Solved – Is a partial F-Test on a model reduced by only one variable valid

f-testmultiple regressionpartial-effect

For a recent project, I used multiple linear regression to model data. I attempted to choose between my initial full model and a reduced model by performing a partial F-test. The models used were the following:

Full model: $\hat{Y}$ = $B_0$ + $B_1$$x_1$ + $B_2$$x_2$ + $B_3$$x_3$

Reduced Model: $\hat{Y}$ = $B_0$ + $B_2$$x_2$ + $B_3$$x_3$

The reduced model has only one covariate removed – $x_1$. The partial F-Test returned a F-value of 2.162 and a P-value greater than 0.1, so I did not reject the null hypothesis and concluded that the addition of $x_1$ in the model may not make any significant linear contribution to the prediction of $Y$.

I was later told by my teacher that I should not run a partial F-Test on a reduced model of only one less variable. The teacher said that not only should I not do such a thing, but also that doing so would be wrong. I was told to instead just compare the two models using each model's $R^2$ values, F-values, and T-values.

I can see how the partial F-test in this case might not provide helpful information, since the null hypothesis it tests is that $B_1 = 0$, which is the same thing that the T-test on the parameter $x_1$ tests (in the full model). [Note: I'm not sure what the exact relationship between this T-value and the partial F-test's F-value is.] Nonetheless, why would it would be considered wrong to use and interpret the results from the partial F-test?

Best Answer

I don't know what course you were taking and what qualifications your teacher has but it sounds like you are doing a likelihood ratio test on nested models and you wind up rejecting the higher order model in favor of the reduced model. This is valid as long as the reduced model is nested within the full model. It doesn't matter if the full model has 1 additional variable or 5. It is also appropriate to take the full model and test whether or not B1 is 0. This would be a t test. Both tests should give consistent results.

If your teacher said what you claim then the teacher is wrong.

Related Solutions

Regression – Estimating $b_1 x_1+b_2 x_2$ Instead of $b_1 x_1+b_2 x_2+b_3x_3$

The issue you need to worry about is called endogeneity. More specifically, it depends on whether $x_3$ is correlated in the population with $x_1$ or $x_2$. If it is, then the associated $b_j$s will be biased. That is because OLS regression methods force the residuals, $u_i$, to be uncorrelated with your covariates, $x_j$s. However, your residuals are composed of some irreducible randomness, $\varepsilon_i$, and the unobserved (but relevant) variable, $x_3$, which by stipulation is correlated with $x_1$ and / or $x_2$. On the other hand, if both $x_1$ and $x_2$ are uncorrelated with $x_3$ in the population, then their $b$s won't be biased by this (they may well be biased by something else, of course). One way econometricians try to deal with this issue is by using instrumental variables.

For the sake of greater clarity, I've written a quick simulation in R that demonstrates the sampling distribution of $b_2$ is unbiased / centered on the true value of $\beta_2$, when it is uncorrelated with $x_3$. In the second run, however, note that $x_3$ is uncorrelated with $x_1$, but not $x_2$. Not coincidentally, $b_1$ is unbiased, but $b_2$ is biased.

library(MASS)                          # you'll need this package below
N     = 100                            # this is how much data we'll use
beta0 = -71                            # these are the true values of the
beta1 = .84                            # parameters
beta2 = .64
beta3 = .34

############## uncorrelated version

b0VectU = vector(length=10000)         # these will store the parameter
b1VectU = vector(length=10000)         # estimates
b2VectU = vector(length=10000)
set.seed(7508)                         # this makes the simulation reproducible

for(i in 1:10000){                     # we'll do this 10k times
  x1 = rnorm(N)
  x2 = rnorm(N)                        # these variables are uncorrelated
  x3 = rnorm(N)
  y  = beta0 + beta1*x1 + beta2*x2 + beta3*x3 + rnorm(100)
  mod = lm(y~x1+x2)                    # note all 3 variables are relevant
                                       # but the model omits x3
  b0VectU[i] = coef(mod)[1]            # here I'm storing the estimates
  b1VectU[i] = coef(mod)[2]
  b2VectU[i] = coef(mod)[3]
}
mean(b0VectU)  # [1] -71.00005         # all 3 of these are centered on the
mean(b1VectU)  # [1] 0.8399306         # the true values / are unbiased
mean(b2VectU)  # [1] 0.6398391         # e.g., .64 = .64

############## correlated version

r23 = .7                               # this will be the correlation in the
b0VectC = vector(length=10000)         # population between x2 & x3
b1VectC = vector(length=10000)
b2VectC = vector(length=10000)
set.seed(2734)

for(i in 1:10000){
  x1 = rnorm(N)
  X  = mvrnorm(N, mu=c(0,0), Sigma=rbind(c(  1, r23),
                                         c(r23,   1)))
  x2 = X[,1]
  x3 = X[,2]                           # x3 is correated w/ x2, but not x1
  y  = beta0 + beta1*x1 + beta2*x2 + beta3*x3 + rnorm(100)
                                       # once again, all 3 variables are relevant
  mod = lm(y~x1+x2)                    # but the model omits x3
  b0VectC[i] = coef(mod)[1]
  b1VectC[i] = coef(mod)[2]            # we store the estimates again
  b2VectC[i] = coef(mod)[3]
}
mean(b0VectC)  # [1] -70.99916         # the 1st 2 are unbiased
mean(b1VectC)  # [1] 0.8409656         # but the sampling dist of x2 is biased
mean(b2VectC)  # [1] 0.8784184         # .88 not equal to .64

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