Nonlinear regression analysis cannot calculate P values for the independent variables in your model. Why not? And, what do you use instead? Those are the topics of this blog post.

Nonlinear regression is an excellent statistical analysis when you need the maximum flexibility for fitting curves in your data. However, just like there are sound reasons for no R-squared values in nonlinear regression, there are valid reasons for why there are no P values for the coefficient estimates.

## Why Are P Values Possible in Linear Regression?

The question above is probably not one that you’ve asked.

P values for the independent variables in linear regression are a valuable statistical tool that seems quite natural. In linear regression, a P value indicates whether the relationship between an independent variable and the dependent variable is statistically significant while controlling for the other variables in the model. For more information, read my post about interpreting P values and regression coefficients.

However, you need to understand why P values are possible in linear regression before you can figure out why they are impossible to calculate for nonlinear regression.

The key point to understand is that a linear regression model is a very restricted form of a model. In a linear regression equation, all terms are either the constant or a parameter multiplied by an independent variable (IV). Then, you build the equation by only adding the terms together. These rules limit the form to just one type:

Dependent variable = constant + parameter * IV + … + parameter * IV

Because of these restrictions, you end up with a consistent form that makes it possible to create a single hypothesis test that is appropriate for all parameter estimates in all linear regression models. Regardless of what an independent variable measures, if the parameter is zero, the value of that term equals zero (0 * IV = 0). This condition indicates that the independent variable has no relationship with the dependent variable because it literally adds nothing to the dependent variable in the equation.

Given the consistent form, the following hypothesis test is valid for all terms in all linear regression models. β_{i} represents the parameter value for an independent variable.

- H
_{0}: β_{i}= 0 - H
_{A}: β_{i}<> 0

The P value for each term measures the amount of evidence against the null hypothesis that the parameter (coefficient) equals zero. If the P value is less than your significance level, reject the null and conclude that the parameter does not equal zero. Changes in the independent variable are related to changes in the dependent variable.

## Why Are P Values Incalculable in Nonlinear Regression?

Conversely, nonlinear regression models can take on virtually an infinite number of forms. There are almost no restrictions on how you can use parameters in a nonlinear regression equation. On the positive side, this flexibility provides nonlinear regression with the most flexible curve-fitting abilities.

However, because there is an incredibly diverse array of potential model forms, it’s impossible to devise a single hypothesis test for all parameters. Instead, the null hypothesis value of each parameter depends on the nonlinear function, the parameter’s location in it, and the research question.

What can you use instead of P values? You’ll need to use your knowledge of both the research area and the nonlinear function to identify the parameter value that corresponds to the null hypothesis. Then, assess the parameter estimates, and particularly the confidence interval of the estimate, to determine whether the variable is statistically significant. If the confidence interval of the estimate excludes the null value, you can conclude that the parameter is statistically significant.

For examples of nonlinear functions, see my post about the differences between linear and nonlinear regression.

To learn about when to use nonlinear regression, read the following:

Marisol Riddell says

Thanks Jim, but now I’m more confused. So my textbook says on page 271 of Introduction to Econometrics by Stock and Watson, that the adjusted R^2 can be used to compare the Log-Lin and log Log models, but not Lin-Log to Log-Log, so the point is that the dependent variable must be the same, but its okay if it’s nonlinear… can you please help me clarify?

Jim Frost says

Hi Marisol,

So, I’m not completely sure why your book is saying that but I might not be fully clear on what it is saying.

It sounds like you’re talking about semi-log models and log-log models and both of those are ways to incorporate nonlinear relationships into the linear model framework. R-squared and F-tests should be valid in these contexts. But, again, I might be missing some detail that your textbook raises.

I do talk about these types of models a bit in posts about fitting curvature and another about log-log plots and models.

Marisol Riddell says

Hi Jim,

Love this website. It has helped with several questions I had. One question I can’t seem to find a straight answer to anywhere is this:

I know I can not use r^2 to determine fit for a nonlinear model… but the F-test is measured with adjusted r^2. Can I use an F test to determine fit of a nonlinear model?

Thanks!

Jim Frost says

Hi Marisol,

That’s a great question. Alas, F-tests are only for linear models, and you can’t use them for nonlinear models. By nonlinear, I’m referring to the statistical sense of the word. You can use F-tests for linear models that use polynomials (and other methods) to model curvature.

Personally, I like to use the standard error of the regression to determine how well a nonlinear model fits the data. You can also assess the residual plots and look for the same things you would when using linear regression.

I hope this helps!

Sameer Kesava says

Hi Jim,

Thank you very much for the blogs, they are very helpful. I have a question with regards to nonlinear data fitting. I have scientific data to which I am fitting linear combination of functions such as Lorentz, Gaussian representing to model a system’s physical properties. The software that measures the data is also used to analyze the data and utilizes Levenberg-Marquardt algorithm to fit the starting models (with starting parameters and range for parameters). After data fitting, it spits out values for parameters with their error estimates and the Mean Squared Error for the fit. And importantly, the correlation matrix for the parameters and no p-values. Some of the parameters have a high correlation (close to 1 or -1) while some have low. The model representing the physical phenomenon can sometimes have as high as 30 parameters. How do I test if some of the parameters are redundant because even if the correlation is high between, lets say, parameters x and y, sometimes both are necessary and sometimes one can be fixed while other is varied. Most important, is there an accepted threshold value for the correlation coefficient below which we can say these parameters are independent. The most important aspect of my analysis is to obtain models representing physical phenomenon which have independent parameters. I apologize for such a long question. I will be eagerly waiting for your reply.

Thank you

Sameer

Vikash says

Nice sir…

Jim Frost says

Thank you, Vikash!

Dwarkesh says

Hello Jim,

Thanks for posting concepts regularly.

I have got a question… When you say that your model is non linear I guess you mean it is nonlinear in parameters, or estimate, because for model which are nonlinear in variable you can transform variable to have a linear relation.

Now the topic mentioned above is for nonlinear in parameter, i guess. And can we use p- value fundamental in nonlinear in variable models? Or it holds for both nonlinear models.

Keep posting great ideas.

Thanks,

Dwarkesh

Jim Frost says

Hi Dwarkesh, yes, I mean a model that is nonlinear in the parameters. That’s a great point and I’ll go back and make it more clear in the post . . . that I’m referring to nonlinear in the parameters. Thank you! And, you’re correct, you can model curvature using a linear model (polynomial, logs, etc) and p-values are quite appropriate there.

Psych n Stats Tutor (@psychnstats) says

Contrasting linear and non-linear was helpful, thanks. Am used to working only with linear

Jim Frost says

You’re very welcome. I’m glad that you found it helpful! I find that it’s easy to forget that linear regression is a very specific special case.

Vishal says

Hi, Jim,

I came across your website on facebook and i find your articles well organized and easy to understand. I have been meaning to ask, this question – why do we have the standard P value as 0.05? Why is it not 0.10 or 0.5 for that matter? When i do ask someone this question, i receive vague answers like, it is as it is, or this was what has been followed for many years.

Jim Frost says

Hi Vishal, first, thanks for the compliments! That makes my day! You’re actually asking about the significance levels rather than P values. The experimenter chooses the significance level, but the P values are calculated by the hypothesis test. Part of the reason we commonly use 0.05 is because of tradition. In fact, both 0.01 and 0.05 were set in place long ago and have persisted over the years. There is some logic behind these values. The significance level is the probability of obtaining a false positive when the null hypothesis is actually correct. So, you know that you want a low value. 10% and 50% error rates were deemed too high. You could also go very low, such as 0.001 and only have a 0.1% error rate. However, lower significance levels also reduce the power of the study. This reduction means that if you use a very low significance level, you might miss some real effects. So, 0.01 and 0.05 were seen as good trade-offs between avoiding false positives while not reducing power too much. However, once they were set in place almost a century ago, they persisted without question largely. However, that is changing right now. Both Bayesian analysis and simulations studies have shown that if you obtain a p-value that is near 0.05, it’s not very strong evidence because it results in a higher than expected error rate. Consequently, there is currently a push to lower the standard significance level from 0.05 to 0.005. We’ll have to see if the proposed standard is adopted or whether we stick with the traditional values!

Vishal says

Hi, Jim,

I appreciate this detailed explanation. Its starting to make sense after reading it over and over. I’m happy with your answer. Thank you!

I will recommend your page to my friends. It is like no other!

Jim Frost says

Hi Vishal, you’re very welcome! Thank you for recommending my website. I really appreciate that!