Simple hypotheses

So far, we have considered hypotheses that involve a single estimator:

1. H_0: \beta_j=\kappa_0

2. H_0: \beta_j\geq \kappa_0

3. H_0: \beta_j\leq \kappa_0

Linear combinations

Today, we consider hypotheses of the form:

H_0: a_0\beta_0 + a_1 \beta_1 + \cdots + a_k \beta_k = \kappa_0

for constants \{a_j\}_{j=0,1,\dots,k}. This is a linear combination of parameters.

Multiple linear combinations

Similarly, we could have q of these statements,

\begin{aligned} H_0: a^1_0\beta_0 + a^1_1 \beta_1 + \cdots + a^1_k \beta_k &= \kappa^1_0 \\ a^2_0\beta_0 + a^2_1 \beta_1 + \cdots + a^2_k \beta_k &= \kappa^2_0 \\ &\vdots \\ a^q_0\beta_0 + a^q_1 \beta_1 + \cdots + a^q_k \beta_k &= \kappa^q_0 \\ \end{aligned}

Single linear combinations

Consider the following example related to environmental and health economics

\ln(PM2.5_\ell) = \beta_0 + \beta_1 EV_\ell + \beta_2 hybrid_\ell + \mathbf{x}_\ell\gamma + u_\ell

where,

• PM2.5 is micrograms of particulate matter (with diameter \leq 2.5 micrometres) per cubic metre in location \ell.

  • standard measure for health effects as absorbable by lungs and bloodstream

• EV and hybrid are number of registered vehicles (/1000) by fuel type in location \ell.

Example: PM2.5

The following estimates use local authority level data for 2019 in the UK.

* Open CSV data
import delimited "$data_dir\pm25.csv", clear

* Estimate univariate linear regression
reg ln_pm25 ev_k hybrid_k petrol_k diesel_k ln_pop

(encoding automatically selected: ISO-8859-1)
(17 vars, 291 obs)

      Source |       SS           df       MS      Number of obs   =       291
-------------+----------------------------------   F(5, 285)       =     33.45
       Model |  2.84542086         5  .569084172   Prob > F        =    0.0000
    Residual |  4.84835963       285  .017011788   R-squared       =    0.3698
-------------+----------------------------------   Adj R-squared   =    0.3588
       Total |  7.69378049       290  .026530278   Root MSE        =    .13043

------------------------------------------------------------------------------
     ln_pm25 | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
        ev_k |   .0779783     .03734     2.09   0.038     .0044811    .1514755
    hybrid_k |   .0322276   .0063524     5.07   0.000      .019724    .0447312
    petrol_k |   .0000559   .0007053     0.08   0.937    -.0013323    .0014442
    diesel_k |   -.003546   .0005762    -6.15   0.000    -.0046802   -.0024119
      ln_pop |    .088135   .0208428     4.23   0.000     .0471096    .1291604
       _cons |   1.264753   .2325884     5.44   0.000     .8069436    1.722562
------------------------------------------------------------------------------

# Load the Stata file
pm25_data <- read.csv(file.path(data_dir, "pm25.csv"))

# Estimate univariate linear regression
model1 <- lm(ln_pm25 ~ ev_k + hybrid_k + petrol_k + diesel_k + ln_pop, data = pm25_data)

# View results
summary(model1)

Call:
lm(formula = ln_pm25 ~ ev_k + hybrid_k + petrol_k + diesel_k + 
    ln_pop, data = pm25_data)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.39959 -0.05714  0.03222  0.08413  0.47799 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  1.265e+00  2.326e-01   5.438 1.16e-07 ***
ev_k         7.798e-02  3.734e-02   2.088   0.0377 *  
hybrid_k     3.223e-02  6.352e-03   5.073 7.06e-07 ***
petrol_k     5.591e-05  7.053e-04   0.079   0.9369    
diesel_k    -3.546e-03  5.762e-04  -6.154 2.55e-09 ***
ln_pop       8.814e-02  2.084e-02   4.229 3.17e-05 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1304 on 285 degrees of freedom
Multiple R-squared:  0.3698,    Adjusted R-squared:  0.3588 
F-statistic: 33.45 on 5 and 285 DF,  p-value: < 2.2e-16

Null hypotheses

From a public policy perspective (if you consider designing subsidies), you want to estimate both \beta_1 and \beta_2, BUT also know if

H_0: \beta_1=\beta_2

That is, should the policy discriminate bewteen electrified fuel types?

• Engineers may be able provide mechanical differences, but they can’t tell you about differences in usage (driver behaviour) which is important for realized pollution mitigation.

Alternative hypothesis

The alternative in this case, may be

H_1: \beta_1\neq \beta_2

But may also be,

H_1: \beta_1<\beta_2

or

H_1: \beta_1>\beta_2

if there is a strong case for ruling out one case (see Wooldridge 2025, 139).

Test statistic

For H_0: \beta_1=\beta_2\Rightarrow H_0:\beta_1-\beta_2=0

\text{test-stat} = \frac{\hat{\beta}_1-\hat{\beta}_2}{\sqrt{Var(\hat{\beta}_1-\hat{\beta}_2)}}

Variance

The key to understanding and executing this test is,¹

Var(\hat{\beta}_1-\hat{\beta}_2) \neq Var(\hat{\beta}_1) + Var(\hat{\beta}_2)

So,

Var(\hat{\beta}_1-\hat{\beta}_2) = Var(\hat{\beta}_1) + Var(\hat{\beta}_2) - \textcolor{red}{2\times}\underbrace{\textcolor{blue}{Cov(\hat{\beta}_1,\hat{\beta}_2)}}_{\neq 0}

Distribution

Under MLR 1-6, the conditional distribution of this test-statistic is t_{n-k-1}

t\text{-stat} = \frac{\hat{\beta}_1-\hat{\beta}_2}{\sqrt{Var(\hat{\beta}_1-\hat{\beta}_2)}} \underset{H_0}{\sim} t_{n-k-1}

under the null.

• You can use the same steps from Lecture 3 to conduct inference: p-value, critical value.

• We tend not to construct confidence intervals in such cases.

• Similarly, you can compute the power under the alternative \beta_1-\beta_2 = \kappa.

Example: PM2.5

* t-test
lincom ev_k - hybrid_k

 ( 1)  ev_k - hybrid_k = 0

------------------------------------------------------------------------------
     ln_pm25 | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         (1) |   .0457507   .0423487     1.08   0.281    -.0376053    .1291066
------------------------------------------------------------------------------

# Load package
library(marginaleffects)

# Estimate linear hypothesis
hypotheses(model1, "ev_k - hybrid_k = 0")

      Hypothesis Estimate Std. Error    z Pr(>|z|)   S   2.5 % 97.5 %
 ev_k-hybrid_k=0   0.0458     0.0423 1.08     0.28 1.8 -0.0373  0.129

Multiple linear hypotheses

Another common form of hypothesis concerns multiple linear combinations of estimators.

\begin{aligned} H_0: a^1_0\beta_0 + a^1_1 \beta_1 + \cdots + a^1_k \beta_k &= \kappa^1_0 \\ a^2_0\beta_0 + a^2_1 \beta_1 + \cdots + a^2_k \beta_k &= \kappa^2_0 \\ &\vdots \\ a^q_0\beta_0 + a^q_1 \beta_1 + \cdots + a^q_k \beta_k &= \kappa^q_0 \\ \end{aligned}

Example: PM2.5

In our example of

PM2.5_\ell = \beta_0 + \beta_1 EV_\ell + \beta_2 hybrid_\ell + \mathbf{x}_\ell\gamma + u_\ell

This could be as simple as,

H_0: \beta_1 = 0\;\text{and}\; \beta_2 = 0

Test

The test is performed by considering two models:

1. Unrestricted model; using our example,

\ln(PM2.5_\ell) = \beta_0 + \beta_1 EV_\ell + \beta_2 hybrid_\ell + \mathbf{x}_\ell\gamma + u_\ell

• storing the SSR_{ur} from this model

Test statistic

The test static that is then given by,

F\text{-stat} = \frac{(SSR_{r}-SSR_{ur})/q}{SSR_{ur}/df_{ur}}

Test statistic

The test static that is then given by,

F\text{-stat} = \frac{(SSR_{r}-SSR_{ur})/q}{SSR_{ur}/df_{ur}}

and,

• q is the number of restrictions (i.e. equations tested) and is given by the difference in residual degrees of freedom

  q = df_r-df_{ur}

Distribution

Under the null, F-stat has an F distribution (hence the name)

F\text{-stat} \underset{H_0}{\sim}F_{q,n-k-1}

We reject H_0 if

F\text{-stat} > F_{q,n-k-1}^{-1}(1-\alpha) = f_{q,n-k-1,1-\alpha}

where F_{q,n-k-1}^{-1}(1-\alpha) is the 1-\alpha percentile of the distribution.

Rejection rule

F\text{-stat} > f_{q,n-k-1,1-\alpha}

• Always a one-sided test.¹

• Doesn’t mean that H_1 is one-sided.

  • You cannot tell which direction the deviation from the null is, because F-stat uses squared-deviations (i.e. sign is lossed) .

• H_1:\; H_0\;\text{false}, or H_1: \beta_1\neq0\;\text{or}\; \beta_2\neq0

Alternative form

Since SSR = (1-\mathbf{R}^2)SST and SST_{r}=SST_{ur},

We can write the F-stat as,

F\text{-stat} = \frac{(\mathbf{R}^2_{ur}-\mathbf{R}^2_{r})/q}{(1-\mathbf{R}^2_{ur})/(n-k-1)}

Note,

• The order switches in the numerator, but remains positive since \mathbf{R}^2_{ur}>\mathbf{R}^2_{r}

• n-k-1=df_{ur}, with k the number of parameters in the unrestricted model

Example: PM2.5

* F-test
test ev_k hybrid_k

 ( 1)  ev_k = 0
 ( 2)  hybrid_k = 0

       F(  2,   285) =   53.88
            Prob > F =    0.0000

# Load package
library(car)

# Estimate linear hypothesis
linearHypothesis(model1, c("ev_k = 0", "hybrid_k = 0"))

Linear hypothesis test:
ev_k = 0
hybrid_k = 0

Model 1: restricted model
Model 2: ln_pm25 ~ ev_k + hybrid_k + petrol_k + diesel_k + ln_pop

  Res.Df    RSS Df Sum of Sq      F    Pr(>F)    
1    287 6.6816                                  
2    285 4.8484  2    1.8332 53.881 < 2.2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Revisiting single linear restriction

You can think of a single linear hypothesis as a special case of multiple where q=1. For example,

H_0: \textcolor{red}{\beta_1} = \textcolor{blue}{\beta_2}

• Restricted model:

  \begin{aligned} \ln(PM2.5_\ell) =& \beta_0 + \textcolor{red}{\beta_1} EV_\ell + \underbrace{\textcolor{red}{\beta_1}}_{=\textcolor{blue}{\beta_2}} hybrid_\ell + \mathbf{x}_\ell\gamma + u_\ell \\ = & \beta_0 + \textcolor{red}{\beta_1} (EV_\ell + hybrid_\ell) + \mathbf{x}_\ell\gamma + u_\ell \end{aligned}

  A regression with a new transformed variable: EV_\ell + hybrid_\ell

Moreover, you can show that for q=1

F\text{-stat} = \frac{SSR_{r}-SSR_{ur}}{SSR_{ur}/(n-k-1)} = (t\text{-stat})^2

Example: PM2.5

* t-test
lincom ev_k - hybrid_k

* F-test
test ev_k=hybrid_k

* check
dis sqrt(r(F))

 ( 1)  ev_k - hybrid_k = 0

------------------------------------------------------------------------------
     ln_pm25 | Coefficient  Std. err.      t    P>|t|     [95% conf. interval]
-------------+----------------------------------------------------------------
         (1) |   .0457507   .0423487     1.08   0.281    -.0376053    .1291066
------------------------------------------------------------------------------


 ( 1)  ev_k - hybrid_k = 0

       F(  1,   285) =    1.17
            Prob > F =    0.2809

1.0803322

# t-test
hypotheses(model1, "ev_k - hybrid_k = 0")

      Hypothesis Estimate Std. Error    z Pr(>|z|)   S   2.5 % 97.5 %
 ev_k-hybrid_k=0   0.0458     0.0423 1.08     0.28 1.8 -0.0373  0.129

# F-test
linearHypothesis(model1, "ev_k = hybrid_k")

Linear hypothesis test:
ev_k - hybrid_k = 0

Model 1: restricted model
Model 2: ln_pm25 ~ ev_k + hybrid_k + petrol_k + diesel_k + ln_pop

  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1    286 4.8682                           
2    285 4.8484  1  0.019855 1.1671 0.2809

Regression significance

A special case of this form of test concerns the joint significance of all the slope parameters in a model:

H_0: \beta_1 = 0,\; \beta_2 = 0,\;\dots,\;\beta_k=0

In this case,

• The restricted model is a regression on a constant (alone) where \Rightarrow \mathbf{R}^2_{r}=0

• And \mathbf{R}^2_{ur} is simply the \mathbf{R}^2 from the model

• The number of restrictions is q=k

We define the F-stat as,

F\text{-stat} = \frac{\mathbf{R}^2/k}{(1-\mathbf{R}^2)/(n-k-1)}

Dummy variables: recap

Recall, when we have the model:

y_i = \beta_0 + \beta_1 d_i + u_i

where d_i\in\{0,1\}. Then,

\beta_1 = E[y_i|d_i=1]-E[y_i|d_i=0]

Dummy variable interactions

Now, suppose the model was

y_i = \beta_0 + \beta_1 x_i + \beta_2 d_i + \beta_3 x_i\times d_i + u_i

where x_i is continuous.

• What is the interpretation of each parameter?

y_i = \beta_0 + \beta_1 x_i + \beta_2 d_i + \beta_3 x_i\times d_i + u_i

• \beta_0=E[y_i|\textcolor{blue}{d_i=0},x_i=0]

• \beta_1 = \frac{dE[y_i |\textcolor{blue}{d_i=0},x_i]}{dx_i}

• \beta_2=E[y_i|\textcolor{red}{d_i=1},x_i=0]-E[y_i|\textcolor{blue}{d_i=0},x_i=0]

• \beta_3 = \frac{dE[y_i |\textcolor{red}{d_i=1},x_i]}{dx_i}-\frac{dE[y_i |\textcolor{blue}{d_i=0},x_i]}{dx_i}

Another way of thinking about it:

• For d_i=0, the model is:

  y_i = \textcolor{blue}{\beta_0} + \textcolor{blue}{\beta_1} x_i + u_i

• For d_i = 1, the model is:

  y_i = \beta_0 + \beta_1 x_i + \beta_2 + \beta_3 x_i + u_i

  Which can be rewritten as:

  y_i = \underbrace{(\beta_0 + \beta_2)}_{\textcolor{red}{\gamma_0}} + \underbrace{(\beta_1 + \beta_3)}_{\textcolor{red}{\gamma_1}} x_i + u_i

We have two models: for d_i=0 and d_i=1.

• \beta_2 is the difference in intercept

• \beta_3 is the difference in slope

Example

The typical example is the returns to education for men and women (as in Wooldridge 2025, 243)

• But, given what we know about the ‘parent penalty’ (Kleven et al. 2019), it seems more pertinent to focus on experience

  \ln(wage_i) = \beta_0 + \beta_1 exp_i + u_i

If we overlay the fitted lines we get

* overlay fitted lines
twoway (lfit lnwage exp if sex==1, lc(blue)) (lfit lnwage exp if sex==2, lc(red) lwidth(*2)), xtitle("Potential Experience (=Age-(Years of Education +6))") legend(order(1 "Male" 2 "Female") r(2) ring(0) pos(5)) ylabel(2(1)6)

wage_data$group <- factor(wage_data$sex, levels = c(1, 2),
                          labels = c("Male", "Female"))

ggplot(wage_data, aes(x = exp, y = lnwage, colour = group)) +
  geom_smooth(data = subset(wage_data, sex == 1),
              method = "lm", se = FALSE, linewidth = 0.75) +
  geom_smooth(data = subset(wage_data, sex == 2),
              method = "lm", se = FALSE, linewidth = 1.5) +
  scale_colour_manual(values = c(Male = "blue", Female = "red")) +
  scale_y_continuous(breaks = 2:6) +
  coord_cartesian(ylim = c(2, 6)) +
  labs(x = "Potential Experience (=Age-(Years of Education +6))",
       y = "lnwage", colour = NULL) +
  theme_minimal() +/
  theme(legend.position = "inside",
        legend.position.inside = c(0.98, 0.02),
        legend.justification = c(1, 0))

Error in parse(text = input): <text>:14:20: unexpected '/'
13:        y = "lnwage", colour = NULL) +
14:   theme_minimal() +/
                       ^

F-test

The F-test can be used to test differences in a model across groups.¹

• For example, whether the Mincer equation (from Tutorial 1) differs by gender.

  \ln(wage_i) = \beta_0 + \beta_1 edu_i + \beta_2 exp_i + u_i

• We estimated this model for the male-only sample

• The quadratic term in exp has been dropped for simplification

• The model has k+1=3 parameters

Test:

We may wish to ask whether:

• the model for men (group = 1; e.g. “Male”)

  \ln(wage_i) = \textcolor{blue}{\beta_0} + \textcolor{blue}{\beta_1} edu_i + \textcolor{blue}{\beta_2} exp_i + u_i

is the same as,

• the model for women (group = 2; e.g. “Female”)?

  \ln(wage_i) = \textcolor{red}{\gamma_0} + \textcolor{red}{\gamma_1} edu_i + \textcolor{red}{\gamma_2} exp_i + u_i

Null hypothesis

This question implies a null hypothesis:

H_0: \textcolor{blue}{\beta_0} = \textcolor{red}{\gamma_0},\;\textcolor{blue}{\beta_1} = \textcolor{red}{\gamma_1},\;\textcolor{blue}{\beta_2} = \textcolor{red}{\gamma_2}

against,

H_1: H_0\;\text{false}

Approach 1

The F-test for this hypothesis has the form:

F\text{-stat} = \frac{(SSR_P - (\textcolor{blue}{SSR_1} + \textcolor{red}{SSR_2}))/(k+1)}{(\textcolor{blue}{SSR_1} + \textcolor{red}{SSR_2})/(n-2(k+1))}

where,

• SSR_P is the “pooled” (P) SSR from a regression of the joint sample.

• \textcolor{blue}{SSR_1} is the SSR from a regression for just group 1.

• \textcolor{red}{SSR_2} is the SSR from a regression for just group 2.

  • together \textcolor{blue}{SSR_1} and \textcolor{red}{SSR_2} represent the unrestricted model.

• n-2(k+1) is the combined degrees of freedom from group 1 and 2.

  • k+1 parameters are estimated in each group.

Distribution

The F-stat is (unsurprisingly) F-distributed,

F\text{-stat} \underset{H_0}{\sim} F_{k+1,n-2(k+1)}

Approach 2

This test has an equivalent form using interactions. Let,

d_i= \mathbf{1}\{\text{`Group 2'}\}

For example,

d_i= \mathbf{1}\{\text{`Female'}\}

Interactions

We can write the unrestricted model as,

\ln(wage_i) = \beta_0 + \beta_1 edu_i+\beta_2 exp_i +\beta_3 d_i + \beta_4 edu_i\times d_i + \beta_5 exp_i\times d_i +u_i

We now have 6 parameters: 2\times (k+1) (for k=2)

As before, the dummy and dummy-interactions tell you about differences,

\beta_3 = E[\ln(wage_i)|\textcolor{red}{d_i=1},edu_i=0,exp_i=0]-E[\ln(wage_i)|\textcolor{blue}{d_i=0},edu_i=0,exp_i=0]

and

\beta_j = \frac{\partial E[\ln(wage_i)|\textcolor{red}{d_i=1},edu_i,exp_i]}{\partial x_{ij}}-\frac{\partial E[\ln(wage_i)|\textcolor{blue}{d_i=0},edu_i,exp_i]}{\partial x_{ij}} \qquad j=4,5

Test

Intuitively, the test for whether the model differs by group is given by,

H_0: \beta_3 = 0,\;\beta_4=0,\;\beta_5=0

where,

• unrestricted model is given by,

\ln(wage_i) = \beta_0 + \beta_1 edu_i+\beta_2 exp_i +\beta_3 d_i + \beta_4 edu_i\times d_i + \beta_5 exp_i\times d_i +u_i

• df_{ur} = n-6

• restricted model is given by,

\ln(wage_i) = \beta_0 + \beta_1 edu_i + \beta_2 exp_i +v_i

• df_{r} = n-3

F-stat

The F-stat is given by,

F\text{-stat} = \frac{(SSR_{r}-SSR_{ur})/3}{SSR_{ur}/(n-6)}

Equivalence

These two test-statics are equivalent because,

1. SSR_P=SSR_r

2. \textcolor{blue}{SSR_1} +\textcolor{red}{SSR_2} = SSR_{ur}