficient estimates.
  2. A t-test on the average of the three estimated coefficientsWhy not: Tests a different hypothesis: the average can be zero while individual coefficients are not.
  3. Keeping whichever model has the higher adjusted R2Why not: A model-comparison heuristic, not a test of the stated hypothesis.
  4. An F-test comparing the restricted and unrestricted regressions - correct

A joint hypothesis on several coefficients is tested with an F-statistic that compares the fit of the model with and without the restrictions. Running individual t-tests ignores covariances between estimators and misstates the significance level.

A harder one

An unrestricted regression with 4 explanatory variables and 65 observations has a residual sum of squares of 420. Imposing 2 zero restrictions raises it to 500. The 5 percent critical value of F(2, 60) is 3.15. What is the F-statistic and the conclusion?

  1. F = 2.86; fail to reject the restrictionsWhy not: Divides the increase in SSR by k instead of by the number of restrictions q.
  2. F = 4.80; reject the restrictionsWhy not: Uses the restricted SSR in the denominator instead of the unrestricted SSR.
  3. F = 5.71; reject the restrictions - correct
  4. F = 0.19; fail to reject the restrictionsWhy not: Omits the degrees-of-freedom adjustments, taking the plain ratio of the SSR increase to SSR.

F = [(SSR_r - SSR_u)/q] / [SSR_u/(n - k - 1)] = [(500 - 420)/2] / [420/60] = 40/7 = 5.71. This exceeds 3.15, so the restrictions are rejected.

The hardest tier

A regression with 6 explanatory variables and 120 observations has R2 = 0.45. Dropping 3 of the variables gives R2 = 0.40. The 5 percent critical value of F(3, 113) is 2.68. To test whether the 3 dropped coefficients are jointly zero, the F-statistic and decision are:

  1. F = 1.71; fail to reject the null that they are jointly zeroWhy not: Divides the R2 difference by all k variables instead of the q restricted ones.
  2. F = 3.42; reject the null that they are jointly zero - correct
  3. F = 3.14; reject the null that they are jointly zeroWhy not: Uses 1 minus the restricted R2 in the denominator instead of 1 minus the unrestricted R2.
  4. F = 0.09; fail to reject the null that they are jointly zeroWhy not: Omits the degrees-of-freedom divisors, using the plain ratio of the R2 change to 1 - R2.

Because the dependent variable is the same in both models, F = [(R2_u - R2_r)/q] / [(1 - R2_u)/(n - k - 1)] = (0.05/3) / (0.55/113) = 3.42, which exceeds 2.68. The dropped variables are jointly significant.

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