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We wish to decide between the hypotheses (H0: mZ 1 Under the hypothesis H0, Z1 z Z2 sZ
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1.96. Using a two-tailed test of the normal distribution, we would reject H0 only if z 1.96 or z Therefore, we cannot reject H0, and we conclude that the results are not significantly different at a 0.05 level.
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CHAPTER 8 Curve Fitting, Regression, and Correlation
Miscellaneous problems 8.52. Prove formula (25), page 269.
For the least-squares line we have, from Problems 8.20 and 8.21, s2 y.x But by definition, a( y n and, by (6) on page 267, a (x a (x s2 xy s2 x s2 c 1 y x)( y # y) # sxy s2 x s2 (1 y r2) y)2 # s2 y a (x x)( y # n y) # sxy a( y n y)2 # b a (x x)( y # n y) #
b Hence, s2 y.x s2 y
x)2 # sxy 2 as s b d x y
An analogous formula holds for the population (see Problem 8.54).
# 8.53. Prove that E[(Y Y )2] E[(Y (b) a least-squares parabola.
We have Then and so E[(Y (Y # Y)2 # Y)2 Y (Y E[(Y
Yest)2]
# Y Yest)2 Yest)2] (Y
[(Yest
Yest) (Yest E[(Yest
# Y)2] for the case of (a) a least-squares line,
(Yest # Y)2 # Y)2] 2 (Y # Y) Yest)(Yest # Y) # Y)]
2E[(Y
Yest)(Yest
The required result follows if we can show that the last term is zero. (a) For linear regression, Yest E[(Y Yest)(Yest a # Y)] bX. Then E[(Y (a 0 because of the normal equations E(Y (Compare Problem 8.3.) (b) For parabolic regression, Yest E[(Y Yest)(Yest # Y )] a E[(Y (a 0 because of the normal equations E(Y a bX gX 2) 0, E[X(Y a bX gX 2)] 0, E[X 2(Y a bX gX 2)] 0 bX a # Y )E(Y gX2. Then bX a a gX2 )(a bX bX bX gX 2) gX 2)] gX2 bE[X(Y # Y )] a bX gX 2)] a bX) 0, E(XY aX bX2) 0 a # Y)E(Y bX)(a a bX bX) # Y )] bE(XY aX bX2)
gE[X 2(Y
Compare equations (19), page 269.
The result can be extended to higher-order least-squares curves.
CHAPTER 8 Curve Fitting, Regression, and Correlation
8.54. Prove that s2 Y.X s2 (1 Y r2) for least-squares regression.
By definition of the generalized correlation coefficient r, together with Problem 8.53, we have for either the linear or parabolic case r2 E[(Yest E[(Y # Y)2] # )2] Y 1 E[(Y E[(Y Yest)2] # Y)2] 1 s2 Y.X s2 Y
and the result follows at once. The relation also holds for higher-order least-squares curves.
8.55. Show that for the case of linear regression the correlation coefficient as defined by (45) reduces to that defined by (40).
The square of the correlation coefficient, i.e., the coefficient of determination, as given by (45) is in the case of linear regression given by (1) # But since Y (2) a # bX, E[(a bX # Y)2] E[b2(X s2 2 XY s s4 X X Then (1) becomes (3) r2 s2 XY s2 s2 X Y or r sXY sX sY # X)2] s2 XY s2 X b2E[(X # X)2] r2 E[(Yest E[(Y # Y)2] # Y)2] E[(a bX s2 Y # Y )2]
as we were required to show. (The correct sign for r is included in sXY.)
8.56. Refer to Table 8-21. (a) Find a least-squares regression parabola fitting the data. (b) Compute the regression values (commonly called trend values) for the given years and compare with the actual values. (c) Estimate the population in 1945. (d) Estimate the population in 1960 and compare with the actual value, 179.3. (e) Estimate the population in 1840 and compare with the actual value, 17.1. Table 8-21 Year U.S. Population (millions) 1850 23.2 1860 31.4 1870 39.8 1880 50.2 1890 62.9 1900 76.0 1910 92.0 1920 1930 1940 1950
105.7 122.8 131.7 151.1
Source: Bureau of the Census.
(a) Let the variables x and y denote, respectively, the year and the population during that year. The equation of a least-squares parabola fitting the data is (1) y a bx cx2
where a, b, and c are found from the normal equations ay (2) a xy
2 ax y
an a ax
b ax
c a x2 c a x3 c a x4
b a x2 b a x3
a a x2
It is convenient to locate the origin so that the middle year, 1900, corresponds to x 0, and to choose a unit that makes the years 1910, 1920, 1930, 1940, 1950 and 1890, 1880, 1870, 1860, 1850 correspond
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