# How to say if the variable is significant looking only at "t value"?

I have the model: Assuming we do not know the "p-value" I'm trying to analyse the significant variable looking only coeff. / std Err and t value.

I'm a bit confused as if I looked beta which has a p=0.1 (90% benchmark according to Gujarati) we can say if it is positively significant. The t of beta is 1.65 which I will take as the greatest value I can use.

Now, I'm looking for the smallest value for t. And here I feel lost. As e.g. NDTS has the t value = -0.18 (which is not significant we know that looking at p) but from another hand, the ROA t is -6.53 but it is still significant....

I need to read these data without knowing "p_value".

• Negative values are interpreted the same as the equivalent positive value but in the other direction, Dec 11 '16 at 14:52
• I'm pretty sure Gujarati gives a rule-of-thumb of |t| >= 2.00 to determine whether or not a variable is statistically significant. Obviously, this is just a rule-of-thumb... and in some cases (probably most of the time) you'll want to entertain variables that don't strictly satisfy this rule. Dec 11 '16 at 15:20

Beforehand: The t distribution changes its shape depending on the degrees of freedom (df). Since in regression $$df= n - p - 1$$ where $$n$$ is the sample size and $$p$$ the number of the predictors, we know that $$df$$ increases if $$n$$ does (assuming $$p$$ stays the same). The important thing here is that the bigger the df the more the t distributions approximates a standard normal distribution. Here is an example from wikipedia that shows how the t distribution changes with increasing df: Here the Density of the t-distribution is red (for 1, 2, 3, 5, 10, and 30 df) compared to the standard normal distribution which is blue. Further, the previous plots shown in green. Thus "You are pretty safe (unless you are interested in the extreme tails) in ignoring the difference between the standard normal and the t-distribution when the degrees of freedom are even moderately large" (source for quote).

We know that in case of a standard normal distribution z< -1.96 and z> 1.96 mark the cutoff for statistical significance if α is set at .05 (see here to learn about the Z-test). So if your sample size is big enough you can say that a t value is significant if the absolute t value is higher or equal to 1.96, meaning $$|t| ≥ 1.96$$. Or if you decide to set α at .01 you would need $$|t| ≥ 2.58$$.

The obvious question is what in this context "large enough" means. I would say it simply depends on how exact you want this rule of thumb to be. But in your case with $$n= 428$$ this should be a reasonable rule of thumb.

When dealing with significance testing, you are deciding between two hypotheses

• $H_0:\beta=0$
• $H_1:\beta\neq0$

A beta coefficient is significant if you are able to reject the null hypothesis for a specified level of significance $\alpha$, that is usually chosen to be $0.05$.

If you use the p-value, you reject the null hypothesis when the p-value is lower than the level of significance $\alpha$. If you use the t-stat, instead, you reject the null hypothesis if the value of the t-stat is greater than the value, which corresponds to the level of significance $\alpha$ on the Normal distribution table. As an example if your level of significance is $0.05$, the correspondent t-stat value is $1.96$, thus when the t-stat reported in the output is higher than $1.96$ you reject the null hypothesis and your coefficient is significant at $5\%$ significance level.

Since, the t-stat is computed as $\beta/s.e.$, if your $\beta$ value is negative, the t-stat will be negative but the comparison has to be made in absolute value, thus in your case $-6.53$ is higher than $1.96$ in absolute value, but it is also higher than $2.58$, the critical value for a $0.01$ ($1\%$) significance level.

• What about the degrees of freedom though? Dec 11 '16 at 15:28