2
$\begingroup$

In a regression $y = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$, how do I use an $F$-test to test the hypothesis $\beta_1+\beta_2=2\beta_3$? The standard $F$-test would test a hypothesis $H_0: \beta_1 = \beta_2 = \beta_3 = 0$

$\endgroup$
  • 2
    $\begingroup$ See discussions of the General Linear Hypothesis, e.g. here $\endgroup$ – Glen_b Mar 21 '15 at 11:59
4
$\begingroup$

@Glen_b already provided a link to the discussion containing the theoretical aspects.

Here is a quick pratical example of how one would do it in R. Please also have a look at these documents which contain the theory as well as examples: Simultaneous Inference in General Parametric Models and Additional multcomp Examples.

We will use the mtcars dataset and build a linear regression model containing three variables: cyl (Number of cylinders), disp (Displacement) and hp (Horsepower) to predict the variable mpg (Miles/Gallon).

Then, we test the following hypothesis: $\beta_{\mathrm{cyl}}+\beta_{\mathrm{disp}}-2\cdot\beta_{\mathrm{hp}} = 0$.

Using the multcomp package, there are two ways of specifying the hypothesis:

  1. As a matrix
  2. by symbolic description

I included both version in the code below. In our example, the matrix would simply be a row vector: $\mathbf{K} = (0, 1, 1, -2)$. The zero at the beginning is necessary because our regression model includes an intercept.

By symbolic description means that you can simply state your hypothesis as a character string. In this case: "cyl + disp - 2*hp = 0".

In this example, the estimate of our hypothesis is $-1.2169$ with little evidence that it differs from $0$. The function confint is used to generate a confidence interval for the estimate: $(-2.86; 0.43)$.

#---------------------------------------------------------------------------------------
# Load "multcomp" package
#---------------------------------------------------------------------------------------

require(multcomp)

#---------------------------------------------------------------------------------------
# Load "mtcars" dataset
#---------------------------------------------------------------------------------------

data(mtcars)

#---------------------------------------------------------------------------------------
# Build linear regression model with three variables
#---------------------------------------------------------------------------------------

lm.mod <- lm(mpg~cyl+disp+hp, data = mtcars)

summary(lm.mod)

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 34.18492    2.59078  13.195 1.54e-13 ***
cyl         -1.22742    0.79728  -1.540   0.1349    
disp        -0.01884    0.01040  -1.811   0.0809 .  
hp          -0.01468    0.01465  -1.002   0.3250 

#---------------------------------------------------------------------------------------
# Define the general hypothesis
#---------------------------------------------------------------------------------------

K <- c("cyl + disp - 2*hp = 0") # As a formula

# K <- rbind(c(0, 1, 1, -2)) # As a contrast matrix
# rownames(K) <- c("cyl + disp - 2hp")
# colnames(K) <- names(coef(lm.mod))

#---------------------------------------------------------------------------------------
# Evaluate the general hypothesis and calculate confidence intervals
#---------------------------------------------------------------------------------------

glht.mod <- glht(lm.mod, linfct = K)

summary(glht.mod)

     Simultaneous Tests for General Linear Hypotheses

Fit: lm(formula = mpg ~ cyl + disp + hp, data = mtcars)

Linear Hypotheses:
                         Estimate Std. Error t value Pr(>|t|)
cyl + disp - 2 * hp == 0  -1.2169     0.8036  -1.514    0.141
(Adjusted p values reported -- single-step method)

confint(glht.mod)

     Simultaneous Confidence Intervals

Fit: lm(formula = mpg ~ cyl + disp + hp, data = mtcars)

Quantile = 2.0484
95% family-wise confidence level

Linear Hypotheses:
                         Estimate lwr     upr    
cyl + disp - 2 * hp == 0 -1.2169  -2.8631  0.4293
$\endgroup$
  • $\begingroup$ This is much appreciated! $\endgroup$ – Manish Agarwal Mar 22 '15 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.