Take the 2-minute tour ×
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

I used Tobit regression and wanted to test if the coefficients are significant. The problem is that when I run the R code I get t-values instead of z-values. I found code on how to calculate p-values for z-values and I'm wondering if you can use the same code for the t-values that you use for the z-values to calculate the p-values.

This is the code for when you have z-values:

ctable <- coef(summary(m))
pvals <- 2 * pt(abs(ctable[, "z value"]), df.residual(m), lower.tail = FALSE)
cbind(ctable, pvals)

This is the code I used for the t-values:

ctable <- coef(summary(m))
pvals <- 2 * pt(abs(ctable[, "t value"]), df.residual(m), lower.tail = FALSE)
cbind(ctable, pvals)

                    Value  Std. Error   t value         pvals
(Intercept):1   209.559678 32.50165463  6.447662  3.327237e-10
(Intercept):2    4.184758  0.05227107 80.058783 7.628269e-246    
read             2.697963  0.62024382  4.349843  1.737355e-05   
math             5.914589  0.70502334  8.389210  8.842612e-16
proggeneral     -12.714341 12.36481457 -1.028268  3.044547e-01
progvocational  -46.143265 13.65615960 -3.378934  8.002457e-04
share|improve this question

1 Answer 1

Computationally this is a correct calculation of a p value for a two sided test of whether there is significant evidence the parameter behind your t value is different from zero, assuming the t statistic does indeed have a t distribution under the null hypothesis.

I'm surprised the summary() method doesn't itself calculate this p value. This makes me wonder whether the author has reason to doubt whether these statistics really do have t distributions and this is a valid test to use. I would investigate this further if I were you.

I would also round your p values (and probably everything else in that table) to two decimal places to avoid giving a false sense of precision.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.