Simple regression, switching X and Y, t-test stays the same? (edited) I did a simple regression where the independent variable was the educational background of the mother. The dependent variable was the language skills score of the child. 
I switched them around and did another regression to see which coefficients change and which didn't. 
The t-test of the gradient was the same for both regressions. I find it difficult to explain why that is the case. The b-value and the standard error, the values that I need to calculate the t-test, were different for each regression. It puzzles me how dividing a different b-value by a different standard error gets the same outcome. 
Edit: I have a related but different question now. If I standardize the two variables, so turn them into z-scores, what would the value of the intercept be? 
I have no idea how to calculate this. I don't see the relation between z-scores and the intercept. 
Maybe the answer is quite simple but I have to admit that learning about z-scores was a long time ago so maybe I'm missing something very obvious. 
 A: $\newcommand{\Cov}{\operatorname{Cov}}\newcommand{\Var}{\operatorname{Var}}$In a simple regression, with centered regressions for simplicity,
$$y_i = bx_i + u_i, i=1,\ldots,n$$
Ordinary Least Squares estimation gives
$$\widehat b = \frac { \widehat \Cov(x,y)}{{\widehat \Var(x)}}$$
While 
$$\widehat \Var(\widehat b) = \frac {\widehat \sigma^2_u}{\widehat \Var(x)}$$
So in reality the t-statistic is
$$t(\widehat b) = \frac{\widehat \Cov(x,y)}{\widehat \sigma_u\cdot \sqrt{\widehat \Var(x)}}$$
In the reverse model 
$$x_i = b'y_i + u'_i, i=1,\ldots,n$$
we will have
$$t(\widehat b') = \frac{\widehat \Cov(x,y)}{\widehat \sigma_{u'}\cdot \sqrt{\widehat \Var(y)}}$$
So what one has to examine is whether
$$\widehat \sigma^2_u\cdot \widehat \Var(x) =\text{??}\;\; \widehat \sigma^2_{u'}\cdot \widehat \Var(y)$$
Switching to matrix representation, these estimated quantities are sums of squared terms divided by some factor ($n$ or $n-1$). Ignore these factors to get
$$(y'M_xy)\cdot (x'x)  =\text{??}\;\; (x'M_yx)\cdot (y'y)$$
where $M_x = I-x(x'x)^{-1}x' = I - \frac {xx'}{x'x}$, and analogously for $M_y$. Expanding we examine whether
$$\left[y'\left(I - \frac {xx'}{x'x}\right)y\right]\cdot (x'x)  =\text{??}\;\; \left[x'\left(I - \frac {yy'}{y'y}\right)x\right]\cdot (y'y)$$
$$\implies (y'y)(x'x) - (y'x)(x'y) = \text{??} \;\; (x'x)(y'y) - (x'y)(y'x)$$
Each term in parenthesis is a scalar. So indeed the two sides are equal, and therefore what the OP obtained is not approximate or a chance event, it is exact and an algebraic property of the OLS estimation in the simple regression:
$$t(\widehat b) = t(\widehat b')$$
