Is it okay to rely on the Residual Standard Error of a zero intercept linear regression (given that R-squared is not reliable in this case)? $y = f(x,...)$
$y$ represents the total effort in hours to complete a mix of tasks (each $x$ is a different mix) of varying complexity. It might be a non-linear relationship, but for now, I only have the count of tasks. No complexity data. I'd like to understand the correlation between total effort and task count.
A simple regression throws this up:
> summary(lm(data=dat, y ~ x))

Call:
lm(formula = y ~ x, data = dat)

Residuals:
     Min       1Q   Median       3Q      Max 
-2912.84  -189.26    12.88   148.09  3138.23 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)      -148.09     102.96  -1.438    0.155    
x                 146.76      13.89  10.568 1.69e-15 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 724.2 on 62 degrees of freedom
Multiple R-squared:  0.643, Adjusted R-squared:  0.6373 
F-statistic: 111.7 on 1 and 62 DF,  p-value: 1.687e-15


I re-ran it with a forced zero intercept.
It now says:
> summary(lm(data=dat, y ~ 0 + x))

Call:
lm(formula = y ~ 0 + x, data = dat)

Residuals:
    Min      1Q  Median      3Q     Max 
-2775.5  -287.4  -118.5     0.0  3389.7 

Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
x                 137.24      12.31   11.15   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 730.3 on 63 degrees of freedom
Multiple R-squared:  0.6635,    Adjusted R-squared:  0.6582 
F-statistic: 124.2 on 1 and 63 DF,  p-value: < 2.2e-16

Question: Which of the two above is a better fit? Should the Residual Standard Error* (RSE) be the guide? Or should I simply go with the second model because a zero-intercept makes sense in this case?
Note: The post shared by Nick Cox states that $R^2$ must not be relied upon for zero-intercept cases. I don't fully understand the math described there but I'm hoping that it is okay to rely on the Residual Standard Error in a zero-intercept scenario. Could someone please confirm this as well?
*Reference: This article says RSE is also a goodness-of-fit measure.
 A: While this remains open (full disclosure: I voted to close) I will offer another answer, which is (to start negatively) that the question is hard and unclear, as it mixes together various general issues with distinctly limited specific details arising from the analysis of one dataset. I will also try to bring together various comments already made.
So far as I can see, nothing in the question really illuminates quite what is best or better as an analysis for the dataset being alluded to or for other datasets on the relation between (if I understand this correctly)
number of tasks completed as a predictor
and
total effort as an outcome.
In abstraction, it does seem that zero effort would imply zero achievement as a limiting condition, but that said

*

*The relevance of this as a constraint is an open question, but even if it seems compelling, it still allows many different functional forms, starting with $y = bx$, $y = bx + cx^2$, $y = ax^b$, and so on. Yet broader is a possibility that no simple algebraic specification will work well.


*To advise well on what to do with the data, we need to see the data, which is in principle easy for a small dataset. (I asked in comments for a scatter plot, but a listing would be even better.)
The characteristic terseness of lm output from R is defensible on various grounds but it is not sufficient to advise here on which model makes most sense (substantively, scientifically) and fits the data well, which in my view should be the main questions here.
The idea that a single summary criterion should be sought to make a decision for you seems utterly misguided here, and in general.
A: Answer from my comments
For a model without an intercept, if you want an r-squared-like measure of fit, you might use a version of Efron's pseudo r-squared. It's defined here:

*

*stats.oarc.ucla.edu/other/mult-pkg/faq/general/faq-what-are-pseudo-r-squareds/
With the caveat, that I wrote the former, there's an implementation in R in the rcompanion package, and in the performance package.
As an example:
A = c(1,2,3,4,5,6,7,8,9)
B = c(2,4,5,3,6,5,7,9,8)

model = lm(B ~ 0 + A)

summary(model)

    ### summary() reports an r-squared of 0.955.
    ###  This is probably not a desirable statistic to use
    ###   for a linear model without an intercept.

library(rcompanion)

accuracy(list(model))

   ### Efron.r.squared
   ###            0.67

library(performance)

r2_efron(model) 

   ### 0.6704986

For an OLS model with an intercept, this Efron's pseudo r-squared will be  equal to r-squared.
A: Your question is a bit of a mixture of issues but the key note sounds like 'In model selection, does the intercept have a special role?'.

*

*The intercept is in principle not different from other regressor variables.
It has no special role that makes that you shouldn't remove it any less in comparison to similar situations with other regressor variables.


*However, in practice we often keep the intercept or get an implicit intercept.
Either an intercept occurs in the model anyway because we add a categorical variable (which implicitly adds an intercept even when it wasn't there before)
Or an intercept is common in many models and it is included as a standard as a sort of prior knowledge.
In relationship to the R² value. That measure often relates to a comparison of the variance by expressing it as a decomposition of the variance into different parts. This makes in particular sense when the models are nested. But, if a model has no intercept then it is not anymore nested. This is because the baseline, the total variance, is the computed relative to the mean of the data, and it is a model that contains by definition an intercept term.

Another part of your question is:

Which of the two above is a better fit?

But that is a very broad topic (even without thinking about the intercept) and the question is asked not very clearly.
How do you determine what is better? It depend on the situation. What is your goal, how do you define better?
AIC
If you have a reasonable assumption about the error distribution, then one way to decide on the better test could be the use of the Akaike information criterion. It is a way to compare likelihood while keeping the number of fitted parameters into account.
In the specific case of Gaussian distributed errors with equal variance (note that this might not need to be your case, this is an example): With a simplified expression for Gaussian distributed error terms (Calculate AIC for both linear and non-linear models) $AIC = 2k + n \log(RSS) - n \log(n) + C$ in your case the $n\log(RSS)$ term changes only a little (namerly 64 * log(724.2*62) ≈ 10.71 versus 64 * log(730.3*63) ≈ 10.74, which is only a small improvement/reduction) and the AIC increases by much more due to the extra parameter, which increases the term 2k.
From this perspective the model without intercept is better.
R²
An advantage or R² is that it is intuitive. However it has some limitations. For instance, unlike AIC mentioned above the R² does not take into account the number of parameters used in fitting a model. The issue with the intercept occurs because there are different methods to define and compute R². For instance it could be 'the correlation between observations and estimated values' and in simple linear regression, y = a + bx, this correlation does not change when we remove the intercept.
