The traditional method of smallest squares. Linear regression
The least squares method (MNC) allows you to evaluate different values \u200b\u200busing the results of a set of measurements containing random errors.
Characteristic of MNK.
The main idea of \u200b\u200bthis method is that as the criterion for the accuracy of solving the problem, the sum of the squares of errors, which they strive to minimize are considered. When using this method, both numerical and analytical approach can be applied.
In particular, as a numerical implementation, the method of smallest squares implies a larger number of unknown measurements as much as possible random variable. Moreover, the more calculations, the more accurate will be the solution. At this set of computing (source data), another set of alleged solutions are obtained, from which the best is then selected. If a plurality of solutions to parameterize, then the method of the smallest squares is reduced to the search for the optimal value of the parameters.
As an analytical approach to the implementation of MNA on a plurality of source data (measurements) and the estimated set of solutions, some (functional) is determined, which can be expressed by the formula obtained as a certain hypothesis that requires confirmation. In this case, the method of least squares is reduced to finding a minimum of this functional on the set of squares of source data errors.
Note that not the errors themselves, namely the squares of errors. Why? The fact is that often deviations of measurements from accurate value There are both positive and negative. When determining the average simple summation, it may lead to an incorrect conclusion about the quality of the assessment, since the mutual destruction of positive and negative values \u200b\u200bwill reduce the power of the sampling of the set of measurements. And, consequently, the accuracy of the assessment.
In order not to happen, and summarize the squares of deviations. Even moreover, to level the dimension of the measured value and the final assessment, from the sum of the squares of the errors
Some MNK applications
MNC is widely used in various fields. For example, in the theory of probability and mathematical statistics, the method is used to determine this characteristic of a random variable, as an average quadratic deviation that determines the width of the range of random variance values.
Least square method Used to estimate the parameters, the regression equation.One of the methods for studying stochastic ties between the signs is a regression analysis.
Regression analysis is the conclusion of the regression equation with which average value A random variable (sign-result), if the value of another (or other) variables (factors) is known. It includes the following steps:
- selection of communication form (type of analytical regression equation);
- estimate of the parameters of the equation;
- assessment of the quality of the analytical regression equation.
In the case of a linear pair bond, the regression equation will take the form: y i \u003d a + b · x i + u i. The parameters of this equation A and B are estimated according to statistical observation X and Y. The result of such an assessment is the equation:, where - estimates of parameters A and B, - the value of the resulting feature (variable) obtained by the regression equation (calculated value).
Most often to estimate parameters use method of least squares (MNC).
The least squares method gives the best (wealthy, efficient and unlocked) estimates of the parameters of the regression equation. But only if certain prerequisites are performed relative to a random term (U) and an independent variable (x) (see the backgrounds of the MNC).
The problem of evaluating the parameters of the linear pair equation by the least squares method It consists in the following: to obtain such estimates of the parameters, at which the sum of the squares of the deviations of the actual values \u200b\u200bof the effective sign - Y I on the calculated values \u200b\u200bis minimal.
Formally criterion MNK. You can write like this: 
.
Classification of least squares methods
- Least square method.
- The maximum truthful method (for a normal classical linear regression model, the normality of regression residues is postponed).
- The generalized method of smaller squares of OMNA is used in case of autocorrelation of errors and in the case of heterosdasticity.
- The method of suspended smallest squares (a special case of OMNA with heter-visasic residues).
We illustrate the essence classic smallest square method graphically. To do this, we construct a point schedule according to observations (x i, y i, i \u003d 1; n) in a rectangular coordinate system (such a point chart is called the correlation field). We will try to choose a straight line that is closest to the points of the correlation field. According to the least squares method, the line is selected so that the sum of the squares of the vertical distances between the points of the correlation field and this line would be minimal. 
Mathematical record of this task: 
.
The values \u200b\u200bof Y i and x i \u003d 1 ... n are known to us, these are observational data. In the function S, they are constants. Variables in this feature are the desired parameter estimates - ,. To find a minimum of the 2-variable functions, it is necessary to calculate the private derivatives of this function for each of the parameters and equate them zero, i.e. 
.
As a result, we get a system of 2 normal linear equations: 
Solving this system, find the desired parameter estimates: 
The correctness of the calculation of the parameters of the regression equation can be tested by comparing amounts (perhaps some discrepancy due to the rounding calculations).
To calculate the parameter estimates, you can build table 1.
Regression coefficient sign indicates communication direction (if b\u003e 0, line is direct, if b<0, то связь обратная). Величина b показывает на сколько единиц изменится в среднем признак-результат -y при изменении признака-фактора - х на 1 единицу своего измерения.
Formally the value of the parameter A is the average value of y with x equal to zero. If the signator does not have and cannot have a zero value, then the above interpretation of the parameter and does not make sense.
Estimation of tightness of communication between signs
It is carried out using the linear pair correlation coefficient - R x, y. It can be calculated by the formula: 
. In addition, the coefficient of linear pairing correlation can be determined through the regression coefficient B: 
.
The area of \u200b\u200bpermissible values \u200b\u200bof the linear coefficient of pair correlation from -1 to +1. The correlation coefficient sign indicates the direction of communication. If R x, y\u003e 0, then the connection is straight; If R x, Y<0, то связь обратная.
If this coefficient is close to one, the connection between the features can be interpreted as quite close linear. If its module is equal to a unit ê R x, y ê \u003d 1, the connection between the signs is functional linear. If the signs x and y are linearly independent, then R x, y is close to 0.
To calculate R x, Y can also use table 1.
To assess the quality of the obtained regression equation, the theoretical determination coefficient is calculated - R 2 YX: 
,
where D 2 is the dispersion of Y; explained by the regression equation;
e 2 - residual (unexplained regression equation) dispersion y;
s 2 Y is a total (complete) dispersion Y.
The determination coefficient characterizes the proportion of variation (dispersion) of the effective sign Y, explained by the regression (and, consequently, the factor X), in the general variation (dispersion) y. The determination coefficient R 2 YX takes values \u200b\u200bfrom 0 to 1. Accordingly, the value of 1-R 2 YX characterizes the fraction of the dispersion Y caused by the influence of other unaccounted factors in the model and the specification errors.
With paired linear regression R 2 YX \u003d R 2 YX.
Example.
Experimental data on variable values H. and W. Led in the table. 
As a result of their alignment, a function was obtained 
Using least square method, approximate this data linear dependence y \u003d 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 experimental data. Make a drawing.
The essence of the least squares method (MNC).
The task is to find the coefficients linear dependence, in which the function of two variables but and b. 
Takes the smallest value. That is, with data but and b. The sum of the squares of the deviations of the experimental data from the direct line will be the smallest. This is the whole essence of the least squares method.
Thus, the example solution comes down to finding the extremum function of two variables.
Displays the formula for finding coefficients.
A system of two equations with two unknowns is compiled and solved. We find private derivatives in variable but and b., equate these derivatives to zero. 
Solve the resulting system of equations by any method (for example for a substitution method or) and we obtain formulas for finding coefficients using the least squares method (MNC). 
With data but and B. function Takes the smallest value. Proof of this fact is given.
That is the whole method of least squares. Formula for finding a parameter a. contains amounts ,, and parameter n. - Number of experimental data. The values \u200b\u200bof these sums are recommended to calculate separately. Coefficient b. Located after calculation a..
It's time to remember about the source example.
Decision.
In our example N \u003d 5.. Fill out a table for the convenience of calculating amounts that are included in the formula of the desired coefficients. 
Values \u200b\u200bin the fourth line of the table are obtained by multiplying the values \u200b\u200bof the 2nd string to the values \u200b\u200bof the 3rd string for each number I..
Values \u200b\u200bin the fifth line of the table are obtained by the construction of the 2nd string values \u200b\u200bfor each number. I..
The values \u200b\u200bof the last column of the table are the sums of values \u200b\u200bby lines.
We use the formulas of the least squares method for finding coefficients but and b.. We substitute the corresponding values \u200b\u200bfrom the last column of the table: 
Hence, y \u003d 0.165x + 2.184 - The desired approximating straight line.
It remains to find out which of the lines y \u003d 0.165x + 2.184 or It is better to approximate the initial data, that is, it is estimated by the method of smallest squares.
Assessment of the error of the least squares method.
This requires to calculate the sums of the squares of the deviations of the source data from these lines. 
and 
A smaller value corresponds to a line that is better in the sense of the smaller square method approximates the source data. 
Since, then straight y \u003d 0.165x + 2.184 Better brings the source data.
Graphic illustration of the least squares method (MNC).
On the charts everything is perfectly visible. The red line is the found straight y \u003d 0.165x + 2.184, blue line is Pink dots are the source data.
What is it necessary for all these approximations?
I personally use to solve the problems of smoothing data, interpolation and extrapolation problems (in the initial example could ask to find the observed value y. for x \u003d 3. or for x \u003d 6. According to the MND method). But let us talk more about this later in another section of the site.
Evidence.
So as for found but and b. The function has taken the smallest value, it is necessary that at this point the matrix of the quadratic form of the second order differential for the function It was positively defined. Show it.
The second order differential is: 
I.e
Consequently, the quadratic form matrix is 
and the values \u200b\u200bof the elements do not depend on but and B..
We show that the matrix is \u200b\u200bpositively defined. To do this, it is necessary that the angular minors are positive.
Corner Minor of the first order 
. The inequality is strict, since the points are mismatched. In the future, we will mean.
Second-order Corner Minor 
We prove that 
method of mathematical induction.
Output: Found values but and B. correspond to the smallest value of the function Therefore, are the desired parameters for the method of smallest squares.
After alignment, we obtain the function of the following form: G (x) \u003d x + 1 3 + 1.
We can approximate this data using the linear dependence y \u003d a x + b, calculate the corresponding parameters. To do this, we will need to apply the so-called least square method. It will also be necessary to make a drawing to check which line will better align experimental data.
What exactly is the MNC (the least squares method)
The main thing that we need to do is to find such coefficients of linear dependence, in which the value of the function of two variables f (a, b) \u003d σ i \u003d 1 n (y i - (a x i + b)) 2 will be the smallest. In other words, at certain values \u200b\u200bof a and b, the sum of the squares of the deviations of the submitted data from the resulting direct will have a minimum value. This is the meaning of the smaller square method. All we need to do to solve the example is to find the extremum function of two variables.
How to output formulas for calculating coefficients
In order to output the formula for calculating the coefficients, it is necessary to compile and solve the system of equations with two variables. To do this, we calculate the private derivatives of expressions f (a, b) \u003d σ i \u003d 1 n (y i - (a x i + b)) 2 by a and b and equate them to 0.
Δ f (a, b) δ a \u003d 0 Δ f (a, b) δ b \u003d 0 ⇔ - 2 σ i \u003d 1 n (yi - (axi + b)) xi \u003d 0 - 2 σ i \u003d 1 n ( yi - (axi + b)) \u003d 0 ⇔ a Σ i \u003d 1 nxi 2 + b Σ i \u003d 1 nxi \u003d σ i \u003d 1 nxiyia σ i \u003d 1 nxi + σ i \u003d 1 nb \u003d σ i \u003d 1 nyi ⇔ a Σ i \u003d 1 nxi 2 + b Σ i \u003d 1 nxi \u003d σ i \u003d 1 nxiyia σ i \u003d 1 nxi + nb \u003d σ i \u003d 1 NYI
To solve the system of equations, you can use any methods, for example, a substitution or a craver method. As a result, we need to obtain formulas with which the coefficients according to the least squares method are calculated.
n Σ i \u003d 1 n x i y i - σ i \u003d 1 n x i Σ i \u003d 1 n y i n Σ i \u003d 1 n - σ i \u003d 1 n x i 2 b \u003d σ i \u003d 1 n y i - a σ i \u003d 1 n x i n
We calculated variable values \u200b\u200bat which function
F (a, b) \u003d σ i \u003d 1 n (y i - (a x i + b)) 2 will take the minimum value. In the third paragraph, we prove why it is precisely the same.
This is the use of the smaller square method in practice. Its formula, which is used to search for parameter A, includes σ i \u003d 1 n x i, σ i \u003d 1 n y i, σ i \u003d 1 n x i i i, σ i \u003d 1 n x i 2, and parameter
N - The number of experimental data is indicated. We advise you to calculate each amount separately. The value of the coefficient B is calculated immediately after a.
Turn again to the original example.
Example 1.
Here we have N is five. To make it easier to calculate the necessary amounts included in the formulas of the coefficients, fill in the table.
| i \u003d 1. | i \u003d 2. | i \u003d 3. | i \u003d 4. | i \u003d 5. | Σ i \u003d 1 5 | |
| X I. | 0 | 1 | 2 | 4 | 5 | 12 |
| Y I. | 2 , 1 | 2 , 4 | 2 , 6 | 2 , 8 | 3 | 12 , 9 |
| x i y i | 0 | 2 , 4 | 5 , 2 | 11 , 2 | 15 | 33 , 8 |
| X i 2. | 0 | 1 | 4 | 16 | 25 | 46 |
Decision
The fourth line includes the data obtained by multiplying the values \u200b\u200bfrom the second row to the third values \u200b\u200bfor each individual i. The fifth line contains data from the second, elevated to the square. The last column summarizes the values \u200b\u200bof individual lines.
We use the least squares method to calculate the coefficients you need and b. To do this, we will substitute the desired values \u200b\u200bfrom the last column and we calculate the amount:
n Σ i \u003d 1 nxiyi - σ i \u003d 1 nxi σ i \u003d 1 nyin σ i \u003d 1 n - σ i \u003d 1 nxi 2 b \u003d σ i \u003d 1 nyi - a σ i \u003d 1 nxin ⇒ a \u003d 5 · 33 8 - 12 · 12, 9 5 · 46 - 12 2 B \u003d 12, 9 - A · 12 5 ⇒ A ≈ 0, 165 B ≈ 2, 184
We needed that the desired approximating straight will look like y \u003d 0, 165 x + 2, 184. Now we need to determine which line will be better to approximate the data - G (x) \u003d x + 1 3 + 1 or 0, 165 x + 2, 184. We will evaluate using the least squares method.
To calculate the error, we need to find the sums of the squares of the data deviations from the direct σ 1 \u003d σ i \u003d 1 n (Yi - (Axi + Bi)) 2 and σ 2 \u003d σ i \u003d 1 n (yi - g (xi)) 2, The minimum value will correspond to a more suitable line.
σ 1 \u003d σ i \u003d 1 n (yi - (axi + bi)) 2 \u003d \u003d σ i \u003d 1 5 (yi - (0, 165 xi + 2, 184)) 2 ≈ 0, 019 σ 2 \u003d σ i \u003d 1 n (yi - g (xi)) 2 \u003d \u003d σ i \u003d 1 5 (yi - (xi + 1 3 + 1)) 2 ≈ 0, 096
Answer: Since σ 1.< σ 2 , то прямой, наилучшим образом аппроксимирующей исходные данные, будет
y \u003d 0, 165 x + 2, 184.
The least squares method is clearly shown in graphic illustration. With the help of the red line, straight g (x) \u003d x + 1 3 + 1, blue - y \u003d 0, 165 x + 2, 184 is marked. The initial data is indicated by pink dots.
Let us explain what exactly the approximation of a similar type is needed.
They can be used in tasks requiring data smoothing, as well as in those where data should be interpolated or extrapolated. For example, in the problem, disassembled above, it would be possible to find the value of the observed value y at x \u003d 3 or at x \u003d 6. Such examples we devoted a separate article.
Proof of MNK method
In order for the function to take the minimum value with the calculated A and B, it is necessary that at this point the matrix of the quadratic form of the differential function of the form F (a, b) \u003d σ i \u003d 1 n (y i - (a x i + b)) 2 was positively defined. Let's show how it should look like.
Example 2.
We have a second order differential:
d 2 f (a; b) \u003d δ 2 f (a; b) δ a 2 d 2 a + 2 δ 2 f (a; b) δ a δ bdadb + δ 2 f (a; b) δ b 2 D 2 B.
Decision
δ 2 f (a; b) δ a 2 \u003d δ δ f (a; b) δ a δ a \u003d \u003d δ - 2 σ i \u003d 1 n (yi - (axi + b)) xi δ a \u003d 2 σ i \u003d 1 n (xi) 2 δ 2 f (a; b) δ a δ b \u003d δ δ f (a; b) δ a δ b \u003d \u003d δ - 2 σ i \u003d 1 n (yi - (axi + b) ) xi Δ b \u003d 2 Σ i \u003d 1 nxi δ 2 f (a; b) δ b 2 \u003d δ δ f (a; b) Δ b δ b \u003d δ - 2 σ i \u003d 1 n (yi - (AXI + b)) Δ b \u003d 2 σ i \u003d 1 n (1) \u003d 2 n
In other words, it can be written as: d 2 f (a; b) \u003d 2 σ i \u003d 1 n (x i) 2 d 2 a + 2 · 2 σ x i i \u003d 1 n d И d b + (2 n) d 2 b.
We obtained the matrix of the quadratic form m \u003d 2 σ i \u003d 1 n (x i) 2 2 σ i \u003d 1 n x i 2 σ i \u003d 1 n x i 2 n.
In this case, the values \u200b\u200bof the individual elements will not vary depending on A and B. Is this matrix positively defined? To answer this question, check if its corner minors are positive.
Calculate the first-order corner of the first order: 2 σ i \u003d 1 n (x i) 2\u003e 0. Since points I do not coincide, the inequality is strict. We will have this in mind for further calculations.
Calculate the angular minor of the second order:
d E T (m) \u003d 2 Σ i \u003d 1 n (x i) 2 2 σ i \u003d 1 n x i 2 Σ i \u003d 1 n x i 2 n \u003d 4 n Σ i \u003d 1 n (x i) 2 - σ i \u003d 1 n x i 2
After that, we turn to the proof of the inequality n Σ i \u003d 1 n (x i) 2 - σ i \u003d 1 n x i 2\u003e 0 using mathematical induction.
- Check whether this inequality will be valid for arbitrary n. Take 2 and calculate:
2 Σ i \u003d 1 2 (xi) 2 - σ i \u003d 1 2 xi 2 \u003d 2 x 1 2 + x 2 2 - x 1 + x 2 2 \u003d x 1 2 - 2 x 1 x 2 + x 2 2 \u003d x 1 + x 2 2\u003e 0
We have faithful equality (if the x 1 and x 2 values \u200b\u200bare not coincided).
- We assume that this inequality will be faithful to N, i.e. n Σ i \u003d 1 n (x i) 2 - σ i \u003d 1 n x i 2\u003e 0 is valid.
- Now we prove justice at n + 1, i.e. which (n + 1) σ i \u003d 1 n + 1 (xi) 2 - σ i \u003d 1 n + 1 xi 2\u003e 0, if the N Σ i \u003d 1 n (xi) 2 is σ i \u003d 1 nxi 2\u003e 0.
Calculate:
(n + 1) Σ i \u003d 1 n + 1 (xi) 2 - σ i \u003d 1 n + 1 xi 2 \u003d (n + 1) σ i \u003d 1 n (xi) 2 + xn + 1 2 - σ i \u003d 1 nxi + xn + 1 2 \u003d n Σ i \u003d 1 n (xi) 2 + n · xn + 1 2 + σ i \u003d 1 n (xi) 2 + xn + 1 2 - - σ i \u003d 1 nxi 2 + 2 xn + 1 Σ i \u003d 1 nxi + xn + 1 2 \u003d σ i \u003d 1 n (xi) 2 - σ i \u003d 1 nxi 2 + n · xn + 1 2 - xn + 1 σ i \u003d 1 nxi + Σ i \u003d 1 n (xi) 2 \u003d \u003d σ i \u003d 1 n (xi) 2 - σ i \u003d 1 nxi 2 + xn + 1 2 - 2 xn + 1 x 1 + x 1 2 + + xn + 1 2 - 2 xn + 1 x 2 + x 2 2 +. . . + xn + 1 2 - 2 xn + 1 x 1 + xn 2 \u003d \u003d n Σ i \u003d 1 n (xi) 2 - σ i \u003d 1 nxi 2 + + (xn + 1 - x 1) 2 + (xn + 1 - x 2) 2 +. . . + (x n - 1 - x n) 2\u003e 0
The expression concluded in curly brackets will be greater than 0 (based on what we assumed in paragraph 2), and the remaining terms will be greater than 0, since they are all squares of numbers. We have proven inequality.
Answer: Found a and b will correspond to the smallest value of the function f (a, b) \u003d σ i \u003d 1 n (y i - (a x i + b)) 2, it means that they are the desired parameters of the least squares method (MNK).
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By choosing a type of regression function, i.e. The type of the dependence model of the dependence y from x (or x from y), for example, the linear model y x \u003d a + bx, it is necessary to determine the specific values \u200b\u200bof the model coefficients.
At different values, A and B, an infinite number of dependences of the form y x \u003d a + bx can be constructed on the coordinate plane there is an infinite number of direct, we also need such a dependence that corresponds to the observed values \u200b\u200bin the best possible way. Thus, the task is reduced to the selection of the best coefficients.
Linear function A + BX We are looking for, based on only some existing observations. To find a function with the best compliance with the observed values, we use the method of the smallest squares.
Denote: Y i - the value calculated by the equation y i \u003d a + bx i. Y i is the measured value, ε i \u003d y i -Y i - the difference between the measured and calculated by the equation values, ε i \u003d y i -a-bx i.
In the least squares method, ε i, the difference between the measured Y i and the values \u200b\u200bcalculated by the equation values \u200b\u200by i was minimal. Therefore, we find the coefficients a and b so that the sum of the squares of the deviations of the observed values \u200b\u200bfrom the values \u200b\u200bon the straight line of the regression turned out to be the smallest:
Exploring this function of arguments A and using derivatives to extremum, it can be proved that the function takes the minimum value if the coefficients A and B are system solutions:
(2)
If we divide both parts of normal equations on n, then we get:
Considering that 
(3)
Receive 
From here, substituting the value A in the first equation, we get:
At the same time, B is called the regression coefficient; A is called a free member of the regression equation and calculate according to the formula:
The resulting direct is an estimate for the theoretical line of regression. We have:
So, 
It is the equation of linear regression.
Regression may be straight (b\u003e 0) and reverse (B Example 1. The measurement results of X and Y values \u200b\u200bare given in the table:
| x I. | -2 | 0 | 1 | 2 | 4 |
| y I. | 0.5 | 1 | 1.5 | 2 | 3 |
Assuming that between X and Y there is a linear dependence Y \u003d A + BX, which method of least squares determine the coefficients a and b.
Decision. Here n \u003d 5
x i \u003d -2 + 0 + 1 + 2 + 4 \u003d 5;
x i 2 \u003d 4 + 0 + 1 + 4 + 16 \u003d 25
x i y i \u003d -2 0.5 + 0 1 + 1 1.5 + 2 2 + 4 3 \u003d 16.5
y i \u003d 0.5 + 1 + 1.5 + 2 + 3 \u003d 8
and the normal system (2) has the form 
Solving this system, we obtain: b \u003d 0.425, a \u003d 1.175. Therefore, y \u003d 1.175 + 0.425x.
Example 2. There is a sample of 10 observations of economic indicators (x) and (y).
| x I. | 180 | 172 | 173 | 169 | 175 | 170 | 179 | 170 | 167 | 174 |
| y I. | 186 | 180 | 176 | 171 | 182 | 166 | 182 | 172 | 169 | 177 |
It is required to find a selective regression equation on X. Construct a selective line of regression Y to X.
Decision. 1. We will organize data on the X I and Y i values. We get a new table:
| x I. | 167 | 169 | 170 | 170 | 172 | 173 | 174 | 175 | 179 | 180 |
| y I. | 169 | 171 | 166 | 172 | 180 | 176 | 177 | 182 | 182 | 186 |
To simplify the calculations, we will make the calculated table into which you bring the necessary numerical values.
| x I. | y I. | x i 2. | x i y i |
| 167 | 169 | 27889 | 28223 |
| 169 | 171 | 28561 | 28899 |
| 170 | 166 | 28900 | 28220 |
| 170 | 172 | 28900 | 29240 |
| 172 | 180 | 29584 | 30960 |
| 173 | 176 | 29929 | 30448 |
| 174 | 177 | 30276 | 30798 |
| 175 | 182 | 30625 | 31850 |
| 179 | 182 | 32041 | 32578 |
| 180 | 186 | 32400 | 33480 |
| Σx i \u003d 1729 | Σy i \u003d 1761 | Σx i 2 299105 | Σx i y i \u003d 304696 |
| x \u003d 172.9 | y \u003d 176.1 | x i 2 \u003d 29910.5 | xy \u003d 30469.6 |
According to formula (4), calculate the regression coefficient
and according to formula (5)
Thus, the regression selective equation has the form y \u003d -59.34 + 1.3804x.
Application on the coordinate plane of the point (X i; Y i) and note the direct regression.
Figure 4.
Figure 4 shows how the observed values \u200b\u200bare located relative to the regression line. For the numerical estimate of the deviations y I from Y i, where Y I is observed, and y I determined by the regression of the value, will be a table:
| x I. | y I. | Y I. | Y i -y i |
| 167 | 169 | 168.055 | -0.945 |
| 169 | 171 | 170.778 | -0.222 |
| 170 | 166 | 172.140 | 6.140 |
| 170 | 172 | 172.140 | 0.140 |
| 172 | 180 | 174.863 | -5.137 |
| 173 | 176 | 176.225 | 0.225 |
| 174 | 177 | 177.587 | 0.587 |
| 175 | 182 | 178.949 | -3.051 |
| 179 | 182 | 184.395 | 2.395 |
| 180 | 186 | 185.757 | -0.243 |
The values \u200b\u200bof Y i are calculated according to the regression equation.
A noticeable deviation of some observed values \u200b\u200bfrom the regression line is explained by a small number of observations. In the study of the degree of linear dependence Y from x, the number of observations is taken into account. The strength of the dependence is determined by the correlation coefficient.
