LSM for a function of two variables. Approximation of experimental data. Least square method. Practical implementation of LSM for linear dependence on a non-programmable calculator

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 options A 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 A And b takes the smallest value. That is, given the data A 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 A 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 the coefficients using the least squares method (LSM).

With data A 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 A And b. We substitute in them the corresponding values from the last column of the table:

Hence, 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 A 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.

It has many applications, as it allows an approximate representation of a given function by other simpler ones. LSM can be extremely useful in processing observations, and it is actively used to estimate some quantities from the results of measurements of others containing random errors. In this article, you will learn how to implement least squares calculations in Excel.

Statement of the problem on a specific example

Suppose there are two indicators X and Y. Moreover, Y depends on X. Since OLS is of interest to us from the point of view of regression analysis (in Excel, its methods are implemented using built-in functions), we should immediately proceed to consider a specific problem.

So, let X be the selling area of a grocery store, measured in square meters, and Y be the annual turnover, defined in millions of rubles.

It is required to make a forecast of what turnover (Y) the store will have if it has one or another retail space. Obviously, the function Y = f (X) is increasing, since the hypermarket sells more goods than the stall.

A few words about the correctness of the initial data used for prediction

Let's say we have a table built with data for n stores.

According to mathematical statistics, the results will be more or less correct if the data on at least 5-6 objects are examined. Also, "anomalous" results cannot be used. In particular, an elite small boutique can have a turnover many times greater than the turnover of large outlets of the “masmarket” class.

The essence of the method

The table data can be displayed on the Cartesian plane as points M 1 (x 1, y 1), ... M n (x n, y n). Now the solution of the problem will be reduced to the selection of an approximating function y = f (x), which has a graph passing as close as possible to the points M 1, M 2, .. M n .

Of course, you can use a high degree polynomial, but this option is not only difficult to implement, but simply incorrect, since it will not reflect the main trend that needs to be detected. The most reasonable solution is to search for a straight line y = ax + b, which best approximates the experimental data, and more precisely, the coefficients - a and b.

Accuracy score

For any approximation, the assessment of its accuracy is of particular importance. Denote by e i the difference (deviation) between the functional and experimental values for the point x i , i.e. e i = y i - f (x i).

Obviously, to assess the accuracy of the approximation, you can use the sum of deviations, i.e., when choosing a straight line for an approximate representation of the dependence of X on Y, preference should be given to the one that has the smallest value of the sum e i at all points under consideration. However, not everything is so simple, since along with positive deviations, there will practically be negative ones.

You can solve the problem using the deviation modules or their squares. The latter method is the most widely used. It is used in many areas, including regression analysis (in Excel, its implementation is carried out using two built-in functions), and has long been proven to be effective.

Least square method

In Excel, as you know, there is a built-in autosum function that allows you to calculate the values of all values located in the selected range. Thus, nothing will prevent us from calculating the value of the expression (e 1 2 + e 2 2 + e 3 2 + ... e n 2).

In mathematical notation, this looks like:

Since the decision was initially made to approximate using a straight line, we have:

Thus, the task of finding a straight line that best describes a specific relationship between X and Y amounts to calculating the minimum of a function of two variables:

This requires equating to zero partial derivatives with respect to new variables a and b, and solving a primitive system consisting of two equations with 2 unknowns of the form:

After simple transformations, including dividing by 2 and manipulating the sums, we get:

Solving it, for example, by Cramer's method, we obtain a stationary point with certain coefficients a * and b * . This is the minimum, i.e. to predict what turnover the store will have for a certain area, the straight line y = a * x + b * is suitable, which is a regression model for the example in question. Of course, it will not allow you to find the exact result, but it will help you get an idea of \u200b\u200bwhether buying a store on credit for a particular area will pay off.

How to implement the least squares method in Excel

Excel has a function for calculating the value of the least squares. It has the following form: TREND (known Y values; known X values; new X values; constant). Let's apply the formula for calculating the OLS in Excel to our table.

To do this, in the cell in which the result of the calculation using the least squares method in Excel should be displayed, enter the “=” sign and select the “TREND” function. In the window that opens, fill in the appropriate fields, highlighting:

range of known values for Y (in this case data for turnover);
range x 1 , …x n , i.e. the size of retail space;
and known and unknown values of x, for which you need to find out the size of the turnover (for information about their location on the worksheet, see below).

In addition, there is a logical variable "Const" in the formula. If you enter 1 in the field corresponding to it, then this will mean that calculations should be carried out, assuming that b \u003d 0.

If you need to know the forecast for more than one x value, then after entering the formula, you should not press "Enter", but you need to type the combination "Shift" + "Control" + "Enter" ("Enter") on the keyboard.

Some Features

Regression analysis can be accessible even to dummies. The Excel formula for predicting the value of an array of unknown variables - "TREND" - can be used even by those who have never heard of the least squares method. It is enough just to know some features of its work. In particular:

If you place the range of known values of the variable y in one row or column, then each row (column) with known values of x will be perceived by the program as a separate variable.
If the range with known x is not specified in the TREND window, then in case of using the function in Excel, the program will consider it as an array consisting of integers, the number of which corresponds to the range with the given values of the variable y.
To output an array of "predicted" values, the trend expression must be entered as an array formula.
If no new x values are specified, then the TREND function considers them equal to the known ones. If they are not specified, then array 1 is taken as an argument; 2; 3; 4;…, which is commensurate with the range with already given parameters y.
The range containing the new x values must have the same or more rows or columns as the range with the given y values. In other words, it must be proportionate to the independent variables.
An array with known x values can contain multiple variables. However, if we are talking about only one, then it is required that the ranges with the given values of x and y be commensurate. In the case of several variables, it is necessary that the range with the given y values fit in one column or one row.

FORECAST function

It is implemented using several functions. One of them is called "PREDICTION". It is similar to TREND, i.e. it gives the result of calculations using the least squares method. However, only for one X, for which the value of Y is unknown.

Now you know the Excel formulas for dummies that allow you to predict the value of the future value of an indicator according to a linear trend.

The least squares method is one of the most common and most developed due to its simplicity and efficiency of methods for estimating the parameters of linear. At the same time, certain caution should be observed when using it, since the models built using it may not meet a number of requirements for the quality of their parameters and, as a result, not “well” reflect the patterns of process development.

Let us consider the procedure for estimating the parameters of a linear econometric model using the least squares method in more detail. Such a model in general form can be represented by equation (1.2):

y t = a 0 + a 1 x 1 t +...+ a n x nt + ε t .

The initial data when estimating the parameters a 0 , a 1 ,..., a n is the vector of values of the dependent variable y= (y 1 , y 2 , ... , y T)" and the matrix of values of independent variables

in which the first column, consisting of ones, corresponds to the coefficient of the model .

The method of least squares got its name based on the basic principle that the parameter estimates obtained on its basis should satisfy: the sum of squares of the model error should be minimal.

Examples of solving problems by the least squares method

Example 2.1. The trading enterprise has a network consisting of 12 stores, information on the activities of which is presented in Table. 2.1.

The company's management would like to know how the size of the annual depends on the sales area of the store.

Table 2.1

Shop number	Annual turnover, million rubles	Trade area, thousand m 2

Least squares solution. Let us designate - the annual turnover of the -th store, million rubles; - selling area of the -th store, thousand m 2.

Fig.2.1. Scatterplot for Example 2.1

To determine the form of the functional relationship between the variables and construct a scatterplot (Fig. 2.1).

Based on the scatter diagram, we can conclude that the annual turnover is positively dependent on the selling area (i.e., y will increase with the growth of ). The most appropriate form of functional connection is − linear.

Information for further calculations is presented in Table. 2.2. Using the least squares method, we estimate the parameters of the linear one-factor econometric model

Table 2.2

Thus,

Therefore, with an increase in the trading area by 1 thousand m 2, other things being equal, the average annual turnover increases by 67.8871 million rubles.

Example 2.2. The management of the enterprise noticed that the annual turnover depends not only on the sales area of the store (see example 2.1), but also on the average number of visitors. The relevant information is presented in table. 2.3.

Table 2.3

Solution. Denote - the average number of visitors to the th store per day, thousand people.

To determine the form of the functional relationship between the variables and construct a scatterplot (Fig. 2.2).

Based on the scatter diagram, we can conclude that the annual turnover is positively related to the average number of visitors per day (i.e., y will increase with the growth of ). The form of functional dependence is linear.

Rice. 2.2. Scatterplot for example 2.2

Table 2.4

In general, it is necessary to determine the parameters of the two-factor econometric model

y t \u003d a 0 + a 1 x 1 t + a 2 x 2 t + ε t

The information required for further calculations is presented in Table. 2.4.

Let us estimate the parameters of a linear two-factor econometric model using the least squares method.

Thus,

Evaluation of the coefficient = 61.6583 shows that, other things being equal, with an increase in the trading area by 1 thousand m 2, the annual turnover will increase by an average of 61.6583 million rubles.

Least square method

Least square method ( MNK, OLS, Ordinary Least Squares) - one of the basic methods of regression analysis for estimating unknown parameters of regression models from sample data. The method is based on minimizing the sum of squares of regression residuals.

It should be noted that the least squares method itself can be called a method for solving a problem in any area, if the solution consists of or satisfies a certain criterion for minimizing the sum of squares of some functions of the unknown variables. Therefore, the least squares method can also be used for an approximate representation (approximation) of a given function by other (simpler) functions, when finding a set of quantities that satisfy equations or restrictions, the number of which exceeds the number of these quantities, etc.

The essence of the MNC

Let some (parametric) model of probabilistic (regression) dependence between the (explained) variable y and many factors (explanatory variables) x

where is the vector of unknown model parameters

- Random model error.

Let there also be sample observations of the values of the indicated variables. Let be the observation number (). Then are the values of the variables in the -th observation. Then, for given values of the parameters b, it is possible to calculate the theoretical (model) values of the explained variable y:

The value of the residuals depends on the values of the parameters b.

The essence of LSM (ordinary, classical) is to find such parameters b for which the sum of the squares of the residuals (eng. Residual Sum of Squares) will be minimal:

In the general case, this problem can be solved by numerical methods of optimization (minimization). In this case, one speaks of nonlinear least squares(NLS or NLLS - English. Non Linear Least Squares). In many cases, an analytical solution can be obtained. To solve the minimization problem, it is necessary to find the stationary points of the function by differentiating it with respect to the unknown parameters b, equating the derivatives to zero, and solving the resulting system of equations:

If the random errors of the model are normally distributed, have the same variance, and are not correlated with each other, the least squares parameter estimates are the same as the maximum likelihood method (MLM) estimates.

LSM in the case of a linear model

Let the regression dependence be linear:

Let y- column vector of observations of the explained variable, and - matrix of observations of factors (rows of the matrix - vectors of factor values in a given observation, by columns - vector of values of a given factor in all observations). The matrix representation of the linear model has the form:

Then the vector of estimates of the explained variable and the vector of regression residuals will be equal to

accordingly, the sum of the squares of the regression residuals will be equal to

Differentiating this function with respect to the parameter vector and equating the derivatives to zero, we obtain a system of equations (in matrix form):

The solution of this system of equations gives the general formula for the least squares estimates for the linear model:

For analytical purposes, the last representation of this formula turns out to be useful. If the data in the regression model centered, then in this representation the first matrix has the meaning of the sample covariance matrix of factors, and the second one is the vector of covariances of factors with dependent variable. If, in addition, the data is also normalized at the SKO (that is, ultimately standardized), then the first matrix has the meaning of the sample correlation matrix of factors, the second vector - the vector of sample correlations of factors with the dependent variable.

An important property of LLS estimates for models with a constant- the line of the constructed regression passes through the center of gravity of the sample data, that is, the equality is fulfilled:

In particular, in the extreme case when the only regressor is a constant, we find that the OLS estimate of a single parameter (the constant itself) is equal to the mean value of the variable being explained. That is, the arithmetic mean, known for its good properties from the laws of large numbers, is also an least squares estimate - it satisfies the criterion for the minimum sum of squared deviations from it.

Example: simple (pairwise) regression

In the case of paired linear regression, the calculation formulas are simplified (you can do without matrix algebra):

Properties of OLS estimates

First of all, we note that for linear models, the least squares estimates are linear estimates, as follows from the above formula. For unbiased OLS estimates, it is necessary and sufficient to fulfill the most important condition of regression analysis: the mathematical expectation of a random error conditional on the factors must be equal to zero. This condition is satisfied, in particular, if

the mathematical expectation of random errors is zero, and
factors and random errors are independent random variables.

The second condition - the condition of exogenous factors - is fundamental. If this property is not satisfied, then we can assume that almost any estimates will be extremely unsatisfactory: they will not even be consistent (that is, even a very large amount of data does not allow obtaining qualitative estimates in this case). In the classical case, a stronger assumption is made about the determinism of factors, in contrast to a random error, which automatically means that the exogenous condition is satisfied. In the general case, for the consistency of the estimates, it is sufficient to fulfill the exogeneity condition together with the convergence of the matrix to some non-singular matrix with an increase in the sample size to infinity.

In order for, in addition to the consistency and unbiasedness, the estimates of the (usual) LSM to be also effective (the best in the class of linear unbiased estimates), it is necessary to fulfill additional properties of a random error:

These assumptions can be formulated for the covariance matrix of the random error vector

A linear model that satisfies these conditions is called classical. The least squares estimators for classical linear regression are unbiased, consistent, and the most efficient estimators in the class of all linear unbiased estimators (the abbreviation blue (Best Linear Unbaised Estimator) is the best linear unbiased estimate; in domestic literature, the Gauss-Markov theorem is more often cited). As it is easy to show, the covariance matrix of the coefficient estimates vector will be equal to:

Generalized least squares

The method of least squares allows for a wide generalization. Instead of minimizing the sum of squares of the residuals, one can minimize some positive definite quadratic form of the residual vector , where is some symmetric positive definite weight matrix. Ordinary least squares is a special case of this approach, when the weight matrix is proportional to the identity matrix. As is known from the theory of symmetric matrices (or operators), there is a decomposition for such matrices. Therefore, the specified functional can be represented as follows, that is, this functional can be represented as the sum of the squares of some transformed "residuals". Thus, we can distinguish a class of least squares methods - LS-methods (Least Squares).

It is proved (Aitken's theorem) that for a generalized linear regression model (in which no restrictions are imposed on the covariance matrix of random errors), the most effective (in the class of linear unbiased estimates) are estimates of the so-called. generalized OLS (OMNK, GLS - Generalized Least Squares)- LS-method with a weight matrix equal to the inverse covariance matrix of random errors: .

It can be shown that the formula for the GLS-estimates of the parameters of the linear model has the form

The covariance matrix of these estimates, respectively, will be equal to

In fact, the essence of the OLS lies in a certain (linear) transformation (P) of the original data and the application of the usual least squares to the transformed data. The purpose of this transformation is that for the transformed data, the random errors already satisfy the classical assumptions.

Weighted least squares

In the case of a diagonal weight matrix (and hence the covariance matrix of random errors), we have the so-called weighted least squares (WLS - Weighted Least Squares). In this case, the weighted sum of squares of the residuals of the model is minimized, that is, each observation receives a "weight" that is inversely proportional to the variance of the random error in this observation: . In fact, the data is transformed by weighting the observations (dividing by an amount proportional to the assumed standard deviation of the random errors), and normal least squares is applied to the weighted data.

Some special cases of application of LSM in practice

Linear Approximation

Consider the case when, as a result of studying the dependence of a certain scalar quantity on a certain scalar quantity (This can be, for example, the dependence of voltage on current strength: , where is a constant value, the resistance of the conductor), these quantities were measured, as a result of which the values \u200b\u200band and their corresponding values. Measurement data should be recorded in a table.

Table. Measurement results.

Measurement No.
1
2
3
4
5
6

The question sounds like this: what value of the coefficient can be chosen to best describe the dependence ? According to the least squares, this value should be such that the sum of the squared deviations of the values from the values

was minimal

The sum of squared deviations has one extremum - a minimum, which allows us to use this formula. Let's find the value of the coefficient from this formula. To do this, we transform its left side as follows:

The last formula allows us to find the value of the coefficient , which was required in the problem.

Story

Until the beginning of the XIX century. scientists did not have certain rules for solving a system of equations in which the number of unknowns is less than the number of equations; Until that time, particular methods were used, depending on the type of equations and on the ingenuity of the calculators, and therefore different calculators, starting from the same observational data, came to different conclusions. Gauss (1795) is credited with the first application of the method, and Legendre (1805) independently discovered and published it under its modern name (fr. Methode des moindres quarres ) . Laplace related the method to the theory of probability, and the American mathematician Adrain (1808) considered its probabilistic applications. The method is widespread and improved by further research by Encke, Bessel, Hansen and others.

Alternative use of MNCs

The idea of the least squares method can also be used in other cases not directly related to regression analysis. The fact is that the sum of squares is one of the most common proximity measures for vectors (the Euclidean metric in finite-dimensional spaces).

One application is "solving" systems of linear equations in which the number of equations is greater than the number of variables

where the matrix is not square, but rectangular.

Such a system of equations, in the general case, has no solution (if the rank is actually greater than the number of variables). Therefore, this system can be "solved" only in the sense of choosing such a vector in order to minimize the "distance" between the vectors and . To do this, you can apply the criterion for minimizing the sum of squared differences of the left and right parts of the equations of the system, that is, . It is easy to show that the solution of this minimization problem leads to the solution of the following system of equations

Example.

Experimental data on the values of variables X And at are given in the table.

As a result of their alignment, the function

The essence of the method of least squares (LSM).

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 functions by variables A And b, we equate these derivatives to zero.

We solve the resulting system of equations by any method (for example substitution method or Cramer's method) and obtain formulas for finding the coefficients using the least squares method (LSM).

With data A And b function takes the smallest value. The proof of this fact is given below the text at the end of the page.

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 A And b. We substitute in them the corresponding values from the last column of the table:

Hence, 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.

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.

In practice, when modeling various processes - in particular, economic, physical, technical, social - one or another method of calculating the approximate values of functions from their known values at some fixed points is widely used.

Problems of approximation of functions of this kind often arise:

when constructing approximate formulas for calculating the values of the characteristic quantities of the process under study according to the tabular data obtained as a result of the experiment;

in numerical integration, differentiation, solving differential equations, etc.;

if it is necessary to calculate the values of functions at intermediate points of the considered interval;

when determining the values of the characteristic quantities of the process outside the interval under consideration, in particular, when forecasting.

If, in order to model a certain process specified by a table, a function is constructed that approximately describes this process based on the least squares method, it will be called an approximating function (regression), and the task of constructing approximating functions itself will be an approximation problem.

This article discusses the possibilities of the MS Excel package for solving such problems, in addition, methods and techniques for constructing (creating) regressions for tabularly given functions (which is the basis of regression analysis) are given.

There are two options for building regressions in Excel.

Adding selected regressions (trendlines) to a chart built on the basis of a data table for the studied process characteristic (available only if a chart is built);

Using the built-in statistical functions of the Excel worksheet, which allows you to get regressions (trend lines) directly from the source data table.

Adding Trendlines to a Chart

For a table of data describing a certain process and represented by a diagram, Excel has an effective regression analysis tool that allows you to:

build on the basis of the least squares method and add to the diagram five types of regressions that model the process under study with varying degrees of accuracy;

add an equation of the constructed regression to the diagram;

determine the degree of compliance of the selected regression with the data displayed on the chart.

Based on the chart data, Excel allows you to get linear, polynomial, logarithmic, exponential, exponential types of regressions, which are given by the equation:

y = y(x)

where x is an independent variable, which often takes the values of a sequence of natural numbers (1; 2; 3; ...) and produces, for example, a countdown of the time of the process under study (characteristics).

1 . Linear regression is good at modeling features that increase or decrease at a constant rate. This is the simplest model of the process under study. It is built according to the equation:

y=mx+b

where m is the tangent of the slope of the linear regression to the x-axis; b - coordinate of the point of intersection of the linear regression with the y-axis.

2 . A polynomial trendline is useful for describing characteristics that have several distinct extremes (highs and lows). The choice of the degree of the polynomial is determined by the number of extrema of the characteristic under study. Thus, a polynomial of the second degree can well describe a process that has only one maximum or minimum; polynomial of the third degree - no more than two extrema; polynomial of the fourth degree - no more than three extrema, etc.

In this case, the trend line is built in accordance with the equation:

y = c0 + c1x + c2x2 + c3x3 + c4x4 + c5x5 + c6x6

where the coefficients c0, c1, c2,... c6 are constants whose values are determined during construction.

3 . The logarithmic trend line is successfully used in modeling characteristics, the values of which change rapidly at first, and then gradually stabilize.

y = c ln(x) + b

4 . The power trend line gives good results if the values of the studied dependence are characterized by a constant change in the growth rate. An example of such a dependence can serve as a graph of uniformly accelerated movement of the car. If there are zero or negative values in the data, you cannot use a power trend line.

It is built in accordance with the equation:

y = cxb

where the coefficients b, c are constants.

5 . An exponential trendline should be used if the rate of change in the data is continuously increasing. For data containing zero or negative values, this kind of approximation is also not applicable.

It is built in accordance with the equation:

y=cebx

where the coefficients b, c are constants.

When selecting a trend line, Excel automatically calculates the value of R2, which characterizes the accuracy of the approximation: the closer the R2 value is to one, the more reliably the trend line approximates the process under study. If necessary, the value of R2 can always be displayed on the diagram.

Determined by the formula:

To add a trend line to a data series:

activate the chart built on the basis of the data series, i.e., click within the chart area. The Chart item will appear in the main menu;

after clicking on this item, a menu will appear on the screen, in which you should select the Add trend line command.

The same actions are easily implemented if you hover over the graph corresponding to one of the data series and right-click; in the context menu that appears, select the Add trend line command. The Trendline dialog box will appear on the screen with the Type tab opened (Fig. 1).

After that you need:

On the Type tab, select the required trend line type (Linear is selected by default). For the Polynomial type, in the Degree field, specify the degree of the selected polynomial.

1 . The Built on Series field lists all the data series in the chart in question. To add a trendline to a specific data series, select its name in the Built on series field.

If necessary, by going to the Parameters tab (Fig. 2), you can set the following parameters for the trend line:

change the name of the trend line in the Name of the approximating (smoothed) curve field.

set the number of periods (forward or backward) for the forecast in the Forecast field;

display the equation of the trend line in the chart area, for which you should enable the checkbox show the equation on the chart;

display the value of the approximation reliability R2 in the diagram area, for which you should enable the checkbox put the value of the approximation reliability (R^2) on the diagram;

set the point of intersection of the trend line with the Y-axis, for which you should enable the checkbox Intersection of the curve with the Y-axis at a point;

click the OK button to close the dialog box.

There are three ways to start editing an already built trendline:

use the Selected trend line command from the Format menu, after selecting the trend line;

select the Format Trendline command from the context menu, which is called by right-clicking on the trendline;

by double clicking on the trend line.

The Format Trendline dialog box will appear on the screen (Fig. 3), containing three tabs: View, Type, Parameters, and the contents of the last two completely coincide with the similar tabs of the Trendline dialog box (Fig. 1-2). On the View tab, you can set the line type, its color and thickness.

To delete an already constructed trend line, select the trend line to be deleted and press the Delete key.

The advantages of the considered regression analysis tool are:

the relative ease of plotting a trend line on charts without creating a data table for it;

a fairly wide list of types of proposed trend lines, and this list includes the most commonly used types of regression;

the possibility of predicting the behavior of the process under study for an arbitrary (within common sense) number of steps forward, as well as back;

the possibility of obtaining the equation of the trend line in an analytical form;

the possibility, if necessary, of obtaining an assessment of the reliability of the approximation.

The disadvantages include the following points:

the construction of a trend line is carried out only if there is a chart built on a series of data;

the process of generating data series for the characteristic under study based on the trend line equations obtained for it is somewhat cluttered: the desired regression equations are updated with each change in the values of the original data series, but only within the chart area, while the data series formed on the basis of the old line equation trend, remains unchanged;

In PivotChart reports, when you change the chart view or the associated PivotTable report, existing trendlines are not retained, so you must ensure that the layout of the report meets your requirements before you draw trendlines or otherwise format the PivotChart report.

Trend lines can be added to data series presented on charts such as a graph, histogram, flat non-normalized area charts, bar, scatter, bubble and stock charts.

You cannot add trendlines to data series on 3-D, Standard, Radar, Pie, and Donut charts.

Using Built-in Excel Functions

Excel also provides a regression analysis tool for plotting trendlines outside the chart area. A number of statistical worksheet functions can be used for this purpose, but all of them allow you to build only linear or exponential regressions.

Excel has several functions for building linear regression, in particular:

TREND;

SLOPE and CUT.

As well as several functions for constructing an exponential trend line, in particular:

LGRFPapprox.

It should be noted that the techniques for constructing regressions using the TREND and GROWTH functions are practically the same. The same can be said about the pair of functions LINEST and LGRFPRIBL. For these four functions, when creating a table of values, Excel features such as array formulas are used, which somewhat clutters up the process of building regressions. We also note that the construction of a linear regression, in our opinion, is easiest to implement using the SLOPE and INTERCEPT functions, where the first of them determines the slope of the linear regression, and the second determines the segment cut off by the regression on the y-axis.

The advantages of the built-in functions tool for regression analysis are:

a fairly simple process of the same type of formation of data series of the characteristic under study for all built-in statistical functions that set trend lines;

a standard technique for constructing trend lines based on the generated data series;

the ability to predict the behavior of the process under study for the required number of steps forward or backward.

And the disadvantages include the fact that Excel does not have built-in functions for creating other (except linear and exponential) types of trend lines. This circumstance often does not allow choosing a sufficiently accurate model of the process under study, as well as obtaining forecasts close to reality. In addition, when using the TREND and GROW functions, the equations of the trend lines are not known.

It should be noted that the authors did not set the goal of the article to present the course of regression analysis with varying degrees of completeness. Its main task is to show the capabilities of the Excel package in solving approximation problems using specific examples; demonstrate what effective tools Excel has for building regressions and forecasting; illustrate how relatively easily such problems can be solved even by a user who does not have deep knowledge of regression analysis.

Examples of solving specific problems

Consider the solution of specific problems using the listed tools of the Excel package.

Task 1

With a table of data on the profit of a motor transport enterprise for 1995-2002. you need to do the following.

Build a chart.

Add linear and polynomial (quadratic and cubic) trend lines to the chart.

Using the trend line equations, obtain tabular data on the profit of the enterprise for each trend line for 1995-2004.

Make a profit forecast for the enterprise for 2003 and 2004.

The solution of the problem

In the range of cells A4:C11 of the Excel worksheet, we enter the worksheet shown in Fig. 4.

Having selected the range of cells B4:C11, we build a chart.

We activate the constructed chart and, using the method described above, after selecting the type of trend line in the Trend Line dialog box (see Fig. 1), we alternately add linear, quadratic and cubic trend lines to the chart. In the same dialog box, open the Parameters tab (see Fig. 2), in the Name of the approximating (smoothed) curve field, enter the name of the trend to be added, and in the Forecast forward for: periods field, set the value 2, since it is planned to make a profit forecast for two years ahead. To display the regression equation and the value of the approximation reliability R2 in the diagram area, enable the check boxes Show the equation on the screen and place the value of the approximation reliability (R^2) on the diagram. For better visual perception, we change the type, color and thickness of the plotted trend lines, for which we use the View tab of the Trend Line Format dialog box (see Fig. 3). The resulting chart with added trend lines is shown in fig. 5.

To obtain tabular data on the profit of the enterprise for each trend line for 1995-2004. Let's use the equations of the trend lines presented in fig. 5. To do this, in the cells of the D3:F3 range, enter textual information about the type of the selected trend line: Linear trend, Quadratic trend, Cubic trend. Next, enter the linear regression formula in cell D4 and, using the fill marker, copy this formula with relative references to the range of cells D5:D13. It should be noted that each cell with a linear regression formula from the range of cells D4:D13 has a corresponding cell from the range A4:A13 as an argument. Similarly, for quadratic regression, the cell range E4:E13 is filled, and for cubic regression, the cell range F4:F13 is filled. Thus, a forecast was made for the profit of the enterprise for 2003 and 2004. with three trends. The resulting table of values is shown in fig. 6.

Task 2

Build a chart.

Add logarithmic, exponential and exponential trend lines to the chart.

Derive the equations of the obtained trend lines, as well as the values of the approximation reliability R2 for each of them.

Using the trend line equations, obtain tabular data on the profit of the enterprise for each trend line for 1995-2002.

Make a profit forecast for the business for 2003 and 2004 using these trend lines.

The solution of the problem

Following the methodology given in solving problem 1, we obtain a diagram with added logarithmic, exponential and exponential trend lines (Fig. 7). Further, using the obtained trend line equations, we fill in the table of values for the profit of the enterprise, including the predicted values for 2003 and 2004. (Fig. 8).

On fig. 5 and fig. it can be seen that the model with a logarithmic trend corresponds to the lowest value of the approximation reliability

R2 = 0.8659

The highest values of R2 correspond to models with a polynomial trend: quadratic (R2 = 0.9263) and cubic (R2 = 0.933).

Task 3

With a table of data on the profit of a motor transport enterprise for 1995-2002, given in task 1, you must perform the following steps.

Get data series for linear and exponential trendlines using the TREND and GROW functions.

Using the TREND and GROWTH functions, make a profit forecast for the enterprise for 2003 and 2004.

For the initial data and the received data series, construct a diagram.

The solution of the problem

Let's use the worksheet of task 1 (see Fig. 4). Let's start with the TREND function:

select the range of cells D4:D11, which should be filled with the values of the TREND function corresponding to the known data on the profit of the enterprise;

call the Function command from the Insert menu. In the Function Wizard dialog box that appears, select the TREND function from the Statistical category, and then click the OK button. The same operation can be performed by pressing the button (Insert function) of the standard toolbar.

In the Function Arguments dialog box that appears, enter the range of cells C4:C11 in the Known_values_y field; in the Known_values_x field - the range of cells B4:B11;

to make the entered formula an array formula, use the key combination + + .

The formula we entered in the formula bar will look like: =(TREND(C4:C11;B4:B11)).

As a result, the range of cells D4:D11 is filled with the corresponding values of the TREND function (Fig. 9).

To make a forecast of the company's profit for 2003 and 2004. necessary:

select the range of cells D12:D13, where the values predicted by the TREND function will be entered.

call the TREND function and in the Function Arguments dialog box that appears, enter in the Known_values_y field - the range of cells C4:C11; in the Known_values_x field - the range of cells B4:B11; and in the field New_values_x - the range of cells B12:B13.

turn this formula into an array formula using the keyboard shortcut Ctrl + Shift + Enter.

The entered formula will look like: =(TREND(C4:C11;B4:B11;B12:B13)), and the range of cells D12:D13 will be filled with the predicted values of the TREND function (see Fig. 9).

Similarly, a data series is filled using the GROWTH function, which is used in the analysis of nonlinear dependencies and works exactly the same as its linear counterpart TREND.

Figure 10 shows the table in formula display mode.

For the initial data and the obtained data series, the diagram shown in fig. eleven.

Task 4

With a table of data on the receipt of applications for services by the dispatching service of a motor transport enterprise for the period from the 1st to the 11th day of the current month, the following actions must be performed.

Obtain data series for linear regression: using the SLOPE and INTERCEPT functions; using the LINEST function.

Retrieve a data series for exponential regression using the LYFFPRIB function.

Using the above functions, make a forecast about the receipt of applications to the dispatch service for the period from the 12th to the 14th day of the current month.

For the original and received data series, construct a diagram.

The solution of the problem

Note that, unlike the TREND and GROW functions, none of the functions listed above (SLOPE, INTERCEPTION, LINEST, LGRFPRIB) are regressions. These functions play only an auxiliary role, determining the necessary regression parameters.

For linear and exponential regressions built using the SLOPE, INTERCEPT, LINEST, LGRFINB functions, the appearance of their equations is always known, in contrast to the linear and exponential regressions corresponding to the TREND and GROWTH functions.

1 . Let's build a linear regression that has the equation:

y=mx+b

using the SLOPE and INTERCEPT functions, with the slope of the regression m being determined by the SLOPE function, and the constant term b - by the INTERCEPT function.

To do this, we perform the following actions:

enter the source table in the range of cells A4:B14;

the value of the parameter m will be determined in cell C19. Select from the Statistical category the Slope function; enter the range of cells B4:B14 in the known_values_y field and the range of cells A4:A14 in the known_values_x field. The formula will be entered into cell C19: =SLOPE(B4:B14;A4:A14);

using a similar method, the value of the parameter b in cell D19 is determined. And its content will look like this: = INTERCEPT(B4:B14;A4:A14). Thus, the values of the parameters m and b, necessary for constructing a linear regression, will be stored, respectively, in cells C19, D19;

then we enter the linear regression formula in cell C4 in the form: = $ C * A4 + $ D. In this formula, cells C19 and D19 are written with absolute references (the cell address should not change with possible copying). The absolute reference sign $ can be typed either from the keyboard or using the F4 key, after placing the cursor on the cell address. Using the fill handle, copy this formula to the range of cells C4:C17. We get the desired data series (Fig. 12). Due to the fact that the number of requests is an integer, you should set the number format on the Number tab of the Cell Format window with the number of decimal places to 0.

2 . Now let's build a linear regression given by the equation:

y=mx+b

using the LINEST function.

For this:

enter the LINEST function as an array formula into the range of cells C20:D20: =(LINEST(B4:B14;A4:A14)). As a result, we get the value of the parameter m in cell C20, and the value of the parameter b in cell D20;

enter the formula in cell D4: =$C*A4+$D;

copy this formula using the fill marker to the range of cells D4:D17 and get the desired data series.

3 . We build an exponential regression that has the equation:

with the help of the LGRFPRIBL function, it is performed similarly:

in the range of cells C21:D21, enter the function LGRFPRIBL as an array formula: =( LGRFPRIBL (B4:B14;A4:A14)). In this case, the value of the parameter m will be determined in cell C21, and the value of the parameter b will be determined in cell D21;

the formula is entered into cell E4: =$D*$C^A4;

using the fill marker, this formula is copied to the range of cells E4:E17, where the data series for exponential regression will be located (see Fig. 12).

On fig. 13 shows a table where we can see the functions we use with the necessary cell ranges, as well as formulas.

Value R 2 called determination coefficient.

The task of constructing a regression dependence is to find the vector of coefficients m of the model (1) at which the coefficient R takes the maximum value.

To assess the significance of R, Fisher's F-test is used, calculated by the formula

Where n- sample size (number of experiments);

k is the number of model coefficients.

If F exceeds some critical value for the data n And k and the accepted confidence level, then the value of R is considered significant. Tables of critical values of F are given in reference books on mathematical statistics.

Thus, the significance of R is determined not only by its value, but also by the ratio between the number of experiments and the number of coefficients (parameters) of the model. Indeed, the correlation ratio for n=2 for a simple linear model is 1 (through 2 points on the plane, you can always draw a single straight line). However, if the experimental data are random variables, such a value of R should be trusted with great care. Usually, in order to obtain a significant R and reliable regression, it is aimed at ensuring that the number of experiments significantly exceeds the number of model coefficients (n>k).

To build a linear regression model, you must:

1) prepare a list of n rows and m columns containing the experimental data (column containing the output value Y must be either first or last in the list); for example, let's take the data of the previous task, adding a column called "period number", numbering the numbers of periods from 1 to 12. (these will be the values X)

2) go to menu Data/Data Analysis/Regression

If the "Data Analysis" item in the "Tools" menu is missing, then you should go to the "Add-Ins" item of the same menu and check the "Analysis Package" box.

3) in the "Regression" dialog box, set:

input interval Y;

input interval X;

output interval - the upper left cell of the interval in which the calculation results will be placed (it is recommended to place it on a new worksheet);