Debe intervalos de confianza para la regresión lineal de los coeficientes se basan en la normal o $t$ distribución?

Question

Vamos a tener algún modelo lineal, por ejemplo, sólo el simple análisis de la VARIANZA:

# data generation
set.seed(1.234)                      
Ng <- c(41, 37, 42)                    
data <- rnorm(sum(Ng), mean = rep(c(-1, 0, 1), Ng), sd = 1)      
fact <- as.factor(rep(LETTERS[1:3], Ng)) 

m1 = lm(data ~ 0 + fact)
summary(m1)

El resultado es el siguiente:

Call:
lm(formula = data ~ 0 + fact)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.30047 -0.60414 -0.04078  0.54316  2.25323 

Coefficients:
      Estimate Std. Error t value Pr(>|t|)    
factA  -0.9142     0.1388  -6.588 1.34e-09 ***
factB   0.1484     0.1461   1.016    0.312    
factC   1.0990     0.1371   8.015 9.25e-13 ***
---
Signif. codes:  0 ‘***' 0.001 ‘**' 0.01 ‘*' 0.05 ‘.' 0.1 ‘ ' 1 

Residual standard error: 0.8886 on 117 degrees of freedom
Multiple R-squared: 0.4816,     Adjusted R-squared: 0.4683 
F-statistic: 36.23 on 3 and 117 DF,  p-value: < 2.2e-16 

Ahora trato de dos diferentes métodos de estimación de intervalo de confianza de estos parámetros

c = coef(summary(m1))

# 1st method: CI limits from SE, assuming normal distribution
cbind(low = c[,1] - qnorm(p = 0.975) * c[,2], 
    high = c[,1] + qnorm(p = 0.975) * c[,2])

# 2nd method
confint(m1)

Preguntas:

1. ¿Cuál es la distribución de la estimación lineal de los coeficientes de regresión? Normal o $t$?
2. ¿Por qué ambos métodos dan lugar a distintos resultados? Suponiendo que la distribución normal y correcto SE, yo esperaría que ambos métodos tienen el mismo resultado.

Muchas gracias!

datos ~ 0 + hecho

EDITAR después de una respuesta:

La respuesta es exacta, esto le dará exactamente el mismo resultado como confint(m1)!

# 3rd method
cbind(low = c[,1] - qt(p = 0.975, df = sum(Ng) - 3) * c[,2], 
    high = c[,1] + qt(p = 0.975, df = sum(Ng) - 3) * c[,2])

Answer 1

(1) Cuando los errores están distribuidos normalmente y su varianza es no conocido, $$\frac{\hat{\beta} - \beta_0}{{\rm se}(\hat{\beta})}$$ has a $t$-distribution under the null hypothesis that $\beta_0$ is the true regression coefficient. The default in R is to test $\beta_0 = 0$, so the $t$-statistics reported there are just $$\frac{\hat{\beta}}{{\rm se}(\hat{\beta})}$$

Note that, under some regularity conditions, the statistic above is always asymptotically normally distributed, regardless of whether the errors are normal or whether the error variance is known.

(2) The reason you're getting different results is that the percentiles of the normal distribution are different from the percentiles of the $t$-distribution. Therefore, the multiplier you're using in front of the standard error is different, which, in turn gives different confidence intervals.

Specifically, recall that the confidence interval using the normal distribution is

$$ \hat{\beta} \pm z_{\alpha/2} \cdot {\rm se}(\hat{\beta}) $$

where $z_{\alpha/2}$ is the $\alpha/2$ quantile of the normal distribution. In the standard case of a $95\$ confidence interval, $%\alfa = .05$ and $z_{\alpha/2} \approx 1.96$. The confidence interval based on the $t$-distribution is

$$ \hat{\beta} \pm t_{\alpha/2,n-p} \cdot {\rm se}(\hat{\beta}) $$

where the multiplier $t_{\alpha/2,n-p}$ is based on the quantiles of the $t$-distribution with $n-p$ degrees of freedom where $n$ is the sample size and $p$ is the number of predictors. When $n$ is large, $t_{\alpha/2,n-p}$ and $z_{\alpha/2}$ are about the same.

Below is a plot of the $t$ multipliers for sample sizes ranging from $5$ to $300$ (I've assumed $p=1$ for this plot, but that qualitatively changes nothing). The $t$-multipliers are larger, but, as you can see below, they do converge to the $z$ (línea negra) multiplicador como el tamaño de la muestra aumenta.