barcode font reporting services Curve Fitting, Regression, and Correlation in Software

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CHAPTER 8 Curve Fitting, Regression, and Correlation
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(a) Plot the points (1, 1), (3, 2), (4, 4), (6, 4), (8, 5), (9, 7), (11, 8), and (14, 9) on a rectangular coordinate system as shown in Fig. 8-6. A straight line approximating the data is drawn freehand in the figure. For a method eliminating the need for individual judgment, see Problem 8.4, which uses the method of least squares. (b) To obtain the equation of the line constructed in (a), choose any two points on the line, such as P and Q. The coordinates of these points as read from the graph are approximately (0, 1) and (12, 7.5). Then from Problem 8.1, y or y 1 0.542x or y 1 0.542x. 1 7.5 12 1 (x 0 0)
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8.3. Derive the normal equations (4), page 267, for the least-squares line.
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Refer to Fig. 8-7. The values of y on the least-squares line corresponding to x1, x2, . . . , xn are a bx1, a bx2, c, a bxn
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Fig. 8-7
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The corresponding vertical deviations are d1 a bx1 c y1, d2 a bx2 y2, c, dn c a bxn yn
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Then the sum of the squares of the deviations is d2 1 or d2 2 d2 n (a bx1 ad
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y1)2 a (a
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bx2 bx
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y2)2 y)2
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yn)2
This is a function of a and b, i.e., F(a, b) g(a bx (or a maximum) is that 'F>'a 0, 'F>'b 0. Since 'F 'a 'F 'b we obtain a (a i.e., ay bx an y) b ax 0 ' a 'a (a ' a 'b (a bx bx y)2 y)2
y)2. A necessary condition for this to be a minimum
a 2(a a 2x(a
bx bx
y) y)
a x(a a xy
bx a ax
b a x2
as required. It can be shown that these actually yield a minimum.
8.4. Fit a least-squares line to the data of Problem 8.2 using (a) x as independent variable, (b) x as dependent variable.
(a) The equation of the line is y a bx. The normal equations are ay a xy an a ax b ax b a x2
CHAPTER 8 Curve Fitting, Regression, and Correlation
The work involved in computing the sums can be arranged as in Table 8-2. Although the last column is not needed for this part of the problem, it has been added to the table for use in part (b). Since there are 8 pairs of values of x and y, n 8 and the normal equations become 8a 56a 56b 524b 40 364
7 11 x
6 7 6 Solving simultaneously, a 11 or 0.545, b 11 or 0.636; and the required least-squares line is y 11 y 0.545 0.636x. Note that this is not the line obtained in Problem 8.2 using the freehand method.
Table 8-2 x 1 3 4 6 8 9 11 14 gx 56 gy y 1 2 4 4 5 7 8 9 40 x2 1 9 16 36 64 81 121 196 g x2 524 g xy xy 1 6 16 24 40 63 88 126 364 gy2 y2 1 4 16 16 25 49 64 81 256
Another method
Q a yR Q a x2 R a na n a xy b na x2 Q a xR x2 Q a xR Q a xR Q a xyR
(40)(524) (56)(364) (8)(524) (56)2
6 11
or 0.545
Q a xR Q a yR
(8)(364) (56)(40) (8)(524) (56)2
7 11
or 0.636
(b) If x is considered as the dependent variable and y as the independent variable, the equation of the leastsquares line is x c dy and the normal equations are ax a xy cn c ay d ay d a y2
Then using Table 8-2, the normal equations become 8c 40c
1 from which c 0.50, d 3 or 1.50. 2 or 2 These values can also be obtained from
40d 256d
56 364
Q a xR Q a y2 R c n a y2 n a xy d n a y2
Q a yR Q a xyR Q a yR
(56)(256) (40)(364) (8)(256) (40)2
Q a xR Q a yR Q a yR
(8)(364) (56)(40) (8)(256) (40)2
CHAPTER 8 Curve Fitting, Regression, and Correlation
Therefore, the required equation of the least-squares line is x Note that by solving this equation for y, we obtain y 0.333 obtained in part (a). 0.50 1.50y. 0.667x, which is not the same as the line
8.5. Graph the two lines obtained in Problem 8.4.
0.500 1.50y, are shown in Fig. 8-8. Note that the The graphs of the two lines, y 0.545 0.636x and x two lines in this case are practically coincident, which is an indication that the data are very well described by a linear relationship. The line obtained in part (a) is often called the regression line of y on x and is used for estimating y for given values of x. The line obtained in part (b) is called the regression line of x on y and is used for estimating x for given values of y.
Fig. 8-8
8.6. (a) Show that the two least-squares lines obtained in Problem 8.4 intersect at point (x, y). (b) Estimate the # # value of y when x 12. (c) Estimate the value of x when y 3. x # ax n 56 8 7, y # ay n 40 8 5
Then point (x, y), called the centroid, is (7, 5). # # (a) Point (7, 5) lies on line y Point (7, 5) lies on line x 0.545
1 2 3 2 y,
0.636x or, more exactly, y 1 3 since 7 2 2 (5).
6 11
7 11 x,
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