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where we have used E(X mX) 0, E(Y mY) 0. Denoting the last expression by F(a, b), we have 'F 'a 2(mY bmX a), 'F 'b 2bs2 X 2sXY 2mX(mY bmX a)
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Setting these equal to zero, which is a necessary condition for F(a, b) to be a minimum, we find mY Therefore, if y a bx, then y mY y or y a b(x mY mY sY bmX mX) or sXY (x s2 X ra x mX) mX sX b bs2 X sXY
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The similarity of the above proof for populations, using expectations, to the corresponding proof for samples, using summations, should be noted. In general, results for samples have analogous results for populations and conversely.
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8.40. The joint density function of the random variables X and Y is f (x, y)
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Find the least-squares regression curve of (a) Y on X, (b) X on Y. (a) The marginal density function of X is f1(x)
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CHAPTER 8 Curve Fitting, Regression, and Correlation
for 0 x 1, and f1(x) 0 otherwise. Hence, for 0 f2( y Z x) f (x, y) f1(x) x cx 0
x 0 y
1, the conditional density of Y given X is y 1 1
2y 1
0 or y
and the least-squares regression curve of Y on X is given by y E(Y Z X x) 3 yf2(y Z x) dy 2y b dy 1 3x 6x 4 6 0 or x 1.
1 x 30 ya x
Neither f2( y Z x) nor the least-squares regression curve is defined when x (b) For 0 y 1, the marginal density function of Y is f2( y) Hence, for 0 y
1 2 30 3 (x
2y) dx
1 (1 3
1, the conditional density of X given Y is f (x, y) f2( y) 2x c1 0 3 4y 4y 0 x x 1 1
f1(x Z y)
0 or x
and the least-squares regression curve of X on Y is given by x E(X Z Y
xf1(x Z y) dx 4y b dx 4y 2 3 6y 12y 12y) are
2x 30 xa 1
Neither f1(x Z y) nor the least-squares regression curve is defined when y 0 or y 1. Note that the two regression curves y (3x 4)>(6x 6) and x (2 6y)>(3 different.
# # 8.41. Find (a) X, (b) Y, (c) s2 , (d) s2 , (e) sXY, (f) r for the distribution in Problem 8.40. X Y
(a) (b) (c) Then (d) Then (e) # X # Y # X2 s2 X # Y2 s2 Y ## XY
1 2 3y 0x c 3 (x 0 1 0 y
2y) d dx dy 2y) d dx dy 2y)d dx dy 5 2 a b 9
5 9 11 18 7 18 13 162 4 9
2 y c (x 3
1 2 2 3y 0x c 3 (x 0
# X2
# X2
7 18
1 2 2 3y 0y c 3 (x 0
2y)d dx dy a 11 2 b 18 23 324
# Y2
# Y2
1 2 3y 0xy c 3 (x 0
2y) d dx dy
CHAPTER 8 Curve Fitting, Regression, and Correlation
5 11 a ba b 9 18
Then (f)
sXY r
XY sXY sXsY
## XY
1 162 0.0818
1>162 !13>162!23>324
Note that the linear correlation coefficient is small. This is to be expected from observation of the leastsquares regression lines obtained in Problem 8.42.
8.42. Write the least-squares regression lines of (a) Y on X, (b) X on Y for Problem 8.40.
(a) The regression line of Y on X is y # Y y sY # Y # x X ra s b or X # X) or y # X 11 18 1>162 ax 13>162 5 b 9
sXY (x s2 X x
(b) The regression line of X on Y is
# y Y ra s b or Y # Y) or x 5 9 1>162 ay 23>324 11 b 18
sXY (y s2 Y
Sampling theory of regression 8.43. In Problem 8.11 we found the regression equation of y on x to be y 35.82 0.476x. Test the hypothesis at a 0.05 significance level that the regression coefficient of the population regression equation is as low as 0.180.
t b b !n sy.x >sx 2 0.476 0.180 !12 1.28>2.66 2 1.95
! x2 x2 2.66 from Problem 8.11. since sy.x 1.28 (computed in Problem 8.22) and sx # # On the basis of a one-tailed test of Student s distribution at a 0.05 level, we would reject the hypothesis that the regression coefficient is as low as 0.180 if t t0.95 1.81 for 12 2 10 degrees of freedom. Therefore, we can reject the hypothesis.
8.44. Find 95% confidence limits for the regression coefficient of Problem 8.43.
b b t !n sy.x as b 2 x t0.975 2.23 for 12 2 10 degrees of
Then 95% confidence limits for b (obtained by putting t freedom) are given by b sy.x 2.23 as b !12 2 x 0.476
2.23 1.28 a b !10 2.66
i.e., we are 95% confident that b lies between 0.136 and 0.816.
8.45. In Problem 8.11, find 95% confidence limits for the heights of sons whose fathers heights are (a) 65.0, (b) 70.0 inches.
Since t0.975 2.23 for 12 2 10 degrees of freedom, the 95% confidence limits for yp are y0 where y0 35.82 2.23 2n sy.x n 2 A 1 n(x0 s2 x 2.66 (Problem 8.43), and n 12. x)2 #
0.476x0 (Problem 8.11), sy.x
1.28, sx
CHAPTER 8 Curve Fitting, Regression, and Correlation
65.0, y0 66.76 inches. Also, (x0 2.23 (1.28) 12 A !10 x)2 # 1 (65.0 12(2.78) (2.66)2 800>12)2 2.78. Then 95% confidence limits
(a) If x0 are
3.80 inches
i.e., we can be about 95% confident that the sons heights are between 63.0 and 70.6 inches. (b) If x0 70.0, y0 69.14 inches. Also, (x0 x)2 (70.0 800>12)2 11.11. Then the 95% confidence # limits are computed to be 69.14 5.09 inches, i.e., we can be about 95% confident that the sons heights are between 64.1 and 74.2 inches. Note that for large values of n, 95% confidence limits are given approximately by y0 1.96 sy.x or y0 2sy.x provided that x0 x is not too large. This agrees with the approximate results mentioned on # page 269. The methods of this problem hold regardless of the size of n or x0 x, i.e., the sampling methods # are exact for a normal population.
8.46. In Problem 8.11, find 95% confidence limits for the mean heights of sons whose fathers heights are (a) 65.0, (b) 70.0 inches.
Since t0.975 2.23 for 10 degrees of freedom, the 95% confidence limits for yp are # y0 where y0 35.82 2.23 sy.x 1 !20 A (x0 s2 x x)2 #
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