Modify test t for linear regression parameters in R

I do not know of any function capable of doing this in the R. In general, what is suggested is to do likelihood ratio tests, which are much more general and solve more sophisticated problems, even in generalized linear models.

However, nothing prevents us from writing our own function. After all, we have to test the hypotheses

H_0: \ beta = \ beta_0

H_1: \ beta! = \ beta_0

where β_0 is the reference value under which we wish to perform the test. By default, the value of \ beta_0 equals zero. The TesteCoeficiente function, defined below, implements this hypothesis test for any value of \ beta_0. Just enter the following values for the function:

reg : The regression model obtained with the function lm

coeficiente : the number of the coefficient in the regression model. For y = \beta_0 + \beta_1*x , 1 means test \beta_0 and 2 means test \beta_1

h0 : value of the null hypothesis to be tested. In general this value is zero

With this set, just run the example below to see how the function is executed:

x <- c(1, 2, 3, 4, 5)
y <- c(1.2, 2.4, 3.3, 4.2, 5.1)
reg <- lm(y~x)

TesteCoeficiente <- function(reg, coeficiente, h0){
  estimativas <- coef(summary(reg))
  estatistica <- (estimativas[coeficiente, 1]-h0)/estimativas[coeficiente, 2]
  2 * pt(abs(estatistica), reg$df.residual, lower.tail = FALSE)
}

TesteCoeficiente(reg, coeficiente=2, h0=0)
[1] 2.771281e+01 1.031328e-04

summary(reg)
Call:
lm(formula = y ~ x)

Residuals:
         1          2          3          4          5 
-1.200e-01  1.200e-01  6.000e-02  1.457e-16 -6.000e-02 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.36000    0.11489   3.133 0.051929 .  
x            0.96000    0.03464  27.713 0.000103 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1095 on 3 degrees of freedom
Multiple R-squared:  0.9961,    Adjusted R-squared:  0.9948 
F-statistic:   768 on 1 and 3 DF,  p-value: 0.0001031

Note that the output of the TesteCoeficiente function is identical to that of the summary function by testing h0 = 0 . Therefore, the function works for this particular case. It is quite reasonable to assume that it works for any value of \ beta_0. Now just change the value of the h0 parameter of the function to perform the desired hypothesis test.

With R base you can ask for the confidence intervals with the confint function:

confint(reg)
                   2.5 %    97.5 %
(Intercept) -0.005635243 0.7256352
x            0.849756826 1.0702432

In the above case, with a 95% interval, any value within the confidence interval is not "rejected" by a hypothesis test with a significance level of 5% (so you would not reject that b1 = 1, for example ).

Similarly, any value outside the range is rejected at the significance level of 5%. To change the confidence level of the range, change the level parameter. for example, if you want 99%, confint(reg, level = 0.99) .

If you prefer to do a specific test instead of having the range, a package that has convenience functions for regression analysis is the car package. Among them, the linearHypothesis function is for hypothesis testing. To test whether the coefficient of x (b1) equals 1:

# se o pacote não estiver instalado rode install.packages("car") antes
library(car) 
linearHypothesis(reg, "x = 1")
Linear hypothesis test

Hypothesis:
x = 1

Model 1: restricted model
Model 2: y ~ x

  Res.Df   RSS Df Sum of Sq      F Pr(>F)
1      4 0.052                           
2      3 0.036  1     0.016 1.3333 0.3318

The p-value in this case is 0.3318 , that is, you do not reject the hypothesis, as we saw with the confidence interval. The linearHypothesis function also accepts any other linear constraint on the parameters. For example, you can test whether the intercept (b0) is equal to the coefficient of x (b1):

linearHypothesis(reg, "(Intercept) = x")
Linear hypothesis test

Hypothesis:
(Intercept) - x = 0

Model 1: restricted model
Model 2: y ~ x

  Res.Df   RSS Df Sum of Sq      F  Pr(>F)  
1      4 0.236                              
2      3 0.036  1       0.2 16.667 0.02655 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Among other tests.

PS: I know that here is not the place for this, but beware of the interpretation of p-values. For example, with respect to "b0 is not different from zero", strictly you do not reject the null hypothesis that b0 is equal to any value within its confidence interval (as 0.6), not just zero specifically. p>