Significance of regression coefficients and their equality Suppose, we want to regress $y$ on $x_1$ and $x_2$, i.e.
$$ y = \alpha + \beta_1 x_1 + \beta_2 x_2 + \varepsilon \hspace{1cm} (1)$$
Is it, in principle, possible that simultaneously:


*

*$\beta_1$ is statistically significant but $\beta_2$ is not while

*$\beta_1$ statistically equals to $\beta_2$


?
 A: Yes.  This answer interprets the question in the following way:


*

*$\beta_1$ is significantly different from zero in the full model
$$y = \alpha + \beta_1 x_1 + \beta_2 x_2 + \varepsilon$$

*$\beta_2$ is not significantly different from zero in the full model.

*Either (a) $\beta_1=\beta_2$ or (b) a test of $H_0:\beta_1=\beta_2$ is not significant.  The latter is equivalent to the full model not being significantly better than the reduced model
$$y = \alpha + \beta(x_1 + x_2) + \varepsilon.$$
Intuitively, $y$ must have a detectable linear relationship with $x_1$ but not with $x_2$, even though the coefficients ("slopes") of those relationships are the same.  This could happen when the spread of $x_1$ in the data is substantially greater than the spread of $x_2$: the wider spread of $x_1$ will induce greater changes in $y$, even when $\beta_1 \approx \beta_2$, making $\beta_1$ more readily detectable than $\beta_2$.
To illustrate, I played around with (a) the amount of data $n$ and (b) the variance of $\varepsilon$ to produce this phenomenon.  The data are
$$(x_1, x_2, y) = ((1, 2, \ldots, 2n), (1,\ldots,1,-1,\ldots,-1), x_1+x_2+\varepsilon)$$
where $\varepsilon$ are independently and identically distributed with a mean of zero and standard deviation of $3$.  As $n$ grows, $x_1$ becomes more spread out (from $1$ through $2n$) while $x_2$ is confined to the interval $[-1,1]$.  The true underlying relationship is $\alpha=0, \beta_1=\beta_2=1$.
The following is R code to generate this example.
n <- 12
x1 <- 1:(2*n)
x2 <- c(rep(-1,n), rep(1,n))
set.seed(17)
y <- x1 + x2 + rnorm(2*n, sd=3)



*

*Here is the fit of the full model.
> summary(fit.full <- lm(y ~ x1+x2))

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  -0.5223     1.8358  -0.284    0.779    
x1            1.1400     0.1416   8.053 7.41e-08 ***
x2            0.4886     0.9800   0.499    0.623    

$\beta_1$ is significant at any reasonable threshold ($p$ is essentially zero), while, $\beta_2$ is not significant at any reasonable threshold ($p=0.623$).

*The full model is not a significant improvement over the full model ($p = 0.5618$):
>fit.partial <- lm(y ~ I(x1+x2))
>anova(fit.partial, fit.full)

Analysis of Variance Table

Model 1: y ~ I(x1 + x2)
Model 2: y ~ x1 + x2
  Res.Df    RSS Df Sum of Sq      F Pr(>F)
1     22 122.36                           
2     21 120.37  1    1.9924 0.3476 0.5618

