Precision in summary output of lm R - r

Question

I am doing some exercises using package r-exams, in which I print a summary from an lm object and ask students things like, “which is the estimated value of the intercept”. The idea is that the student copies the values of the summary output and use that value as the correct answer. The issue here is that I use the values from coef() function as the correct answers, but this is not a good idea since the precision of these values are quite different from the precision of the values shown in the summary output. Here is an example:
set.seed(123)
library(tidyverse)
## DATA GENERATION
xbreaks<-c(runif(1,4,4.8),runif(1,6,6.9),runif(1,7.8,8.5),runif(1,9,10))
ybreaks<-c(runif(1,500,1000),runif(1,1800,4000),runif(1,200,800))
b11<-(ybreaks[2]-ybreaks[1])/(xbreaks[2]-xbreaks[1])
b10<-ybreaks[1]-b11*xbreaks[1]
b31<-(ybreaks[3]-ybreaks[2])/(xbreaks[4]-xbreaks[3])
b30<-ybreaks[2]-b31*xbreaks[3]
points_df<-data.frame(x=xbreaks,y=ybreaks[c(1,2,2,3)])
n<-rpois(3,120)
x1<-runif(n[1],xbreaks[1],xbreaks[2])
x2<-runif(n[2],xbreaks[2],xbreaks[3])
x3<-runif(n[3],xbreaks[3],xbreaks[4])
y<-c(b10+b11*x1+rnorm(n[1],0,200),
ybreaks[2]+rnorm(n[2],0,200),
b30+b31*x3+rnorm(n[3],0,200))
z0_aw<-data.frame(ph=c(x1,x2,x3),UFC=y,case=factor(c(rep(1,n[1]),rep(2,n[2]),rep(3,n[3]))))
mean_x<-z0_aw$ph%>% mean %>% round(2)
caserng<-sample(1:4,1)
modrng<-sample(1:2,1)
if(caserng!=4){
z0_aw<-z0_aw[z0_aw$case == caserng,]
}
if(modrng==1){
m0<-lm(UFC~ph,data=z0_aw)
}else{
cl <- call("lm", formula = UFC ~ I(ph - mean_x), data = as.name("z0_aw"))
cl$formula[[3]][[2]][[3]] <- mean_x
m0<-eval(cl)
}
summary(m0)
#>
#> Call:
#> lm(formula = UFC ~ I(ph - 7.2), data = z0_aw)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -555.53 -121.98 5.46 115.38 457.08
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 2726.86 57.33 47.57 <2e-16 ***
#> I(ph - 7.2) -840.05 31.46 -26.70 <2e-16 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 182.7 on 116 degrees of freedom
#> Multiple R-squared: 0.8601, Adjusted R-squared: 0.8589
#> F-statistic: 713.1 on 1 and 116 DF, p-value: < 2.2e-16
coef(m0)
#> (Intercept) I(ph - 7.2)
#> 2726.8605 -840.0515
Created on 2021-05-14 by the reprex package (v2.0.0)
Suppose that extol: 0.0001 in r-exams is set, and the student is asked to give the estimated value of the intercept. The student will get a wrong answer since he will answer 2726.86 but the correct answer from coef is 2726.8605 .
As can be seen, output of summary uses 2 decimals, whereas coef() values has quite more precision. I want to know how many decimals is summary using in order to apply the same format to values produced by coef(). This will ensure that the answer provided by the student is the same as the summary output.
I just want to do this:
answers<-coef(m0) %>% format(digits=dsum) %>% as.numeric()
where dsum is the number of digits used also by the summary output.
Note: retain a precision of 4 decimals is needed since I also ask students about the R-squared value provided in the same summary output, so it is not a good idea to set extol: 0.01 for example. Also the problems are generated at random and the magnitude of the estimated coefficients changes, as I have noted that this is directly related to the precision used in summary output.

Answer 1

Some useful information for such questions in R/exams:
The extol can also be a vector so that you can set different tolerances for coefficients and R-squared etc.
When asking about the R-squared, though, I typically ask for it "in percent". Then the same tolerance may be suitable as for the coefficients.
I would recommend to control the size of the coefficients suitably so that digits and extol can be set accordingly.
Personally, I typically store the exsolution at a higher precision than I request from the students. For example, exsolution could be 12.345678 while I only set extol to 0.01. This makes sure that when the correct answer is rounded to two decimal places it is inside the correct interval determined by exsolution and extol.
Details on formatting of the coefficients in the summary:
It is not obvious where exactly the formatting happens: The summary() method for lm objects returns an object of class summary.lm which has its own print() method which in turn calls printCoefmat(). The latter is the function that does the actual formatting.
When setting the digits in these functions, this controls the number of significant digits and not the number of decimal places. This is particularly important when the coefficients become relatively large (say, in the thousands or more).
The coefficients are not formatted individually but jointly with the corresponding standard errors. The details depend on the digits, the size of both coefficients and standard errors, and whether any coefficients are aliased or exactly zero etc.
Without aliased/zero coefficients the formatting from summary(m0) can be replicated using format_coef(m0) as defined below. That's essentially the boiled-down code from printCoefmat().
format_coef <- function(object, digits = max(3L, getOption("digits") - 2L)) {
coef_se <- summary(object)$coefficients[, 1L:2L]
digmin <- 1L + floor(log10(range(abs(coef_se))))
format(round(coef_se, max(1L, digits - digmin)), digits = digits)[, 1L]
}