Derive z- or t-statistic from p-values of regression coefficients from a probit/logit model I have the results from an empirical study reporting the results for a probit and logit model. They just report the p-values of the regression coefficients. I want to derive the corresponding $t$-/$z$-statistics.
Is it just the inverse of the $t$-distribution with the corresponding degrees of freedom and the given $p$-value? E.g., for the probit if $\beta_2=0.02$, $df=96$, $p=0.023$ the inverse of the $t$-distribution would be $2.31$ (using for example EXCEL's TINV(0.023;96) function).
Can I approximate the $z$-statistic in the same way if the degrees of freedom are large enough (say at least 30)? Or is there another way to get the value for the $z$-statistic from the reported results (sample size, df, and p-values)?
Thanks for your help!
 A: Let $r$ be the degrees of freedom of the residuals and $F$ be the cdf for the t-distribution. For a two-sided test:
$$ p = 2\left( 1 - F\left( \mathopen| t \mathclose|,r\right)\right) $$
Reversing this and using the fact that $b$ and $t$ will have same sign:


*

*If $b > 0$, then $t = F^{-1}\left(1 - \frac{p}{2} , r \right)$

*If $b < 0$, then $t = F^{-1}\left(\frac{p}{2} , r \right)$


Or equivalently:
$$ t = \mathrm{sign}(b) \cdot F^{-1}\left(1 - \frac{p}{2}, r \right) $$
In your case, the calculation would be:
$$\begin{align*} t &= F^{-1}\left(1 - \frac{.023}{2}, 96 \right) \\ &\approx 2.31 \end{align*}$$
You appear to have used the Excel function T.INV.2T which automatically handles the two-sided test math. Observe that:


*

*Excel function T.INV(.023, 96) returns -2.02.

*Excel function T.INV.2T(.023, 96) returns the correct answer of 2.31.

*Excel function T.INV(1 - .023 / 2, 96) returns 2.31.


To calculate the z-statistic from the pvalues of the z-statistic, you would do essentially do the same procedure but with the normal inverse CDF rather than the t inverse CDF.
