I have a dataset in which the dependent variable (y) has known and substantial error, and yet the observations happen to line up quite well along a line when plotted against the independent variable (x). Fitting a linear regression seems to substantially overestimate the precision of the slope estimate for y vs. x.

How can one appropriately propagate the known error in y through to the estimate of the slope?

I think there is part of an answer here, but it assumes the point fit a linear regression exactly: Calculate uncertainty of linear regression slope based on data uncertainty

As a reproducible example in R:

# the data
dat <- data.frame(x = 0:8, y = seq(0,16, length.out=9)+rnorm(9, 0, 0.5), y.se = 3)

# fit a naive model, not considering error in y
mod <- lm(y ~ x, dat)
preds <- predict(mod, se.fit = TRUE)

plot(dat$x, dat$y, ylim=c(-7,22))
arrows(dat$x, dat$y-1.96*dat$y.se, dat$x, dat$y+1.96*dat$y.se, length=0)

# plot the confidence interval on the linear regression
polygon(c(dat$x, rev(dat$x)), c(preds$fit+preds$se.fit, rev(preds$fit-preds$se.fit)), col = 'grey')

enter image description here

The slope is estimated very precisely near 2.0:

            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.04670    0.32783   0.142    0.891    
x            1.97546    0.06886  28.689 1.61e-08 ***

Visually, however, a slope as low as 0.7 or as high as 3.3 would still fit through the error bounds of y quite well.

  • $\begingroup$ since the line goes through the observations so nicely maybe the error is overestimated? $\endgroup$
    – Aksakal
    Jan 6, 2022 at 16:47
  • $\begingroup$ In this case, the error is not overestimated. Thanks. $\endgroup$ Jan 6, 2022 at 16:56
  • $\begingroup$ you can easily repurpose your code example to simulate different slopes (like bootstrapping) to see that the uncertainty should be very small. it is extremely unlikely for your experiment to produce a slope of 0.7 or 3.3 $\endgroup$
    – Aksakal
    Jan 6, 2022 at 18:44
  • $\begingroup$ One must be clear about the difference between mean-response and predicted-response errors. The variance of a mean response estimate is determined by the coefficient covariance matrix, giving the confidence bands you show. Many cases or a wide span of x values can counterbalance high variance in individual y values and give precise slope estimates, implicit in the answer from @user347168. The variance of predicted response for a single new case adds to that the MSE for y. That's more related to what your vertical bars show. $\endgroup$
    – EdM
    Jan 19, 2022 at 20:43

2 Answers 2


This can be handled with structural equation modelling (SEM)


code = '
yhat ~ x     # Latent variable predicted by x
yhat =~ 1*y  # y is the single indicator of yhat
y ~~ 9*y     # y has error variance of 9 (3^2)

fit = lavaan(code, dat)
lavaan 0.6-7 ended normally after 3 iterations

  Estimator                                         ML
  Optimization method                           NLMINB
  Number of free parameters                          1
  Number of observations                             9
Model Test User Model:
  Test statistic                                24.572
  Degrees of freedom                                 1
  P-value (Chi-square)                           0.000

Parameter Estimates:

  Standard errors                             Standard
  Information                                 Expected
  Information saturated (h1) model          Structured

Latent Variables:
                   Estimate  Std.Err  z-value  P(>|z|)
  yhat =~                                             
    y                 1.000                           

                   Estimate  Std.Err  z-value  P(>|z|)
  yhat ~                                              
    x                 1.975    0.387    5.101    0.000

                   Estimate  Std.Err  z-value  P(>|z|)
   .y                 9.000                           
   .yhat              0.000   

Plots can also be useful...

semPaths(fit, what = 'path', whatLabels = 'est', layout = 'circle2')

enter image description here

  • $\begingroup$ Could you point me towards some background reading on this approach? It looks promising. $\endgroup$ Jan 7, 2022 at 17:08
  • $\begingroup$ I think I learned about this from Kaplan's book, but this blog post might also be a good starting point. $\endgroup$
    – Eoin
    Jan 7, 2022 at 18:52

I'm not quite sure what your problem actually is. If there's random error in y but not x, the regression of y on x will be unbiased. The uncertainty due to the error will be properly reflected in the standard error of the parameter estimate.

In your example, the y.se variable is only used to draw the "arrows" in the graph. It's not used to calculate y, so it's not reflected in the lm results. Your y and x are almost perfectly correlated (r>.99), so one can predict the other almost perfectly.


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