# t test of individual coefficient and wald test of euqality of two coefficients

I've run a regression $Y = a + \beta_1X_1 + \beta_2X_2 + \epsilon$, and I have an interest in testing which coefficient of $X_1$ and $X_2$ has a stronger impact on $Y$.

Here are the parameter estimates: $\hat{\beta_1} = 0.086$, $se(\hat{\beta_1}) = 0.019$, $\hat{\beta_2} = 0.068$, $se(\hat{\beta_2}) = 0.051$. $cov(\hat{\beta_1}, \hat{\beta_2}) < -0.001$.

Obviously, by t-test, $\hat{\beta_1}$ is significant (p. < 0.001) whereas $\hat{\beta_2}$ is insignificant (p. > 0.1). However, Wald test of equality of two coefficients fails to reject the null. So, we cannot conclude that $X_1$ has a stronger impact than $X_2$ on $Y$.

Questions is, how should we understand the difference in significance (by t-test) between $\hat{\beta_1}$ and $\hat{\beta_2}$, when Wald test fails to reject that $\hat{\beta_1}$ and $\hat{\beta_2}$ are equal?

• Relations like "significant difference" or "insignificant difference" are not transitive; you are treating them like you expect they should be. Also see for example, the discussion here: stat.columbia.edu/~gelman/research/published/signif4.pdf Mar 28, 2019 at 23:48

When performing the t-test, your hypothesis of interest is $$H_0: \beta_i = 0\: \text{ vs }\: H_1: \beta_i\ne0\:\:i\in\{1,2\}$$ whereas in the Wald test you're testing $$H_0: \beta_1=\beta_2\: \text{ vs }\: H_1: \beta_1\ne\beta_2$$ However, based on the way you formulated your question, it appears to me that your null hypothesis is something of the form: $$H_0: \begin{pmatrix}1&0\\0&1\\1&-1\end{pmatrix}\begin{pmatrix}\beta_1\\\beta_2\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$ and therefore the Wald Test you need to perform for the above's null hypothesis.
A final word regarding your example at hand, sometimes your inference may be flawed because you did not take into account certain features of your data when 1) estimating the parameters of interest, or 2) when using the correct critical values. For example, the normal approximation (or $\chi^2$) is not a great one if you have very few observations since it's an asymptotic approximation.