Example.
Experimental data on the values of variables X And at are given in the table. 
As a result of their alignment, the function 
Using least square method, approximate these data with a linear dependence y=ax+b(find parameters but And b). Find out which of the two lines is better (in the sense of the least squares method) aligns the experimental data. Make a drawing.
The essence of the method of least squares (LSM).
The problem is to find the linear dependence coefficients for which the function of two variables but And b 
takes the smallest value. That is, given the data but And b the sum of the squared deviations of the experimental data from the found straight line will be the smallest. This is the whole point of the least squares method.
Thus, the solution of the example is reduced to finding the extremum of a function of two variables.
Derivation of formulas for finding coefficients.
A system of two equations with two unknowns is compiled and solved. Finding partial derivatives of a function with respect to variables but And b, we equate these derivatives to zero. 
We solve the resulting system of equations by any method (for example substitution method or ) and obtain formulas for finding coefficients using the least squares method (LSM). 
With data but And b function takes the smallest value. The proof of this fact is given.
That's the whole method of least squares. Formula for finding the parameter a contains the sums , , , and the parameter n- amount of experimental data. The values of these sums are recommended to be calculated separately. Coefficient b found after calculation a.
It's time to remember the original example.
Solution.
In our example n=5. We fill in the table for the convenience of calculating the amounts that are included in the formulas of the required coefficients. 
The values in the fourth row of the table are obtained by multiplying the values of the 2nd row by the values of the 3rd row for each number i.
The values in the fifth row of the table are obtained by squaring the values of the 2nd row for each number i.
The values of the last column of the table are the sums of the values across the rows.
We use the formulas of the least squares method to find the coefficients but And b. We substitute in them the corresponding values from the last column of the table: 
Consequently, y=0.165x+2.184 is the desired approximating straight line.
It remains to find out which of the lines y=0.165x+2.184 or better approximates the original data, i.e. to make an estimate using the least squares method.
Estimation of the error of the method of least squares.
To do this, you need to calculate the sums of squared deviations of the original data from these lines 
And 
, a smaller value corresponds to a line that better approximates the original data in terms of the least squares method. 
Since , then the line y=0.165x+2.184 approximates the original data better.
Graphic illustration of the least squares method (LSM).
Everything looks great on the charts. The red line is the found line y=0.165x+2.184, the blue line is , the pink dots are the original data.
What is it for, what are all these approximations for?
I personally use to solve data smoothing problems, interpolation and extrapolation problems (in the original example, you could be asked to find the value of the observed value y at x=3 or when x=6 according to the MNC method). But we will talk more about this later in another section of the site.
Proof.
So that when found but And b function takes the smallest value, it is necessary that at this point the matrix of the quadratic form of the second-order differential for the function was positive definite. Let's show it.
(see picture). It is required to find the equation of a straight line
The smaller the number in absolute value, the better the straight line (2) is chosen. As a characteristic of the accuracy of the selection of a straight line (2), we can take the sum of squares
The minimum conditions for S will be
 |
(6) |
 |
(7) |
Equations (6) and (7) can be written in the following form:
 |
(8) |
 |
(9) |
From equations (8) and (9) it is easy to find a and b from the experimental values x i and y i . The line (2) defined by equations (8) and (9) is called the line obtained by the least squares method (this name emphasizes that the sum of squares S has a minimum). Equations (8) and (9), from which the straight line (2) is determined, are called normal equations.
It is possible to indicate a simple and general way of compiling normal equations. Using experimental points (1) and equation (2), we can write down the system of equations for a and b
| y 1 \u003d ax 1 +b, | |||
| y 2 \u003dax 2 +b, ... |
(10) | ||
| yn=axn+b, | |||
Multiply the left and right parts of each of these equations by the coefficient at the first unknown a (i.e. x 1 , x 2 , ..., x n) and add the resulting equations, resulting in the first normal equation (8).
We multiply the left and right sides of each of these equations by the coefficient of the second unknown b, i.e. by 1, and add the resulting equations, resulting in the second normal equation (9).
This method of obtaining normal equations is general: it is suitable, for example, for the function
is a constant value and it must be determined from experimental data (1).
The system of equations for k can be written:
Find the line (2) using the least squares method.
Solution. We find:
x i =21, y i =46.3, x i 2 =91, x i y i =179.1.
We write equations (8) and (9)
From here we find
Estimating the accuracy of the least squares method
Let us give an estimate of the accuracy of the method for the linear case when equation (2) takes place.
Let the experimental values x i be exact, and the experimental values y i have random errors with the same variance for all i.
We introduce the notation
| (16) |
Then the solutions of equations (8) and (9) can be represented as
| (17) | |
| (18) | |
| where | |
| (19) | |
| From equation (17) we find | |
| (20) | |
| Similarly, from equation (18) we obtain | |
 |
(21) |
| because | |
 |
(22) |
| From equations (21) and (22) we find | |
 |
(23) |
Equations (20) and (23) give an estimate of the accuracy of the coefficients determined by equations (8) and (9).
Note that the coefficients a and b are correlated. By simple transformations, we find their correlation moment.
From here we find
0.072 at x=1 and 6,
0.041 at x=3.5.
Literature
Shore. I WOULD. Statistical Methods analysis and quality control and reliability. M.: Gosenergoizdat, 1962, p. 552, pp. 92-98.
This book is intended for a wide range of engineers (research institutes, design bureaus, test sites and factories) involved in determining the quality and reliability of electronic equipment and other mass industrial products (machine building, instrument making, artillery, etc.).
The book gives an application of methods mathematical statistics to the issues of processing and evaluating test results, in which the quality and reliability of the tested products are determined. For the convenience of readers, the necessary information from mathematical statistics is given, as well as a large number of auxiliary mathematical tables that facilitate the necessary calculations.
The presentation is illustrated a large number examples taken from the field of radio electronics and artillery technology.
The essence of the least squares method is in finding the parameters of the trend model that best describes the development trend of any random phenomenon in time or space (a trend is a line that characterizes the trend of this development). The task of the least squares method (OLS) is to find not just some trend model, but to find the best or optimal model. This model will be optimal if the sum of the squared deviations between the observed actual values and the corresponding calculated trend values is minimal (smallest):
where is the standard deviation between the observed actual value
and the corresponding calculated trend value,
The actual (observed) value of the phenomenon under study,
Estimated value of the trend model,
The number of observations of the phenomenon under study.
MNC is rarely used on its own. As a rule, most often it is used only as a necessary technique in correlation studies. It should be remembered that the information basis of the LSM can only be a reliable statistical series, and the number of observations should not be less than 4, otherwise, the smoothing procedures of the LSM may lose their common sense.
The OLS toolkit is reduced to the following procedures:
First procedure. It turns out whether there is any tendency at all to change the resulting attribute when the selected factor-argument changes, or in other words, whether there is a connection between " at " And " X ».
Second procedure. It is determined which line (trajectory) is best able to describe or characterize this trend.
Third procedure.
Example. Suppose we have information on the average sunflower yield for the farm under study (Table 9.1).
Table 9.1
|
Observation number |
||||||||||
|
Productivity, c/ha |
Since the level of technology in the production of sunflower in our country has not changed much over the past 10 years, it means that, most likely, the fluctuations in yield in the analyzed period depended very much on fluctuations in weather and climate conditions. Is it true?
First MNC procedure. The hypothesis about the existence of a trend in the change in sunflower yield depending on changes in weather and climate conditions over the analyzed 10 years is being tested.
In this example, for " y » it is advisable to take the yield of sunflower, and for « x » is the number of the observed year in the analyzed period. Testing the hypothesis about the existence of any relationship between " x " And " y » can be done in two ways: manually and using computer programs. Of course, with the availability of computer technology, this problem is solved by itself. But, in order to better understand the OLS tools, it is advisable to test the hypothesis about the existence of a relationship between " x " And " y » manually, when only a pen and an ordinary calculator are at hand. In such cases, the hypothesis of the existence of a trend is best checked visually by the location of the graphic image of the analyzed time series - the correlation field:
The correlation field in our example is located around a slowly increasing line. This in itself indicates the existence of a certain trend in the change in sunflower yield. It is impossible to speak about the presence of any trend only when the correlation field looks like a circle, a circle, a strictly vertical or strictly horizontal cloud, or consists of randomly scattered points. In all other cases, the hypothesis of the existence of a relationship between " x " And " y and continue research.
Second MNC procedure. It is determined which line (trajectory) is best able to describe or characterize the trend in sunflower yield changes for the analyzed period.
With the availability of computer technology, the selection of the optimal trend occurs automatically. With "manual" processing, the choice of the optimal function is carried out, as a rule, in a visual way - by the location of the correlation field. That is, according to the type of chart, the equation of the line is selected, which is best suited to the empirical trend (to the actual trajectory).
As you know, in nature there is a huge variety of functional dependencies, so it is extremely difficult to visually analyze even a small part of them. Fortunately, in real economic practice, most relationships can be accurately described either by a parabola, or a hyperbola, or a straight line. In this regard, with the "manual" option for selecting the best function, you can limit yourself to only these three models.
|
Hyperbola: |
||
|
|
|
Parabola of the second order: 
:
It is easy to see that in our example, the trend in sunflower yield changes over the analyzed 10 years is best characterized by a straight line, so the regression equation will be a straight line equation.
Third procedure. The parameters of the regression equation that characterizes this line are calculated, or in other words, an analytical formula is determined that describes the best trend model.
Finding the values of the parameters of the regression equation, in our case, the parameters and , is the core of the LSM. This process is reduced to solving a system of normal equations.
(9.2)
This system of equations is quite easily solved by the Gauss method. Recall that as a result of the solution, in our example, the values of the parameters and are found. Thus, the found regression equation will have the following form: 
I am a computer programmer. I made the biggest leap in my career when I learned to say: "I do not understand anything!" Now I am not ashamed to tell the luminary of science that he is giving me a lecture, that I do not understand what it, the luminary, is talking to me about. And it's very difficult. Yes, it's hard and embarrassing to admit you don't know. Who likes to admit that he does not know the basics of something-there. By virtue of my profession, I have to attend a large number of presentations and lectures, where, I confess, in the vast majority of cases I feel sleepy, because I do not understand anything. And I don’t understand because the huge problem of the current situation in science lies in mathematics. It assumes that all students are familiar with absolutely all areas of mathematics (which is absurd). To admit that you do not know what a derivative is (that this is a little later) is a shame.
But I've learned to say that I don't know what multiplication is. Yes, I don't know what a subalgebra over a Lie algebra is. Yes, I do not know why quadratic equations are needed in life. By the way, if you are sure that you know, then we have something to talk about! Mathematics is a series of tricks. Mathematicians try to confuse and intimidate the public; where there is no confusion, no reputation, no authority. Yes, it is prestigious to speak in the most abstract language possible, which is complete nonsense in itself.
Do you know what a derivative is? Most likely you will tell me about the limit of the difference relation. In the first year of mathematics at St. Petersburg State University, Viktor Petrovich Khavin me defined derivative as the coefficient of the first term of the Taylor series of the function at the point (it was a separate gymnastics to determine the Taylor series without derivatives). I laughed at this definition for a long time, until I finally understood what it was about. The derivative is nothing more than just a measure of how much the function we are differentiating is similar to the function y=x, y=x^2, y=x^3.
I now have the honor of lecturing students who fear mathematics. If you are afraid of mathematics - we are on the way. As soon as you try to read some text and it seems to you that it is overly complicated, then know that it is badly written. I argue that there is not a single area of mathematics that cannot be spoken about "on fingers" without losing accuracy.
The challenge for the near future: I instructed my students to understand what a linear-quadratic controller is. Don't be shy, waste three minutes of your life, follow the link. If you do not understand anything, then we are on the way. I (a professional mathematician-programmer) also did not understand anything. And I assure you, this can be sorted out "on the fingers." On the this moment I do not know what it is, but I assure you that we will be able to figure it out.
So, the first lecture that I am going to give to my students after they come running to me in horror with the words that a linear-quadratic controller is a terrible bug that you will never master in your life is least squares methods. Can you solve linear equations? If you are reading this text, then most likely not.
So, given two points (x0, y0), (x1, y1), for example, (1,1) and (3,2), the task is to find the equation of a straight line passing through these two points:
illustration
This straight line should have an equation like the following:
Here alpha and beta are unknown to us, but two points of this line are known:
You can write this equation in matrix form:
Here we should make a lyrical digression: what is a matrix? A matrix is nothing but a two-dimensional array. This is a way of storing data, no more values should be given to it. It is up to us how exactly to interpret a certain matrix. Periodically, I will interpret it as a linear mapping, periodically as a quadratic form, and sometimes simply as a set of vectors. This will all be clarified in context.
Let's replace specific matrices with their symbolic representation:
Then (alpha, beta) can be easily found:
More specifically for our previous data:
Which leads to the following equation of a straight line passing through the points (1,1) and (3,2):
Okay, everything is clear here. And let's find the equation of a straight line passing through three points: (x0,y0), (x1,y1) and (x2,y2):
Oh-oh-oh, but we have three equations for two unknowns! The standard mathematician will say that there is no solution. What will the programmer say? And he will first rewrite the previous system of equations in the following form:
In our case vectors i,j,b are three-dimensional, hence (in the general case) there is no solution to this system. Any vector (alpha\*i + beta\*j) lies in the plane spanned by the vectors (i, j). If b does not belong to this plane, then there is no solution (equality in the equation cannot be achieved). What to do? Let's look for a compromise. Let's denote by e(alpha, beta) how exactly we did not achieve equality:
And we will try to minimize this error:
Why a square?
We are looking not just for the minimum of the norm, but for the minimum of the square of the norm. Why? The minimum point itself coincides, and the square gives a smooth function (a quadratic function of the arguments (alpha,beta)), while just the length gives a function in the form of a cone, non-differentiable at the minimum point. Brr. Square is more convenient.
Obviously, the error is minimized when the vector e orthogonal to the plane spanned by the vectors i And j.
Illustration
In other words: we are looking for a line such that the sum of the squared lengths of the distances from all points to this line is minimal:
UPDATE: here I have a jamb, the distance to the line should be measured vertically, not orthographic projection. This commenter is correct.
Illustration
In completely different words (carefully, poorly formalized, but it should be clear on the fingers): we take all possible lines between all pairs of points and look for the average line between all:
Illustration
Another explanation on the fingers: we attach a spring between all data points (here we have three) and the line that we are looking for, and the line of the equilibrium state is exactly what we are looking for.
Quadratic form minimum
So, given the vector b and the plane spanned by the columns-vectors of the matrix A(in this case (x0,x1,x2) and (1,1,1)), we are looking for a vector e with a minimum square of length. Obviously, the minimum is achievable only for the vector e, orthogonal to the plane spanned by the columns-vectors of the matrix A:In other words, we are looking for a vector x=(alpha, beta) such that:
I remind you that this vector x=(alpha, beta) is the minimum of the quadratic function ||e(alpha, beta)||^2:
Here it is useful to remember that the matrix can be interpreted as well as the quadratic form, for example, the identity matrix ((1,0),(0,1)) can be interpreted as a function of x^2 + y^2:
quadratic form
All this gymnastics is known as linear regression.
Laplace equation with Dirichlet boundary condition
Now the simplest real problem: there is a certain triangulated surface, it is necessary to smooth it. For example, let's load my face model:
The original commit is available. To minimize external dependencies, I took the code of my software renderer, already on Habré. To solve the linear system, I use OpenNL , it's a great solver, but it's very difficult to install: you need to copy two files (.h + .c) to your project folder. All smoothing is done by the following code:
For (int d=0; d<3; d++) {
nlNewContext();
nlSolverParameteri(NL_NB_VARIABLES, verts.size());
nlSolverParameteri(NL_LEAST_SQUARES, NL_TRUE);
nlBegin(NL_SYSTEM);
nlBegin(NL_MATRIX);
for (int i=0; i<(int)verts.size(); i++) {
nlBegin(NL_ROW);
nlCoefficient(i, 1);
nlRightHandSide(verts[i][d]);
nlEnd(NL_ROW);
}
for (unsigned int i=0; i
X, Y and Z coordinates are separable, I smooth them separately. That is, I solve three systems of linear equations, each with the same number of variables as the number of vertices in my model. The first n rows of matrix A have only one 1 per row, and the first n rows of vector b have original model coordinates. That is, I spring-tie between the new vertex position and the old vertex position - the new ones shouldn't be too far away from the old ones.
All subsequent rows of matrix A (faces.size()*3 = the number of edges of all triangles in the grid) have one occurrence of 1 and one occurrence of -1, while the vector b has zero components opposite. This means I put a spring on each edge of our triangular mesh: all edges try to get the same vertex as their starting and ending points.
Once again: all vertices are variables, and they cannot deviate far from their original position, but at the same time they try to become similar to each other.
Here is the result:
Everything would be fine, the model is really smoothed, but it moved away from its original edge. Let's change the code a little:
For (int i=0; i<(int)verts.size(); i++) { float scale = border[i] ? 1000: 1; nlBegin(NL_ROW); nlCoefficient(i, scale); nlRightHandSide(scale*verts[i][d]); nlEnd(NL_ROW); }
In our matrix A, for the vertices that are on the edge, I add not a row from the category v_i = verts[i][d], but 1000*v_i = 1000*verts[i][d]. What does it change? And this changes our quadratic form of the error. Now a single deviation from the top at the edge will cost not one unit, as before, but 1000 * 1000 units. That is, we hung a stronger spring on the extreme vertices, the solution prefers to stretch others more strongly. Here is the result:
Let's double the strength of the springs between the vertices:
nlCoefficient(face[ j ], 2); nlCoefficient(face[(j+1)%3], -2);
It is logical that the surface has become smoother:
And now even a hundred times stronger:
What's this? Imagine that we have dipped a wire ring in soapy water. As a result, the resulting soap film will try to have the least curvature as possible, touching the same border - our wire ring. This is exactly what we got by fixing the border and asking for a smooth surface inside. Congratulations, we have just solved the Laplace equation with Dirichlet boundary conditions. Sounds cool? But in fact, just one system of linear equations to solve.
Poisson equation
Let's have another cool name.Let's say I have an image like this:
Everyone is good, but I don't like the chair.
I cut the picture in half: 
And I will select a chair with my hands: 
Then I will drag everything that is white in the mask to the left side of the picture, and at the same time I will say throughout the whole picture that the difference between two neighboring pixels should be equal to the difference between two neighboring pixels of the right image:
For (int i=0; i Here is the result: I have a number of photographs of fabric samples like this one: My task is to make seamless textures from photos of this quality. First, I (automatically) look for a repeating pattern: If I cut out this quadrangle right here, then because of the distortions, the edges will not converge, here is an example of a pattern repeated four times: Hidden text Here is a fragment where the seam is clearly visible: Therefore, I will not cut along a straight line, here is the cut line: Hidden text And here is the pattern repeated four times: Hidden text And its fragment to make it clearer: Already better, the cut did not go in a straight line, bypassing all sorts of curls, but still the seam is visible due to uneven lighting in the original photo. This is where the least squares method for the Poisson equation comes to the rescue. Here's the final result after lighting alignment: The texture turned out perfectly seamless, and all this automatically from a photo of a very mediocre quality. Do not be afraid of mathematics, look for simple explanations, and you will be lucky in engineering. If some physical quantity depends on another quantity, then this dependence can be investigated by measuring y at different values of x. As a result of measurements, a series of values is obtained: x 1 , x 2 , ..., x i , ... , x n ; y 1 , y 2 , ..., y i , ... , y n . Based on the data of such an experiment, it is possible to plot the dependence y = ƒ(x). The resulting curve makes it possible to judge the form of the function ƒ(x). However, the constant coefficients that enter into this function remain unknown. They can be determined using the least squares method. The experimental points, as a rule, do not lie exactly on the curve. The method of least squares requires that the sum of the squared deviations of the experimental points from the curve, i.e. 2 was the smallest. In practice, this method is most often (and most simply) used in the case of a linear relationship, i.e. when y=kx or y = a + bx. Linear dependence is very widespread in physics. And even when the dependence is non-linear, they usually try to build a graph in such a way as to get a straight line. For example, if it is assumed that the refractive index of glass n is related to the wavelength λ of the light wave by the relation n = a + b/λ 2 , then the dependence of n on λ -2 is plotted on the graph. Consider the dependence y=kx(straight line passing through the origin). Compose the value φ - the sum of the squared deviations of our points from the straight line The value of φ is always positive and turns out to be the smaller, the closer our points lie to the straight line. The method of least squares states that for k one should choose such a value at which φ has a minimum The calculation shows that the root-mean-square error in determining the value of k is equal to Let us now consider a somewhat more difficult case, when the points must satisfy the formula y = a + bx(a straight line not passing through the origin). The task is to find the best values of a and b from the given set of values x i , y i . Again we compose a quadratic form φ equal to the sum of the squared deviations of the points x i , y i from the straight line and find the values a and b for which φ has a minimum The joint solution of these equations gives The root-mean-square errors of determining a and b are equal When processing measurement results by this method, it is convenient to summarize all data in a table in which all sums included in formulas (19)–(24) are preliminarily calculated. The forms of these tables are shown in the examples below. Example 1 The basic equation of the dynamics of rotational motion ε = M/J (a straight line passing through the origin) was studied. For various values of the moment M, the angular acceleration ε of a certain body was measured. It is required to determine the moment of inertia of this body. The results of measurements of the moment of force and angular acceleration are listed in the second and third columns tables 5. By formula (19) we determine: To determine the root-mean-square error, we use formula (20) 0.005775kg-one · m -2 . By formula (18) we have SJ = (2.996 0.005775)/0.3337 = 0.05185 kg m 2. Given the reliability P = 0.95 , according to the table of Student coefficients for n = 5, we find t = 2.78 and determine the absolute error ΔJ = 2.78 0.05185 = 0.1441 ≈ 0.2 kg m 2. We write the results in the form: J = (3.0 ± 0.2) kg m 2; Example 2 We calculate the temperature coefficient of resistance of the metal using the least squares method. Resistance depends on temperature according to a linear law R t \u003d R 0 (1 + α t °) \u003d R 0 + R 0 α t °. The free term determines the resistance R 0 at a temperature of 0 ° C, and the angular coefficient is the product of the temperature coefficient α and the resistance R 0 . The results of measurements and calculations are given in the table ( see table 6). By formulas (21), (22) we determine R 0 = ¯ R- α R 0 ¯ t = 1.4005 - 0.002645 85.83333 = 1.1735 Ohm. Let us find an error in the definition of α. Since , then by formula (18) we have: Using formulas (23), (24) we have 0.014126 Ohm. Given the reliability P = 0.95, according to the table of Student's coefficients for n = 6, we find t = 2.57 and determine the absolute error Δα = 2.57 0.000132 = 0.000338 deg -1. α = (23 ± 4) 10 -4 hail-1 at P = 0.95. Example 3 It is required to determine the radius of curvature of the lens from Newton's rings. The radii of Newton's rings r m were measured and the numbers of these rings m were determined. The radii of Newton's rings are related to the radius of curvature of the lens R and the ring number by the equation r 2 m = mλR - 2d 0 R, where d 0 is the thickness of the gap between the lens and the plane-parallel plate (or lens deformation), λ is the wavelength of the incident light. λ = (600 ± 6) nm; then the equation will take the form y = a + bx. The results of measurements and calculations are entered in table 7.
Real life example
I deliberately did not do licked results, because. I just wanted to show exactly how you can apply the least squares methods, this is a training code. Let me now give an example from life: 
or
(19)
, (20)
where – n is the number of measurements.
;
.
(21)
(23)
. (24)Table 5
n
M, N m
ε, s-1
M2
M ε
ε - kM
(ε - kM) 2
1
1.44
0.52
2.0736
0.7488
0.039432
0.001555
2
3.12
1.06
9.7344
3.3072
0.018768
0.000352
3
4.59
1.45
21.0681
6.6555
-0.08181
0.006693
4
5.90
1.92
34.81
11.328
-0.049
0.002401
5
7.45
2.56
55.5025
19.072
0.073725
0.005435
∑
123.1886
41.1115
0.016436
.
Table 6
n
t°, s
r, Ohm
t-¯t
(t-¯t) 2
(t-¯t)r
r-bt-a
(r - bt - a) 2,10 -6
1
23
1.242
-62.8333
3948.028
-78.039
0.007673
58.8722
2
59
1.326
-26.8333
720.0278
-35.581
-0.00353
12.4959
3
84
1.386
-1.83333
3.361111
-2.541
-0.00965
93.1506
4
96
1.417
10.16667
103.3611
14.40617
-0.01039
107.898
5
120
1.512
34.16667
1167.361
51.66
0.021141
446.932
6
133
1.520
47.16667
2224.694
71.69333
-0.00524
27.4556
∑
515
8.403
8166.833
21.5985
746.804
∑/n
85.83333
1.4005
.
;
r 2 m = y;
m = x;
λR = b;
-2d 0 R = a,Table 7
n
x = m
y \u003d r 2, 10 -2 mm 2
m-¯m
(m-¯m) 2
(m-¯m)y
y-bx-a, 10-4
(y - bx - a) 2, 10 -6
1
1
6.101
-2.5
6.25
-0.152525
12.01
1.44229
2
2
11.834
-1.5
2.25
-0.17751
-9.6
0.930766
3
3
17.808
-0.5
0.25
-0.08904
-7.2
0.519086
4
4
23.814
0.5
0.25
0.11907
-1.6
0.0243955
5
5
29.812
1.5
2.25
0.44718
3.28
0.107646
6
6
35.760
2.5
6.25
0.894
3.12
0.0975819
∑
21
125.129
17.5
1.041175
3.12176
∑/n
3.5
20.8548333