406 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 7. Fit a parabola y = a + bx + cx 2 to the folliwng data: 24 6 810 3.07 12.85 31.47 57.38 91.29 x y [Ans. 2 0.34 0.78 0.99 yxx=− + ] 8. Determine the constants a and b by the method of least squares such that bx yae = fits the following data: 24 6 810 4.077 11.084 30.128 81.897 222.62 x y [Ans. 0.50001 1.49989 x ye = ] 9. Fit a least square geometric curve b yax = to the following data: 1234 5 0.524.5812.5 x y [Ans. 1.9977 0.5012 yx= ] 10. A person runs the same race track for five consecutive days and is timed as follows: () () 1 2345 15.3 15.1 15 14.5 14 Day x Time y Make a least square fit to the above data using a function 2 bc a x x ++ . [Ans. 2 6.7512 4.4738 13.0065 y x x =++ ] 11. Use the method of least squares to fit the curve 0 1 c ycx x =+ to the following table of values: 0.1 0.2 0.4 0.5 1 2 21 11 7 6 5 6 x y [Ans. 1.97327 3.28182yx x =+ ] 12. Using the method of least square to fit a parabola 2 yabxcx=+ + in the following data: () ( )()()() − , : 1,2 , 0,0 , 0,1 , 1,2xy [Ans. 2 13 22 yx=+ ] CURVE FITTING 407 13. The pressure of the gas corresponding to various volumes V is measured, given by the following data: () () 3 2 50 60 70 90 100 64.7 51.3 40.5 25.9 78 Vcm pkgcm − Fit the data to the equation .pV c γ = [Ans. 0.28997 167.78765 pV = ] 9.3 REGRESSION We know that in a functional relation between two variables, if we know the value of one variable, then the corresponding value of the other variable can be determined exactly. But, in a statistical relationship between the two variables, when the value of one variable is known, we can simply estimate the corresponding value of another variable. Regression analysis is the method used for estimating the unknown values of one variable corresponding to the known values of another variable. 9.3.1 Dependent and Independent Variables Suppose there is a relation between two variables. The variable, whose values are known, is known as independent variable, while another one is called the dependent variable. 9.3.2 Line of Regression Let {,}:1 ii xy i n ≤≤ and 1}jn≤≤ be a bivariate distribution. If we plot the corresponding values of x and ,y taking the values of x along x-axis and the values of y along y-axis, we obtain a collection of dots, called the scatter-diagram. If the scatter diagram indicates some relationship between x and y, then the dots of the scatter diagram will be concentrated round a line, called the line of regression or the line of best fit. 9.3.3 Regression Line of y on x If we have to predict the values of y from given values of x, then the line of regression has an equation of the form .yabx=+ This is called the regression line of y on x. 9.3.4 Regression Line of x on y If we have to predict the values of x from given values of y, then the line of regression has an equation of the form x = a + by. This is called the regression line of x on y. 9.3.5 To obtain the Equation of Line of Regression of y on x Suppose that the line approximating the set of point ()()() 11 22 33 ,,,,, xy xy xy , , () , nn xy has the equation: y = a + bx (1) 408 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Then, ii yabx =+ and 2 ii i i xy ax bx =+ for each 1, 2, ,in= therefore ii ynabx =+ ∑∑ (2) and 2 ii i i xy a x b x =+ ∑∑∑ (3) Equations () 2 and () 3 are normal equations for this line. Solving () 2 and () 3 for a and b and putting these values in () 1 , we obtain the required equation of the line of regression of y on .x 9.3.6 To obtain the Equation of Line of Regression of xx xx x on yy yy y Suppose that the line approximating the set of points ()()()() 11 22 33 , , , , , , , , nn xy xy xy xy has the equation: xaby=+ (1) Then, ii xaby =+ and 2 ii i i xy ay by =+ for each 1, 2, ,in= therefore ii xnaby =+ ∑∑ (2) and 2 ii i i xy a y b y =+ ∑∑∑ (3) Equations () 2 and () 3 are normal equations for this line. Solving () 2 and () 3 for a and b and putting these values in (1), we obtain the required equation of the line of regression of x on .y Example 1. Find the line of regression of y on x for the following data: x 10987643 y 812710896 Sol. Here 7n = . Now form the table given below: i x i y 2 i x ii xy 10 8 100 80 9 12 81 108 8 7 64 56 71049 70 6 8 36 48 4 9 16 36 36918 i x = ∑ 47 i y = ∑ 60 2 i x = ∑ 352 416 ii xy = ∑ Let the required equation be yabx=+ (1) Then, ii yabx =+ and 2 ii i i xy ax bx =+ for each i. CURVE FITTING 409 Therefore the normal equations are: ii ynabx =+ ∑∑ (2) 2 ii i i xy a x b x =+ ∑∑∑ (3) Putting the values from the table in () 2 and () 3 , we get 60 7 47ab =+ 416 47 355ab=+ Solving these equations, we get 8.582a = and 1.094.b = Putting these values in (1) the required equation is y = 8.582 + 1.094x Ans. Example 2. Find the line of regression of x on y for the following data: x 621048 y 911587 Sol. Here 5n = . Now, form the table given below : i x i y 2 i y ii xy 6 9 81 54 2 11 121 22 10 5 25 50 4 8 64 32 8 7 49 56 i x = ∑ 30 40 i y = ∑ 2 i y = ∑ 340 ii xy = ∑ 214 Let the required line be, xaby=+ (1) Then ii xaby =+ and 2 ii i i xy ay by =+ for each .i Therefore the normal equations are: i i xnaby =+ ∑∑ (2) 2 ii i i xy a y b y =+ ∑∑∑ (3) Putting the values from the table in () 2 and () 3 , we get 30 5 40ab=+ ⇒ 86ab+= 214 40 340ab=+ ⇒ 20 170 107ab+= On solving these equations we get 16.4a = and 1.3.b =− Therefore the requried equation is, 16.4 1.3 .xy=− Ans. 410 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 3. Prove that arithmetic mean of the coefficient of regression is greater than the coefficient of correlation. Sol. Coefficients of regression are r y x σ σ , r x y σ σ We have to prove that .AM r> or 1 2 2 y x xy rr σ σ +> σσ or 1 1 2 y x xy σ σ +> σσ or 20 y x xy σ σ +−> σσ or 22 1 [2]0 xy xy xy σ+σ−σσ > σσ or 2 1 [] xy xy σ−σ σσ which is true. Proved. Example 4. Find the regression line of y on x for the following data: x 1346891114 y 12445789 Estimate the value of ,y when 10.x = Sol. S.No. x yxy 2 x 1 11 1 1 2 32 6 9 3 4 4 16 16 4 6 4 24 36 5 8 5 40 64 6 9 7 63 81 7 11 8 88 121 8 14 9 126 196 Total 56 40 364 524 Let y = a + bx be the line of regression of y on x . Therefore normal equations are : ii ynabx =+ ∑∑ ⇒ 40 8 56ab=+ (1) 2 ii i i xy a x b x =+ ∑∑∑ ⇒ 364 56 524ab=+ (2) On solving (1) and (2) we get 6 11 a = and 7 11 b = The equation of the required line is 67 11 11 yx=+ or 71160xy−+= If 0,x = () 6 7 76 10 10 6 11 11 11 11 y =+ == . Ans. CURVE FITTING 411 Example 5. In a study between the amount of rainfall and the quantity of air pollution removed the following data were collected. Daily Rainfall in 0.01cm 4.3 4.5 5.9 5.6 6.1 5.2 3.8 2.1 Pollution Removed (mg/m 3 ) 12.6 12.1 11.6 11.8 11.4 11.8 13.2 14.1 Find the regression line of y on x. Sol. () xmetre yxy 2 x 4.3 12.6 54.18 18.49 4.5 12.1 54.45 20.25 5.9 11.6 68.44 34.81 5.6 11.8 66.08 31.36 6.1 11.4 69.54 37.21 5.2 11.8 61.36 27.04 3.8 13.2 50.16 14.44 2.1 14.1 29.61 4.41 37.5 98.6 453.82 188.01 Let yabx=+ be the equation of the line of regression of y on .x ∴ Normal equations are: ii ynabx =+ ∑∑ ⇒ 98.6 8 37.5ab=+ 2 ii i i xy a x b x =+ ⇒ ∑∑∑ 453.82 37.5 188.01ab=+ After solving these normal equations we get 15.49a = and 0.675b =− . The equation of the line of regression is y = 15.49 – 0.675x. Ans. 9.3.7 Another Form of Equations of Lines of Regression Theorem 1: Show that the equation of the line of regression of y on x is given by () . y x yyr xx σ −= − σ , where x and y are the means of x-series and y-series respectively; r is the coefficient of correlation between x and ;y x σ and y σ are the standard deviations of x-series and the y-series respectively. Proof: Suppose that the line approximating the set of points ()() 11 22 ,,, xy xy , , () , nn xy has the equation yabx=+ (1) Then ii yabx =+ and 2 ii i i xy ax bx =+ for each 1, 2, , .in= ∴ ii ynabx =+ ∑∑ (2) 412 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 2 ii i i xy a x b x =+ ∑∑∑ (3) From (2), we have ii yx ab nn =+ ∑∑ or yabx=+ (4) Thus, it follows that () ,xy lies on the line. Shifting the origin to () ,xy () 2 becomes () () ii yy nab xx −=+ − ∑∑ or 0a = 3 xx yy ii −= −= ∑∑ bg bg 0 Shifting the origin to () ,xy and taking a = 0, (1) becomes ()() yy bxx −= − (5) (3) becomes ()() () 2 ii i xxyy b xx −−= − ∑∑ (6) From () 6 , we have b = xx y y xx ii i did i di − − ∑ ∑ 2 = dx dy dx ii i bg bg ∑ ∑ 2 = dx dy n ii x . bg bg ∑ σ 2 = r y x σ σ 3 () () , ii xy dx dy r n = σσ ∑ Putting this values of b in () 5 , the required equation of the line if regression of y on x i () () . y x yy r xx σ −= − σ Coefficient of Regression of y on x: The real number . y x br σ = σ is called the coefficient of regression of y on x and is denoted by yx b . Thus y yx x br σ = σ . Theorem 2: The equation of the line of regression of x on y is given by xx– di = ryy x y .– σ σ di Proof: Proceed as in theorem 1. Coefficient of Regression of x on y: The real number b = r x y . σ σ is called the coefficient of regression of x on y and is denoted by b xy . Thus b xy = r x y . σ σ . CURVE FITTING 413 Theorem 3: Prove that: (i) ()() () 2 2 ii ii yx i i xy xy n b x x n − − = − ∑∑ ∑ ∑ ∑ (ii) ()() () 2 2 ii ii xy i i xy xy n b y y n − = − ∑∑ ∑ ∑ ∑ Proof: (i) By definition, we have () 2 . yyx yx x x br r σσσ == σ σ () () 2 cov , x xy = σ [3 rσ y σ x = cov(x, y)] ()() () 2 2 ii ii i i xy xy n x x n − = − ∑∑ ∑ ∑ ∑ Similarly, ()ii can be proved. Example 6. Find the regression coefficient yx b between x and y for the following data: x=24 ∑ , 22 y = 44, xy = 306, x = 164, y = 574 ∑∑ ∑ ∑ and n = 4. Sol. The given data may be written as 24, i x = ∑ 44, i y = ∑ 306, ii xy = ∑ 2 164, i x = ∑ 2 574 i y = ∑ and n = 4. b yx = xy xy n x x n ii ii i i ∑ ∑∑ ∑ ∑ R S | T | U V | W | – – didi di 2 2 = 306 24 44 4 164 24 4 2 – – × af = 306 264 164 144 – – bg = 42 20 = 2.1. Ans. Example 7. Find the regression coefficient b xy between x and y for the following data: x ∑ = 30, y ∑ = 42, xy ∑ = 199, x 2 ∑ = 184, y 2 ∑ = 318 and n = 6. 414 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. The given data may be given as under: x i ∑ = 30, y i ∑ = 42, xy ii ∑ = 199, x i ∑ 2 = 184, y i ∑ 2 = 318 and n = 6. ∴ ()() () 2 2 ii ii xy i i xy xy n b y y n − = − ∑∑ ∑ ∑ ∑ 30 42 199 199 210 11 6 0.46. 42 42 318 294 24 318 6 × − −− ====− × − − Ans. Example 8. For the following observations (x, y), find the regression coefficient b yx and b xy and hence find the correlation coefficient between x and y: (1, 2), (2, 4), (3, 8), (4, 7), (5, 10), (6, 5), (7, 14), (8, 16), (9, 2), (10, 20). Sol. Here n = 10. We may prepare the table, given below: 22 22 12 1 4 2 24 4 16 8 38 9 64 24 4 7 16 49 28 5 10 25 100 50 6 5 36 25 30 7 14 4 196 98 8 16 64 256 128 92 81 4 18 10 20 100 400 200 55 88 385 1114 586 ii i i ii iii i ii xy x y xy xyx y xy == = = = ∑∑∑ ∑ ∑ ()() () () 2 2 2 55 88 586 102 10 1.24 82.5 55 385 10 ii ii yx i i xy xy n b x x n × −− ==== − − ∑∑ ∑ ∑ ∑ And ()() () () 2 2 2 (55 88) 586 102 10 0.30 339.6 88 1114 10 ii ii xy i i xy xy n b y y n × −− ==== − − ∑∑ ∑ ∑ ∑ CURVE FITTING 415 Now, 2 ··· y x yx xy xy bb r r r σ σ == σσ , where r is the coefficient of correlation. ∴ · 1.24 0.30 0.609. yx xy rbb ==×= Thus, 1.24, 0.30 yx xy bb == and 0.609r = . Ans. 9.3.8 Some Properties of Regression Coefficients Let, the regression coefficient of y on x is yx b ; the regression coefficient of x on y is xy b ; and, the correlation coefficient between x and y is r . Then, we have the following results. Theorem 1: Prove that yx xy rbb =⋅ . Proof: We have: y yx x br σ = σ and . x xy y br σ = σ Therefore, b yx b xy = r 2 or r = bb yx xy Remark: Clearly we can say that, correlation coefficient is the geometric mean between the two regression coefficients. Theorem 2: Prove that r, b yx and b xy are of the same sign. Proof: We know that y yx x br σ = σ and x xy y br σ = σ . Since x σ and y σ are both positive, it follows from the two equations, given above that yx b and xy b have the same sign as .r Hence , yx rb and xy b are always of the same sign. Theorem 3: Prove that the arithmetic mean of regression coefficient is greater than the correlation coefficient. Proof: Clearly, the required result is true, If () 1 2 yx xy bb r +> i.e., if 1 . 2 y x xy rr r σ σ +> σσ i.e., if 22 2 yx xy σ+σ>σσ i.e., if 222 ()20 yx xy σ−σ −σσ> i.e., if 2 ()0, yx σ−σ > which is true. Hence the required result is true. Proved Theorem 4: Let θ be the angle between the regression line of y on x and the regression line of x on .y Then, prove that () () 2 22 1 , tan . . xy xy r r − σσ θ= σ+σ Proof: The equation of the line of regression of x on y is () () . x y xx r yy σ −= − σ (i) . 1.3 .xy=− Ans. 410 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 3. Prove that arithmetic mean of the coefficient of regression is greater than the coefficient of correlation. Sol of y along y-axis, we obtain a collection of dots, called the scatter-diagram. If the scatter diagram indicates some relationship between x and y, then the dots of the scatter diagram will. corresponding value of another variable. Regression analysis is the method used for estimating the unknown values of one variable corresponding to the known values of another variable. 9.3.1 Dependent and