pvalues = zero?

The pvalues of the coefficients and model are equal to zero. Are they too significant to be true?

mdl = 

Nonlinear regression model:
    Emod ~ [Nonlinear formula with 7 coefficients and 9 predictors]

Estimated Coefficients:
          Estimate        SE        tStat       pValue   
          ________    __________    ______    ___________

      b1     4.5028        0.05506     81.78              0
      b2    0.99989     0.00011413    8761.1              0
      b3     4.3701       0.048136    90.786              0
      b4    0.99986     3.0081e-05     33239              0
      b5     1.0002        0.00125    800.15              0
      b6     1.0246       0.004444    230.55              0
      b7     4.4177       0.065778     67.16    8.3514e-279

Number of observations: 597, Error degrees of freedom: 591
Root Mean Squared Error: 3.7
R-Squared: 0.348,  Adjusted R-Squared 0.342
F-statistic vs. zero model: 7.69e+03, p-value = 0

Answer 1

[Star Strider]

Looking at the t-statistics, the p-values are simply too small to be calculated. Very small p-values are characteristically the result of a very large data set (here 597).

So yes they are all highly statistically significantly different from zero, meaning that they are all required in the regression, and all help to explain the result.

[wesleynotwise]

Thanks for your reply, Star Strider. So the higher the t-stat values, the smaller the p-values? But the adj R-squared is 0.342, and the p-value of the model is again equal to zero, aren't these two values contradicted with each other?

[Star Strider]

My pleasure.

‘So the higher the t-stat values, the smaller the p-values?’

Yes. The t-statistic is the ‘distance’ (in units of the standard deviation) of a particular value from the mean of the distribution. So the greater the value of the t-statistic, the lower the p-value.

‘But the adj R-squared is 0.342, and the p-value of the model is again equal to zero, aren't these two values contradicted with each other?’

The p-value of the model compares the model fit to the mean of the dependent variable. The better the model fit (the better the model is at explaining the dependent variable relation to the independent variable), the more significant it is, so the F-statistic is high and the p-value is low.

In the immortal words of the Wikipedia article on Coefficient of determination (link), ‘In statistics, the coefficient of determination, denoted R² or r² and pronounced "R squared", is a number that indicates the proportion of the variance in the dependent variable that is predictable from the independent variable(s).’ So here, 34.2% are. It is informative, and while it conveys no specific probability information on how good the fit is, it is calculated similarly to the F-statistic, so the two are related.

[wesleynotwise]

You blew my mind. I wish you work with Matlab and make it more user-friendly.

If t-stat measures the distance from the mean, isn't the greater the t-value, the larger the distance, i.e. not good?

Given that both the p-values are calculated for both the coefficient and model, I assume the former is meant for individual predictor (or variable), and the latter is the overall/global p-value.

[Star Strider]

Thank you! I appreciate your compliment!

‘If t-stat measures the distance from the mean, isn't the greater the t-value, the larger the distance, i.e. not good?’

No, because the t-distribution is essentially the normal distribution, so the mean is also the mode, i.e. the most frequently observed values, creating a ‘bell-shaped’ curve. The lowest probabilities (corresponding to the greatest absolute t-values) are at the tails.

‘Given that both the p-values are calculated for both the coefficient and model, I assume the former is meant for individual predictor (or variable), and the latter is the overall/global p-value.’

Correct (1) for the estimated coefficient of the predictor variable, and (2) for the model explaining more than the mean value of the dependent variable.