I want to know if I am correctly implementing the answer given by Michael Chirico here.

The dataset can obtained using the following code in R:

data = fread(paste0("https://www1.aucegypt.edu/faculty/hadi/RABE5/Data5/P060.txt", "P060.txt"))

I want to test $$H_0: \beta_1 = \beta_3 =0.5$$ using the model $$Y = \beta_0 + \beta_1 X_1 + \beta_3 X_3 + e.$$

In the answer from the link above, we can obtain an equivalent null hypothesis and a new model,

$$H_0: \alpha_1 = \alpha_3 = 0,$$

where the new model is

$$\begin{align*} Y - 0.5(X_1 + X_3) &= \beta_0 + (\beta_1 - 0.5)X_1 + (\beta_3 - 0.5)X_3 + e \\ &= \alpha_0 + \alpha_1 X_1 + \alpha_3 X_3 + e \end{align*}$$

With the new hypothesis and model, this becomes more familiar to me. I can use the partial F-test to determine whether we reject the null hypothesis or not.

In R I do the following:

m.null = lm(Y - 0.5*(X1+X3) ~ 1,     data=supdata)
m.alt  = lm(Y - 0.5*(X1+X3) ~ X1+X3, data=supdata)
anova(m.null, m.alt)

I can obtain the F-statistics and use its p-value to make a decision but I would first like to make sure that my implementation is correct.

  • 2
    $\begingroup$ Yes, it is correct. $\endgroup$ Commented Mar 26, 2018 at 2:48
  • $\begingroup$ Lets say model null is likely to be correct. How you gonna interpret that in terms of beta1=beta3=0.5 ? $\endgroup$
    – Jak Jac
    Commented Mar 26, 2018 at 17:53
  • $\begingroup$ Note that for a general case it's also a good idea to use Likelihood Ratio test instead of F test lmtest::lrtest(m.null, m.alt) due to less restrictions for the asymptotics to hold. Although, for OLS, as in your case, both tests are equivalent. $\endgroup$
    – runr
    Commented May 14, 2019 at 23:07

1 Answer 1


First create the model

data = fread(paste0("http://www1.aucegypt.edu/faculty/hadi/RABE5/Data5/", "P060.txt"))
model <- lm(data = data, Y ~ X1 + X3)

Then you can use the following code:

linearHypothesis(model, c("X1=X3", "X1=0.5"))

You will get the same output with less code and hassle.


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