x{\displaystyle x} y{\displaystyle y} y′{\displaystyle y^{\prime }} y−y′{\displaystyle y-y^{\prime }} (y−y′)2{\displaystyle (y-y^{\prime })^{2}} Note that the table shown in the image above performs the opposite subtractions, y′−y{\displaystyle y^{\prime }-y}. The more standard order, however, is y−y′{\displaystyle y-y^{\prime }}. Because the values in the final column are squared, the negative is not problematic and will not change the outcome. Nevertheless, you should recognize that the more standard calculation is y−y′{\displaystyle y-y^{\prime }}.
The order of the data and the pairing is important for these calculations. You need to be careful to keep your paired data points together in order. For the sample calculations shown above, the data pairs are as follows: (1,2) (2,4) (3,5) (4,4) (5,5)
For this article, it is assumed that you will have the regression line equation available or that it has been predicted by some prior means. For the sample data set in the image above, the regression line is y′=0. 6x+2. 2{\displaystyle y^{\prime }=0. 6x+2. 2}.
Using the equation of the regression line, calculate or “predict” values of y′{\displaystyle y^{\prime }} for each value of x. Insert the x-value into the equation, and find the result for y′{\displaystyle y^{\prime }} as follows: y′=0. 6x+2. 2{\displaystyle y^{\prime }=0. 6x+2. 2} y′(1)=0. 6(1)+2. 2=2. 8{\displaystyle y^{\prime }(1)=0. 6(1)+2. 2=2. 8} y′(2)=0. 6(2)+2. 2=3. 4{\displaystyle y^{\prime }(2)=0. 6(2)+2. 2=3. 4} y′(3)=0. 6(3)+2. 2=4. 0{\displaystyle y^{\prime }(3)=0. 6(3)+2. 2=4. 0} y′(4)=0. 6(4)+2. 2=4. 6{\displaystyle y^{\prime }(4)=0. 6(4)+2. 2=4. 6} y′(5)=0. 6(5)+2. 2=5. 2{\displaystyle y^{\prime }(5)=0. 6(5)+2. 2=5. 2}
For the data in the sample set, these calculations are as follows: y(x)−y′(x){\displaystyle y(x)-y^{\prime }(x)} y(1)−y′(1)=2−2. 8=−0. 8{\displaystyle y(1)-y^{\prime }(1)=2-2. 8=-0. 8} y(2)−y′(2)=4−3. 4=0. 6{\displaystyle y(2)-y^{\prime }(2)=4-3. 4=0. 6} y(3)−y′(3)=5−4=1{\displaystyle y(3)-y^{\prime }(3)=5-4=1} y(4)−y′(4)=4−4. 6=−0. 6{\displaystyle y(4)-y^{\prime }(4)=4-4. 6=-0. 6} y(5)−y′(5)=5−5. 2=−0. 2{\displaystyle y(5)-y^{\prime }(5)=5-5. 2=-0. 2}
For the sample data set, these calculations are as follows: −0. 82=0. 64{\displaystyle -0. 8^{2}=0. 64} 0. 62=0. 36{\displaystyle 0. 6^{2}=0. 36} 12=1. 0{\displaystyle 1^{2}=1. 0} −0. 6=0. 36{\displaystyle -0. 6=0. 36} −0. 2=0. 04{\displaystyle -0. 2=0. 04}
For this sample data set, this calculation is as follows: 0. 64+0. 36+1. 0+0. 36+0. 04=2. 4{\displaystyle 0. 64+0. 36+1. 0+0. 36+0. 04=2. 4}
If the measured data represents an entire population, then you will find the average by dividing by N, the number of data points. However, if you are working with a smaller sample set of the population, then substitute N-2 in the denominator. For the sample data set in this article, we can assume that it is a sample set and not a population, just because there are only 5 data values. Therefore, calculate the Standard Error of the Estimate as follows: σ=2. 45−2{\displaystyle \sigma ={\sqrt {\frac {2. 4}{5-2}}}} σ=2. 43{\displaystyle \sigma ={\sqrt {\frac {2. 4}{3}}}} σ=0. 8{\displaystyle \sigma ={\sqrt {0. 8}}} σ=0. 894{\displaystyle \sigma =0. 894}
With this small sample set, the standard error score of 0. 894 is quite low and represents well organized data results.
