Standard Error of the Estimate

Learning Objectives

Make judgments about the size of the standard error of the estimate from a scatter plot
Compute the standard error of the estimate based on errors of prediction
Compute the standard error using Pearson's correlation
Estimate the standard error of the estimate based on a sample

Figure 1 shows two regression examples. You can see that in Graph A, the points are closer to the line than they are in Graph B. Therefore, the predictions in Graph A are more accurate than in Graph B.

Figure 1. Regressions differing in accuracy of prediction.

The standard error of the estimate is a measure of the accuracy of predictions. Recall that the regression line is the line that minimizes the sum of squared deviations of prediction (also called the sum of squares error). The standard error of the estimate is closely related to this quantity and is defined below:

$\sigma_{e x}=\sqrt{\frac{\sum\left(Y-Y^{\prime}\right)^{2}}{N}}$

where σest is the standard error of the estimate,

$Y$ is an actual score,

$Y'$ is a predicted score, and

$N$ is the number of pairs of scores. The numerator is the sum of squared differences between the actual scores and the predicted scores.

Note the similarity of the formula for σest to the formula for

$σ$ . It turns out that σest is the standard deviation of the errors of prediction (each

$Y$ -

$Y'$ is an error of prediction).

Assume the data in Table 1 are the data from a population of five

$X$ ,

$Y$ pairs.

Table 1. Example data.

	$X$	$Y$	$Y'$	$Y-Y'$	$(Y-Y')^2$
	1.00	1.00	1.210	-0.210	0.044
	2.00	2.00	1.635	0.365	0.133
	3.00	1.30	2.060	-0.760	0.578
	4.00	3.75	2.485	1.265	1.600
	5.00	2.25	2.910	-0.660	0.436
Sum	15.00	10.30	10.30	0.000	2.791

The last column shows that the sum of the squared errors of prediction is 2.791. Therefore, the standard error of the estimate is

$\sigma_{e x}=\sqrt{\frac{2.791}{5}}-0.747$

There is a version of the formula for the standard error in terms of Pearson's correlation:

$\sigma_{e \alpha}=\sqrt{\frac{\left(1-\rho^{2}\right) S S Y}{N}}$

where

$ρ$ is the population value of Pearson's correlation and

$SSY$ is

$S S Y=\sum\left(Y-\mu_{Y}\right)^{2}$

For the data in Table 1,

$μ_y = 2.06$ ,

$SSY = 4.597$ and

$ρ= 0.6268$ . Therefore,

$\sigma_{e x}=\sqrt{\frac{\left(1-0.6268^{2}\right)(4.597)}{5}}=\sqrt{\frac{2.791}{5}}=0.747$

which is the same value computed previously.

Similar formulas are used when the standard error of the estimate is computed from a sample rather than a population. The only difference is that the denominator is

$N-2$ rather than

$N$ . The reason

$N-2$ is used rather than

$N-1$ is that two parameters (the slope and the intercept) were estimated in order to estimate the sum of squares. Formulas for a sample comparable to the ones for a population are shown below.

R code

                x=c(1,2,3,4,5)

                y= c(1,2,1.3,3.75,2.25)

                summary(lm(y~x))

                Call:

                lm(formula = y ~ x)


                Residuals:

                     1      2      3      4      5

                -0.210  0.365 -0.760  1.265 -0.660


                Coefficients:

            Estimate Std. Error t value Pr(>|t|)

(Intercept)    0.785      1.012   0.776    0.494

x              0.425      0.305   1.393    0.258


                Residual standard error: 0.9645 on 3 degrees of freedom

Multiple R-squared:  0.3929,    Adjusted R-squared:  0.1906

F-statistic: 1.942 on 1 and 3 DF,  p-value: 0.2578

Source: David M. Lane , https://onlinestatbook.com/2/regression/accuracy.html
This work is in the Public Domain.