# Interpreting the log-likelihood in ordinary least squares linear regression

I have the following OLS regression results. The F-statistic is 57.95 which on a $$F_{1,22}$$ distribution results in p-value $$1.33 \times 10^{-7}$$. The p-value of $$F_{1,22,0.99} = 7.95$$. From this I deduce that we deem the regression to be significant at the 99% level.

The regression quality, $$R^2$$, is not bad either at 0.725. However, looking at the plot of the data against fitted values, we can see that this is clearly a non-linear relationship and so the regression does not really fit.

Without plotting the fit and inspecting it visually, is there anything in the regression statistics that would have told me this is a poor model? The only thing that stands out is the positive Log-Likelihood at 11.717.

                            OLS Regression Results
==============================================================================
Dep. Variable:                      y   R-squared:                       0.725
Method:                 Least Squares   F-statistic:                     57.95
Date:                Thu, 25 Feb 2021   Prob (F-statistic):           1.33e-07
Time:                        10:34:36   Log-Likelihood:                 11.717
No. Observations:                  24   AIC:                            -19.43
Df Residuals:                      22   BIC:                            -17.08
Df Model:                           1
Covariance Type:            nonrobust
==============================================================================
coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
Intercept      0.1779      0.037      4.868      0.000       0.102       0.254
x           8.348e-05    1.1e-05      7.612      0.000    6.07e-05       0.000
==============================================================================
Omnibus:                        8.432   Durbin-Watson:                   0.384
Prob(Omnibus):                  0.015   Jarque-Bera (JB):                2.289
Skew:                           0.276   Prob(JB):                        0.318
Kurtosis:                       1.591   Cond. No.                     3.85e+03
==============================================================================

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 3.85e+03. This might indicate that there are
strong multicollinearity or other numerical problems.

MSr: 1.180728086543658

MSe: 0.1551049709039136


• The positive log likelihood happens because the error variance is small, which does not tell you anything about model fit. (E.g., if you divide all $Y$ values by 100 you do not affect the model fit, but the log likelihood changes a lot.) – BigBendRegion Feb 25 at 13:05

No there is nothing from that output that could have told you about the possibility of model misspecification.

The formal way to see if there is misspecification (test if the model is indeed nonlinear) would have been to run a "Ramsay reset test".

This mean to estimate the following: $$1: y = a_0 + a_1 x + e$$ calculate: $$\hat{y} = a_0 + a_1 x$$ and estimate: $$2: y = b_0 + b_1 x + b_2 \hat{y}^2 + b_3 \hat{y}^3+e$$

If the $$b_2$$ and or $$b_3$$ are significant, then you had a problem of misspecification (trying to fit a linear model to a nonlinear relationship)

HTH

• Or any of the other bajillion tests for nonlinearity. But since all but the most trivial relationships are nonlinear, the alternative hypothesis is known to be true, a priori. Therefore, tests should not be used to evaluate the linearity (or any other) model assumption, except as a minor adjunct to see whether the patterns shown in the graphs and summary statistics are explainable by chance alone. – BigBendRegion Feb 25 at 13:05
• True, most relationships are in essence nonlinear. The question one answers with this type of test is if a linear function is a sufficiently close approximation to the true functional form so that you can make some generalizations in terms of interpreting the results. – Fcold Feb 25 at 13:32
• Not true. The results are sample site dependent. The linear approximation may be a "sufficiently close approximation," but rejected simply because of a large sample size. – BigBendRegion Feb 26 at 15:06
• Conversely, the linear approximation can be horrible, yet the test "accepts" linearity because of a very small sample size. – BigBendRegion Feb 26 at 15:12