 Hello and welcome to the session. In this session first of all we will discuss a property of a regression line. And that is the regression line y bar that is the mean. Let us discuss the case one independent variable. Now we know that the regression line is also called the least square line. So we can write the equation of the regression line that is the least square line is given as y is equal to mx plus c where m is the slope of the line and c is a constant. Now a normal equation for the least square summation y is equal to n through summation x where m is the number of observations. Now let this be equation number one and this be equation number two. Now dividing of equation number two by n we get summation y over n is equal to c plus now we know that summation y over n is equal to the mean value of y which is denoted by y bar is equal to c plus m into summation x over n is the mean value of x which is denoted by x bar. So this will be equal to the equation as y bar is equal to c plus m into x bar. Now let this be equation number three and this is the equation number three. Now subtracting or you can write the least regression line is given by y minus y bar is equal to m into x minus x bar the whole that is y minus y bar is equal to b y x into x minus x bar the whole where b y x is the regression coefficient of y or x and this is the regression line of y on x and this is the line to the point x bar y bar that is now let us discuss the second case when y is the independent variable. Now proceeding case one the regression line x bar is equal to b x y into y minus y bar the whole where x bar is the mean value of x y bar is the mean value of y and b x y is the regression coefficient of x on y and this is the regression equation of x on y which shows that the line to the point y bar that is the mean value of forms of regression equation. Now we can find out the alternative forms of regression equation by finding the deviations and y series respect now let be the arithmetic mean of x series and y bar be the arithmetic mean of y series then that is the correlation coefficient r is equal to summation x bar minus summation x into summation y over n the whole summation y square minus summation y whole square written as x y minus into n into y bar minus into summation y square by n minus summation r is equal to summation x y minus into sigma x into sigma y because we know that is the standard summation of x is equal to summation x by n now the normal equation i is equal to n to summation x which implies dividing throughout by n it will be n is equal to c plus m into summation x over summation y over n is y r is equal to c plus m into x bar which is equal to y bar minus m into x bar is given as is equal to c into summation x plus m into summation x square now putting the value that it be equation x is equal to y bar minus m into x bar the whole equal to y bar into summation x minus into summation x plus m into summation x square summation x y minus y bar into summation x is equal to m into summation x square minus x bar into summation x into summation x, what happens? Minus x bar into summation x. Therefore, summation x bar minus x bar into y bar summation x by n x bar into y bar. So, it will be equal to summation x square by n minus summation x by n into x bar into y bar into sigma y. Now multiply the numerator with sigma y and multiply the denominator with sigma x. The correlation coefficient is equal to into sigma y over sigma x. It is equal to b by x. That means this is the regression coefficient of y or x. The equation of the regression line y bar is equal to b by x into x minus x bar the whole. That is, y minus y bar is equal to r sigma x into x minus x bar the whole. Similarly, we can find the equation of regression equal to b x y into y minus y bar the whole, where b x y is the regression coefficient of x that is x minus x bar is equal to and here b x y will be equal to over sigma y into y minus y bar the whole. Now that d x is equal to x minus x bar and d y is equal to y minus y bar with the deviations of the variables x and y respectively from the arithmetic means of the series. Now we have obtained r is equal to summation x y minus into y bar 4 upon 2 sigma x into sigma y. Now for the values of d x and d y summation d x into d y that is summation of d x into d y by n into summation d y square by n into d x into d y d x square into summation d y square. The regression equations will be y minus y bar is equal to summation d x into d y that is summation of d x into d y over summation d x square into x minus x bar the whole. Well this is the value of summation d x into summation d x square equation of regression line. The regression coefficient d y x is equal to summation d x into d y summation d x square that is the regression equation of x from y will be given as summation d x into d y over summation d y square into y minus y bar the whole summation d x into d y and summation d y square. Well the regression that is d x y is equal to summation d x into d y over summation d y square. Now we can also find of the regression equation by finding the deviations which are taken from the assumed mean. Now if the epic the deviations from the assumed mean. Now let u is equal to x minus a and v is equal to the assumed means that is of the x and y series respectively and here are the deviations taken from the assumed mean. Then the regression coefficient of y on x that is b y x is equal to r into summation u v minus summation u into summation v over n minus summation u whole square over n v square minus summation v whole square into that is the value of sigma y which will be equal to square root of summation v square over n minus summation v by n. Now sigma x is equal to square root of over n minus summation u over n which will give b y x is equal to summation u v minus summation u into summation v over n v x y will be equal to square minus summation v whole square of regression is given as this y bar is equal to b y x into x minus x bar the whole of regression is given as equal to b x y into y minus y bar the whole. There this is the value of b y x and this is the value of b x y whenever the deviations are taken from the assumed mean. So in this session we have learnt about the property of regression line that is the regression line always passes through the mean of regression equation. So this is the session.