Why should we compare estimates of generalized linear model with its corresponding standard errors?

Question

There is one concept in Statistics that I don't feel clear, and I could not find it in textbooks. Why sometimes do people compare coefficient estimates with corresponding standard errors? Here is the context:

I am reading the book An Introduction to Categorical Data Analysis by Alan Agresti (2nd edition, the thin version). In Chapter 5 section 5.1.2, it talks about an example for logistic regression with multiple predictors. After getting the results, the author was explaining how to interpret coefficient estimates. Two predictors are continuous variables (weight & width). The author says "The estimates for weight and width are only slightly larger than their SE values." Then the author starts to explain other coefficient estimates. So what does it say? "The estimates for weight and width are only slightly larger than their SE values." --- What does it say exactly? Is there any rule for comparing coefficient estimates with their corresponding standard errors? Thank you!

P.S.(edited) By saying "The estimates for weight and width are only slightly larger than their SE values", the author is indicating width and weight are weak effects. I don't understand Why they are weak effects. -- Only because their magnitudes are slightly larger than their SE values?

Answer 1

By saying "the estimates for weight and width are only slightly larger than their SE values" I believe the authors are underlining the (lack of) precision.

For example, let's imagine the coefficient is 100 and the SE is 4. Then, the 95% confidence interval will be 92.16 - 107.84.

On the other hand, let's imagine the same coefficient, but the SE is 40. Then, the 95% confidence interval will be 21.6 - 178.4.

I hope to have made it clearer.

Answer 2

We have point estimates and measures of dispersion. In this particular example, the coefficients are point estimates and the standard errors convey the dispersion. What is more, both are in the same unit (example, meters, pounds, etc.), and in the same scale. In other words, if we have a given coefficient (say, 1.35) with a short standard error (say, 0.3), this is different from having the same coefficient and a much larger standard error (say, 0.95).

Usually, the larger the sample size, the lower the standard error. Also, too much "noise" in the measurements will produce dispersion, i.e., lack of precision, hence a bigger SE.

Saying it in different words, the standard errors convey the precision of the estimates.

Last but not least, the SEs will influence the calculation of confidence intervals (+- 1.96*SE) as well as the p-values.

Answer 3

Agresti says that there is a "small P-value for the overall [likelihood-ratio] test", i.e. "extremely strong evidence that at least one predictor has an effect", but "lack of significance for individual effects." Indeed, if a confidence interval is $\pm 1.96\cdot\mathrm{SE}$, confidence intervals for weight and width are: $$[0.826-1.96\times0.704,0.826+1.96\times0.704]=[-0.55,2.21]$$ and $$[0.264-1.96\times 0.195,0.264+1.96\times 0.195]=[-0.12,0.65]$$ i.e. their effects could be null or even negative.

This is "a warning sign of multicollinearity," Agresty says. That means that if two or more predictors are correlated ("weight and width have a strong correlation (0.887)"), their compound effect is significative, but you can't estimate the single effect of each predictor.

"For practical purposes they are equally good predictors, but it is nearly redundant to use them both," this is why "Our further analysis uses width (W ) with color (C) and spine condition (S) as predictors."