The plot of our population of data suggests that the college entrance test scores for each subpopulation have equal variance. We denote the value of this common variance as \(\sigma^{2}\).
That is, \(\sigma^{2}\) quantifies how much the responses (y) vary around the (unknown) mean population regression line \(\mu_Y=E(Y)=\beta_0 + \beta_1x\).
Why should we care about \(\sigma^{2}\)? The answer to this question pertains to the most common use of an estimated regression line, namely predicting some future response.
Suppose you have two brands (A and B) of thermometers, and each brand offers a Celsius thermometer and a Fahrenheit thermometer. You measure the temperature in Celsius and Fahrenheit using each brand of thermometer on ten different days. Based on the resulting data, you obtain two estimated regression lines — one for brand A and one for brand B. You plan to use the estimated regression lines to predict the temperature in Fahrenheit based on the temperature in Celsius.
Will this thermometer brand (A) yield more precise future predictions …?
… or this one (B)?
As the two plots illustrate, the Fahrenheit responses for the brand B thermometer don't deviate as far from the estimated regression equation as they do for the brand A thermometer. If we use the brand B estimated line to predict the Fahrenheit temperature, our prediction should never really be too far off from the actual observed Fahrenheit temperature. On the other hand, predictions of the Fahrenheit temperatures using the brand A thermometer can deviate quite a bit from the actual observed Fahrenheit temperature. Therefore, the brand B thermometer should yield more precise future predictions than the brand A thermometer.
To get an idea, therefore, of how precise future predictions would be, we need to know how much the responses (y) vary around the (unknown) mean population regression line \(\mu_Y=E(Y)=\beta_0 + \beta_1x\). As stated earlier, \(\sigma^{2}\) quantifies this variance in the responses. Will we ever know this value \(\sigma^{2}\)? No! Because \(\sigma^{2}\)is a population parameter, we will rarely know its true value. The best we can do is estimate it!
To understand the formula for the estimate of \(\sigma^{2}\)in the simple linear regression setting, it is helpful to recall the formula for the estimate of the variance of the responses, \(\sigma^{2}\), when there is only one population.
The following is a plot of the (one) population of IQ measurements. As the plot suggests, the average of the IQ measurements in the population is 100. But, how much do the IQ measurements vary from the mean? That is, how "spread out" are the IQs?
 Sample Variance

\(s^2=\dfrac{\sum_{i=1}^{n}(y_i\bar{y})^2}{n1}\)
The sample variance estimates \(\sigma^{2}\), the variance of one population. The estimate is really close to being like an average. The numerator adds up how far each response \(y_{i}\) is from the estimated mean \(\bar{y}\) in squared units, and the denominator divides the sum by n1, not n as you would expect for an average. What we would really like is for the numerator to add up, in squared units, how far each response \(y_{i}\)is from the unknown population mean \(\mu\). But, we don't know the population mean \(\mu\), so we estimate it with \(\bar{y}\). Doing so "costs us one degree of freedom". That is, we have to divide by n1, and not n because we estimated the unknown population mean \(\mu\).
Now let's extend this thinking to arrive at an estimate for the population variance \(\sigma^{2}\) in the simple linear regression setting. Recall that we assume that \(\sigma^{2}\) is the same for each of the subpopulations. For our example on college entrance test scores and grade point averages, how many subpopulations do we have?
There are four subpopulations depicted in this plot. In general, there are as many subpopulations as there are distinct x values in the population. Each subpopulation has its own mean \(\mu_{Y}\), which depends on x through \(\mu_Y=E(Y)=\beta_0 + \beta_1x\). And, each subpopulation mean can be estimated using the estimated regression equation \(\hat{y}_i=b_0+b_1x_i\).
 Mean square error

\(MSE=\dfrac{\sum_{i=1}^{n}(y_i\hat{y}_i)^2}{n2}\)
The mean square error estimates \(\sigma^{2}\), the common variance of the many subpopulations.
How does the mean square error formula differ from the sample variance formula? The similarities are more striking than the differences. The numerator again adds up, in squared units, how far each response \(y_{i}\) is from its estimated mean. In the regression setting, though, the estimated mean is \(\hat{y}_i\). And, the denominator divides the sum by n2, not n1, because in using \(\hat{y}_i\) to estimate \(\mu_{Y}\), we effectively estimate two parameters — the population intercept \(\beta_{0}\) and the population slope \(\beta_{1}\). That is, we lose two degrees of freedom.
In practice, we will let statistical software, such as Minitab, calculate the mean square error (MSE) for us. The estimate of \(\sigma^{2}\) shows up indirectly on Minitab's "fitted line plot." For example, for the student height and weight data (Student Height Weight data), the quantity emphasized in the box, \(S = 8.64137\), is the square root of MSE. That is, in general, \(S=\sqrt{MSE}\), which estimates \(\sigma\) and is known as the regression standard error or the residual standard error. The fitted line plot here indirectly tells us, therefore, that \(MSE = 8.64137^{2} = 74.67\).
The estimate of \(\sigma^{2}\) shows up directly in Minitab's standard regression analysis output. Again, the quantity S = 8.64137 is the square root of MSE. In the Analysis of Variance table, the value of MSE, 74.67, appears appropriately under the column labeled MS (for Mean Square) and in the row labeled Error.
Analysis of Variance
Source  DF  SS  MS  F  P 

Constant  1  5202.21  5202.21  69.67  0.000 
Error  8  597.39  74.67  
Total  9  5799.60 
Model Summary
S  Rsq  Rsq(adj)  Rsq(pred) 

8.64137  89.7%  88.4%  87.84% 
Regression Equation
wt =266.5 + 6.138 ht