# Standard error of the residuals for a non-linear model

Hi I am new to R and statistics and used to linear models. Can you please explain the output? I used it to make a growth curve.

Formula: length ~ a * (1 - exp(-c * est_age))

Parameters:
Estimate Std. Error t value Pr(>|t|)
a 1.097e+03  1.026e+01 106.966  < 2e-16 ***
c 1.539e-01  1.982e-02   7.765 2.33e-09 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 41.74 on 38 degrees of freedom
Number of iterations to convergence: 6
Achieved convergence tolerance: 7.32e-07

• Spelling out which R function you used is always good practice. – Nick Cox Oct 13 '13 at 11:41
• More information would help. Would you understand the meaning of the column headings if it were a linear regression model rather than a nonlinear model? Would you understand some but not others? – Glen_b Oct 14 '13 at 2:26

It appears that you're using nls.

By typing

?summary.nls


Estimates and standard errors are estimated by the Gauss-Newton algorithm (if the nls defaults are used)

The P-values are the results of a two sided test of whether the parameters are zero or not.

You can check the exact calculations used to create the output shown by typing:

stats:::summary.nls

• Didn't you mean to type "two-sided" instead of "one-sided"? – whuber Oct 13 '13 at 16:25
• Yes I did, thanks, well spotted. @whuber, you really pop up everywhere with good responses – sqrt Oct 13 '13 at 17:22
• The model fitted is not logistic. It grows from 0 and approaches an upper asymptote exponentially. – Nick Cox Oct 13 '13 at 18:28
• Is it not? I labelled it logistic because it is a specific version of the generalised logistic function – sqrt Oct 13 '13 at 18:50
• Really? A glance at the algebra suggests otherwise. In any case, I'd assert that as usually understood in growth modelling a logistic curve has an inflexion. – Nick Cox Oct 13 '13 at 19:07