chapter 11: regression - pepperdine university

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 11: Regression Section 11.2: The Method of Least Squares

11.2.1 b = 1 1 12 2

2

1 1

15(20,127.47) (249.8)(1,200.6)

15(4200.56) (249.8)

n n n

i i i ii i i

n n

i ii i

n x y x y

n x x

= = =

= =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ −

=−⎛ ⎞ ⎛ ⎞

−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ = 3.291

a = 1 1 1,200.6 3.291(249.8)

15

n n

i ii i

y b x

n= =

−−=

∑ ∑ = 25.234

Then y = 25.234 + 3.291x; y(18) = 84.5°F

11.2.2 b = 1 1 12 2

2

1 1

10(3973.35) (36.5)(1070)

10(204.25) (36.5)

n n n

i i i ii i i

n n

i ii i

n x y x y

n x x

= = =

= =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ −

=−⎛ ⎞ ⎛ ⎞

−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ = 0.9953

a = 1 1 1070 0.9953(36.5)

10

n n

i ii i

y b x

n= =

−−=

∑ ∑ = 103.367

11.2.3 1 1 12 2

2

1 1

9(24,628.6) (234)(811.3)

9(10,144) (234)

n n n

i i i ii i i

n n

i ii i

n x y x y

b

n x x

= = =

= =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ −

= =−⎛ ⎞ ⎛ ⎞

−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ = 0.8706

a = 1 1 811.3 0.8706(234)

9

n n

i ii i

y b x

n= =

−−=

∑ ∑ = 67.5088

160 Chapter 11: Regression


As an example of calculating a residual, consider x2 = 4. Then the corresponding residual is y2 − 2y = 71.0 − [67.5088 + 0.8706(4)] = 0.0098. The complete set of residuals, rounded to two

decimal places is

ix ˆi iy y−

0 −0.81 4 0.01

10 0.09 15 0.03 21 −0.09 29 0.14 36 0.55 51 1.69 68 −1.61

A straight line appears to fit these data. 11.2.4 In the first graph, all of the residuals are positive. The residuals in the second graph alternate

from positive to negative. Neither graph would normally occur from linear models. 11.2.5 The value 12 is too “far” from the data observed 11.2.6 The problem here is the gap in x values, leaving some doubt as to the x−y relationship.

11.2.7 b = 2

26(31,402) (360)(2256.6)

26(5365.08) (360)

−−

= 0.412

a = 2256.6 0.412(360)

26

− = 81.088

The least squares line is 81.088 + 0.412x. The plot of the data and least square line is:

Section 11.2: The Method of Least Squares 161


11.2.8 (a) The sums needed are 13

1

12.91ii

x=

=∑ , 13

2

1

15.6171ii

x=

=∑ , 13

1

25.29ii

y=

=∑ , 13

1

29.8762i ii

x y=

=∑

Then b = 2

13(29.8762) (12.91)(25.29)

13(15.6171) (12.91)

−−

= 1.703; a = 1 1.703

(25.29) (12.91) 0.25513 9

− =

The least squares line is y = 0.255 + 1.703x. (b) The residuals do not show a strong pattern, suggesting that a straight line fit is appropriate.



11.2.9 b = 2

9(7,439.37) (41.56)(1,416.1)

9(289.4222) (41.56)

−−

= 9.23

a = (1,416.1) 9.23(41.56)

9

− = 114.72

A linear relationship seems reasonable. 11.2.10 The x values spread evenly across their range, and the scatter diagram has a linear trend. Fitting

this data with a straight line seems appropriate.

11.2.11 b = 2

11(1141) (111)(100)

11(1277) (111)

−−

= 0.84

a = 1072 0.84(111)

11

− = 0.61

The least squares line is y = 0.61 + 0.84x. The residuals given in the table below are large relative

to the x values, which suggests that the linear fit is inadequate.

ix ˆi iy y−

7 −3.5 13 −1.5 14 −1.4

6 −0.7 14 2.6 15 1.8

4 3.0 8 2.7 7 −2.5 9 0.8

14 −1.4

Section 11.2: The Method of Least Squares 163


11.2.12 Using Cramer’s rule we obtain

b =

1

1 1 1 1 1

2

1 1 11

2

1 1

n

ii

n n n n n

i i i i i i ii i i i i

n n nn

i i iii i ii

n n

i ii i

n y

x x y n x y x y

n x x xn x

x x

=

= = = = =

= = ==

= =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=⎛ ⎞ ⎛ ⎞ ⎛ ⎞

−⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑

∑ ∑ ∑ ∑ ∑

∑ ∑ ∑∑

∑ ∑

which is essentially the form of b in Theorem 11.2.1. The first row of the matrix equation is

1 1

n n

i ii i

na x b y= =

⎛ ⎞+ =⎜ ⎟⎝ ⎠∑ ∑ . Solving this equation for a in terms of b gives the expression in Theorem

11.2.1 for a. 11.2.13 When x is substituted for x in the least-squares line equation, we obtain y = a + bx y bx bx y= − + =

11.2.14 The desired b is that value minimizing the equation L = 2

1

( )n

i ii

y bx=

−∑ .

1

2( )( )n

i i ii

dLy bx x

db =

= − −∑ , and setting dL

db = 0 gives 2

1

( )n

i i ii

x y bx=

−∑ = 0. The solution of this

equation is 1

2

1

n

i ii

n

ii

x y

b

x

=

=

=∑

∑.

11.2.15 For these data 1

n

i ii

d v=∑ = 95,161.2, and 2

1

n

ii

d=∑ = 2,685,141.

Then H = 1

2

1

n

i ii

n

ii

d v

d

=

=

∑

∑ =

95,161.2

2,685,141 = 0.03544.



11.2.16 (a) We seek the a value that minimizes the equation L = * 2

1

( )n

i ii

y a b x=

− −∑ .

*

1

2( )( 1)n

i ii

dLy a b x

da =

= − − −∑

Setting dL

da = 0 gives *

1

( )n

i ii

y a b x=

− −∑ = 0.

The solution of this equation is * *1 1

n n

i ii i

y x

a b y b xn n

= == − = −∑ ∑

(b) We seek the b that minimizes the equation

L = * 2

1

( )n

i ii

y a bx=

− −∑

*

1

2( )( )n

i i ii

dLy a bx x

db =

= − − −∑

Setting dL

db = 0 gives * 2

1

( )n

i i i ii

x y a x bx=

− −∑ = 0.

The solution of this equation is b =

*

1 1

2

1

n n

i i ii i

n

ii

x y a x

x

= =

=

−∑ ∑

∑.

11.2.17 b =

*

1 1

2

1

1513 100(45)

575.5

n n

i i ii i

n

ii

x y a x

x

= =

=

−−=

∑ ∑

∑ = −5.19, so y = 100 − 5.19x.

11.2.18 (a) The sums needed are 13

2

1i

i

x=∑ = 54,437,

13

1i i

i

x y=∑ = 3,329.4.

Then b = 3329.4

54,437 = 0.0612.

(b) y(120) = 0.612($120) = $7.34 million

11.2.19 L = 2

1

( sin )n

i i ii

y a bx c x=

− − −∑ . To find the a, b, and c, solve the following set of equations.

(1) dL

da=

1

2( sin )( 1)n

i i ii

y a bx c x=

− − − −∑ = 0 or

na + 1 1 1

sinn n n

i i ii i i

x b x c y= = =

⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∑ ∑ ∑

Section 11.2 The Method of Least Squares 165


(2) dL

db=

1

2( sin )( )n

i i i ii

y a bx c x x=

− − − −∑ = 0 or

2

1 1 1 1

sinn n n n

i i i i i ii i i i

x a x b x x c x y= = = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑

(3) dL

dc=

1

2( sin )( cos )n

i i i ii

y a bx c x x=

− − − −∑ = 0 or

1 1 1 1

cos cos (cos )(sin ) cosn n n n

i i i i i i ii i i i

x a x x b x x c y x= = = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑

11.2.20 (a) One choice for the model is y = bxae . Then ln y is linear with x. Using Theorem 11.2.1 on the pairs (xi, ln yi) gives

b = 1 1 12

2

1 1

ln lnn n n

i i i ii i i

n n

i ii i

n x y x y

n x x

= = =

= =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ = 1

2

10 137.97415 (35)(41.35720)

10(169) 35

n

i=

−

−

∑ = −0.14572

ln a = 1 1

lnn n

i ii i

y b x

n= =

−∑ ∑ =

41.35720 ( 0.14572)(35)

10

− − = 4.64574.

Then a rounded to three decimal places is e4.64574 = 104.140. The desired exponential fit is y = 104.140e−0.146x. This model fits the data well. However, note that the initial percentage by this model is 104.141, when we know it must be 100. This discrepancy suggests using

Question 11.2.16 where *a = 100. In this case,

b =

*

1 1

2

1

ln ln137.97415 4.60517(35)

169

n n

i i ii i

n

ii

x y a x

x

= =

=

⎛ ⎞− ⎜ ⎟⎝ ⎠ −=

∑ ∑

∑ = −0.13732.

This model is y = 100e−0.137x. (b) For the first model, the half life is the solution to 50 = 104.140e−0.146x, or ln(50/104.140) = −0.146x, so x = 5.025. For the second model, the half life is the solution to 0.5 = e−0.137x or ln 0.5 = −0.137x, so x = 5.059.

11.2.21 (a) To fit the model y = bxae , we note that ln y is linear with x. Then

2

11(126.33786) (66)(20.16825)

11(506) 66b

−=

− = 0.0484

20.16825 (0.0484)(66)

ln11

a−= = 1.5431

Then 1.5431a e= = 4.6791, and the model is 0.04844.6791 xy e= .



(b) The model predicts 0.0484(12)4.6791y e= = $8.363 trillion, short of the projected figure.

(c) The pattern of the residuals suggests a poor fit.

11.2.22 (a) 2

10(491.332) (55)(90.862)

10(385) 55b

−=

− = −0.1019

90.862 ( 0.1019)(55)

ln10

a− −= = 9.9467

Then 9.9467a e= = 15470. 6506, and the model is 0.101915470. 6506 xy e−=

(b) 0.1019(11)15470. 6506y e−= = $5043 (c) The exponential curve, which fits the data very well, predicts that a car 0 years old will have a value of $15471, but the selling price is $16,200. The difference $16,200 - $15471 = $729 could be considered initial depreciation.

11.2.23 b = 2

10(133.68654) (55)(22.78325)

10(385) 55

−−

= 0.1285

ln a = 158.58560 (0.12847)(190)

20

− = 6.70885

Then 6.70885a e= = 819.4, and the model is 0.1285819.4 xy e= .

Section 11.2 The Method of Least Squares 167


11.2.24 (a) If dy

dx = by, then

1 dy

y dx= b. Integrate both sides of the latter equality with respect to x:

1 dy

dx bdxy dx

=∫ ∫ , which implies that ln y = bx + C.

Now apply the function ex to both sides to get y = 1 1x xce e aeβ β= , where a = ec. (b) x on the abscissa, ln y on the ordinate

11.2.25 b = 2

7(0.923141) ( 0.067772)(7.195129)

7(0.0948679) ( 0.067772)

− −− −

= 10.538;

log a = 1 10.538

(7.195129)7 7

− (−0.067772) = 1.1299

Then a = 101.1299 = 13.487. The model is 13.487x10.538. The table below gives a comparison of the model values and the observed yi’s.

xj yj Model 0.98 25.000 10.901 0.74 0.950 0.565 1.12 200.000 44.522 1.34 150.000 294.677 0.87 0.940 3.109 0.65 0.090 0.144 1.39 260.000 433.516

11.2.26 (a) b = 1 1 12

2

1 1

log log log log

log log

n n n

i i ii i i

n n

i ii i

n x y x y

n x x

= = =

= =

⎛ ⎞ ⎛ ⎞⋅ − ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑

= 2

15(156.03811) (41.77441)(52.79857)

15(126.6045) 41.77441

−−

= 0.87644

log a = 1 1

log logn n

i ii i

y b x

n= =

⎛ ⎞− ⎜ ⎟⎝ ⎠∑ ∑

= 52.79857 0.87644(41.77441)

15

− = 1.07905

a = 11.99637, and the model is y = 11.99637x0.87644 (b) y(2500) = 11.99637(2500)0.87644 = 11,406

11.2.27 b = 2

4(36.95941) (11.55733)(12.08699)

4(34.80999) 11.55733

−−

= 1.43687

log a = 12.08699 1.43687(11.55733)

4

− = −1.12985

a = 10−1.12985 = 0.07416. The model is y = 0.07416x1.43687



11.2.28 b = 1 1 12

2

1 1

(1/ ) 1/

(1/ ) 1/

n n n

i i i ii i i

n n

i ii i

n x y x y

n x x

= = =

= =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ =

2

7(435.625) (8.01667)(169.1)

7(21.35028) 8.01667

−−

= 19.82681

a = 1 1

1/169.7 19.82681(8.01667)

7

n n

i ii i

y b x

n= =

⎛ ⎞− ⎜ ⎟⎝ ⎠ −=

∑ ∑ = 1.53643

One quarter mile = 0.25(5,280) = 1,320 feet. y(1.32) = 1.53643 + (19.82681)(1/1.32) = 16.557, or $16,557

11.2.29 (d) If y = 1

a bx+, then

1

y = a + bx and 1/y is linear with x.

(e) If y = x

a bx+, then

1 1a bxb a

y x x

+= = + , and 1/y is linear with 1/x.

(f) If y = 1 − /bx ae− , then 1 − y = /bx ae− , and 1

1 y− = /bx ae . Taking ln of both sides gives

1ln /

1bx a

y=

−. Taking ln again yields

ln 1

ln ln1

ay= − +

−b ln x, and ln

1ln

1 y− is linear with ln x.

11.2.30 Let y′ = 1055

lny

y

⎛ ⎞−⎜ ⎟⎝ ⎠

. We find the linear relationship between x and y′. The needed sums are

10

1i

i

x=∑ = 55,

102

1i

i

x=∑ = 385,

10

1i

i

y=

′∑ = −17.28636, 10

1i i

i

x y=

′∑ = −201.76600.

b = 2

10( 201.76600) 55( 17.28636)

10(385) 55

− − −−

= −1.29322

a = 17.28636 ( 1.29322)(55)

10

− − − = 5.38407

Section 11.3: The Linear Model 169


11.2.31 Let y′ = 0

lny

y

⎛ ⎞6 −⎜ ⎟⎝ ⎠

. We find the linear relationship between x and y′. The needed sums are

8

1i

i

x=∑ = 352,

82

1i

i

x=∑ = 16160,

8

1i

i

y=

′∑ = −2.39572, 1

n

i ii

x y=

′∑ = −194.88216.

b = 2

8( 194.88216) 352( 2.39572)

8(16160) 352

− − −−

= −0.13314

a = 2.39572 ( 0.13314)(352)

8

− − − = 5.55870

Section 11.3: The Linear Model

11.3.1 β1 = 2

4(93) 10(40.2)

4(30) 10

−−

= −1.5

β0 = (40.2) ( 1.5)(10)

4

− − = 13.8

Thus, y = 13.8 − 1.5x. t = 0

1 1

42

1

ˆ

/ ( )ii

s x x

β β

=

−

−∑ =

1.5 0

2.114 / 5

− − = −1.59.

Since −t.025,2 = −4.3027 < t = −1.59 < 4.3027 = t.025,2, accept H0.



11.3.2 (a) The radius of the confidence interval = .025,24 262

1

11.7884812.0639

380.464615( )i

i

st

x x=

=

−∑ = 1.247

The center is β1 = 0.412, and the confidence interval is (-0.835, 1.659) (b) Since 0 is in the confidence interval, we cannot reject H0 at the 0.05 level of significance. (c) See the solution to Question 11.2.7. The linear fit for 11 ≤ x ≤ 14 is not very good, suggesting a search for other contributing variables in that x range.

11.3.3 t = 0

1 1

152

1

3.291 0

3.829 / 40.55733/ ( )i

i

s x x

β β

=

− −=

−∑ = 5.47.

Since t = 5.47 > t0.005,13 = 3.0123, reject H0.

11.3.4 To minimize the width of the interval, we must maximize 2

1

( )n

ii

x x=

−∑ . To accomplish this, take

half of the xi to be 0 and half to be +5.

11.3.5 9

2 21

1

ˆVar( ) / ( )ii

x xβ σ=

= −∑ = 45/60 = 0.75. The standard deviation of 1ˆ 0.75β = = 0.866.

( ) 1 1

1 1

ˆ1.5ˆ 1.5

0.866 0.866P P

β ββ β

⎛ ⎞−⎜ ⎟− < = <⎜ ⎟⎝ ⎠

= P(|Z| < 1.73) for the standard normal random

variable Z. P(Z > 1.73) = 1 − 0.9582 = 0.0418, so P(⏐Z⏐ < 1.73) = 1 − 2(0.0418) = 0.9164

11.3.6 2 20 1 1 1

1 1

ˆ ˆ ˆ ˆ( ) [ ( ) ]n n

i i i ii i

Y x Y Y x xβ β β β= =

− − = − − −∑ ∑ = 21

1

ˆ[( ) ( )]n

i ii

Y Y x xβ=

− − −∑

= 2 2 21 1

1 1 1

ˆ ˆ( ) ( ) 2 ( )( )n n n

i i i ii i i

Y Y x x x x Y Yβ β= = =

− + − − − −∑ ∑ ∑

= 2 211 1

21 1 1

1

( )( )ˆ ˆ( ) ( ) 2 ( )( )

( )

n

i in n ni

i i i ini i i

ii

x x Y Y

Y Y x x x x Y Y

x x

β β=

= = =

=

− −− + − − − −

−

∑∑ ∑ ∑

∑

= 21

1 1

ˆ( ) ( )( )n n

i i ii i

Y Y x x Y Yβ= =

− − − −∑ ∑ = 2 21

1 1 1 1

1ˆn n n n

i i i i ii i i i

Y nY x Y x Yn

β= = = =

⎡ ⎤⎛ ⎞ ⎛ ⎞− − −⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

∑ ∑ ∑ ∑

= 2 21 1

1 1 1 1

1ˆ ˆn n n n

i i i i ii i i i

Y x Y nY x Yn

β β= = = =

⎛ ⎞ ⎛ ⎞− − + ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑ = ( )2

1 11 1 1

ˆ ˆn n n

i i i ii i i

Y x Y Y x Yβ β= = =

⎛ ⎞− − − ⎜ ⎟⎝ ⎠∑ ∑ ∑

= 21 0

1 1 1

ˆ ˆn n n

i i i ii i i

Y x Y Yβ β= = =

− −∑ ∑ ∑



11.3.7 The radius of the confidence interval is

2

1.05,7

2

1

(0.959) 101441.8946

9 40609 ( )

n

ii

n

ii

s x

t

x x

=

=

=

−

∑

∑ = 0.957.

The center of the interval is 0β = 67.508. The interval = (66.551, 68.465). 11.3.8 Since there is no reason to believe that radioactivity decreases cancer rates, the test should be H0: β1 = 0 versus H1: β > 0.

t = 0

1 1

92

1

ˆ 9.23 0

14.010 / 97.508/ ( )i

i

s x x

β β

=

− −=

−∑ = 6.51. Since t = 6.51 > t.05,7 = 1.8946, reject H0.

11.3.9 t = 0

1 1

112

1

ˆ 0.84 0

2.404 / 156.909/ ( )i

i

s x x

β β

=

− −=

−∑ = 4.38. Since t = 4.38 > t.025,9 = 2.2622, reject H0.

11.3.10 0 1 0 1 0 11 1 1

1 1 1 1( ) ( ) ( )

n n n

i i i ii i i

E Y E Y x x n x xn n n n

β β β β β β= = =

= = + = + = +∑ ∑ ∑

11.3.11 By Theorem 11.3.2, 0ˆ( )E β = β0, and

2

10

2

1

ˆVar( )

( )

n

ii

n

ii

x

n x x

σβ =

=

=−

∑

∑.

Now, 0 0 0ˆ ˆ( ) / Var( )β β β− is normal, so ( )/ 2 0 0 0 / 2

ˆ ˆ( ) / Var( )P z zα αβ β β− < − < = 1 − α.

Then the confidence interval is 0 / 2 0 0 / 2 0ˆ ˆ ˆ ˆ( Var( ), Var( )z zα αβ β β β− + , or

1 10 / 2 0 / 2

2 2

1 1

ˆ ˆ,

( ) ( )

n n

i ii i

n n

i ii i

x x

z z

n x x n x x

α α

σ σβ β= =

= =

⎛ ⎞⎜ ⎟⎜ ⎟− +⎜ ⎟⎜ ⎟− −⎜ ⎟⎝ ⎠

∑ ∑

∑ ∑

11.3.12 Refer to the four assumptions in the subsection “A Special Case”. (1) Normality of the data cannot be assessed from the scatter plot (2) The standard deviation does not appear to be the same for the three data sets. (3) The means could be collinear (4) Independence of the underlying random variables cannot be assessed from the scatter plot.



11.3.13 Reject the null hypothesis if the statistic is < 2 2 2 2/ 2, 2 .025,22 1 / 2, 2 .975,2210.982 or >n nα αχ χ χ χ− − −= = =

= 36.781. The observed chi square is 2

20

( 2) (24 2)(18.2)

12.6

n s

σ− −

= = 31.778, so do not reject H0.

11.3.14 Case Study 11.3.1 provides the value of s2 = 2181.66. Then the confidence interval for σ2 is

2 2

2 2 2 21 / 2, 2 / 2, 2 .95,19 .05,19

( 2) ( 2) (19)(2181.66) (19)(2181.66) 41,451.54 41,451.54, , ,

30.144 10.177n n

n s n s

α αχ χ χ χ− − −

⎛ ⎞ ⎛ ⎞− − ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

= (1375.12, 4097.22) 11.3.15 The value of is given to be 2.31, so s2 = 5.3361. Then the confidence interval for σ2 is

2 2

2 2 2 21 / 2, 2 / 2, 2 .95,7 .05,7

( 2) ( 2) (6)(5.3361) (6)(5.3361) 32.0166 32.0166, , ,

14.067 2.167n n

n s n s

α αχ χ χ χ− − −

⎛ ⎞ ⎛ ⎞− − ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

= (2.276, 14.775)

11.3.16 (a) The radius of the confidence interval is 2

.25,16 182

1

1 ( )

( )ii

x xt s

nx x

=

−+

−∑

= 2.1199(0.202)21 (14.0 15.0)

18 96.38944

−+ = 0.110

The center is 0 1ˆ ˆy xβ β= + = −0.104 + 0.988(14.0) = 13.728.

The confidence interval is (13.62, 13.84). (b) The radius of the prediction interval is

2

.025,16 182

1

1 ( )1

( )ii

x xt s

nx x

=

−+ +

−∑ = 2.1199(0.202)

21 (14.0 15.0)1

18 96.38944

−+ + = 0.442

The center is 0 1ˆ ˆy xβ β= + = −0.104 + 0.988(14.0) = 13.728. The confidence interval is

(13.29, 14.17).

11.3.17 The radius of the 95% confidence interval is 21 (2.750 2.643)

2.0687(0.0113)25 0.0367

−+ = 0.0139.

The center is 0 1ˆ ˆy xβ β= + = 0.308 + 0.642(2.750) = 2.0735. The confidence interval is

(2.0596, 2.0874)

11.3.18 The radius of the 99% confidence interval is 21 (2500 2148.095)

2.8609(46.708)21 13056523.81

−+

= 31.932. The center is 0 1ˆ ˆy xβ β= + = 15.771 + 0.060(2500) = 165.771. The confidence interval

is (133.839, 197.703). If the official were interested in a specific country, the prediction interval would be of more use.



11.3.19 The radius of the 95% confidence interval for ( |102)E Y is 21 (102 89.75)

2.1448(19.601)16 901

−+

= 20.12. The center is 0 1ˆ ˆy xβ β= + = -107.91 + 2.96(102) = 194.01. The 95% confidence

interval is (194.01 - 20.12, 194.01 + 20.12) = (173.89, 214.13)

The radius of the prediction interval is 21 (102 89.75)

2.1448(19.601) 116 901

−+ + = 46.61

The prediction interval is (194.01 – 46.61, 194.01 + 46.61) = (147.40, 240.62) Thus, the prediction interval contains the Harvard median salary of 215,000, while the confidence

interval does not.

11.3.20 The radius of the 95% confidence interval for ( | 9.00)E Y is 21 (9 4.618)

2.3646(14.010)9 97.507

−+

= 18.387. The center is 0 1ˆ ˆy xβ β= + = 114.72 + 9.23(9) = 197.79. The confidence interval is

(197.79 – 18.387, 197.79 + 18.387) = (179.40, 216.18).

The radius of the 95% prediction interval is 21 (9 4.618)

2.3646(14.010) 19 97.507

−+ + = 37.888.

The prediction interval is (197.79 – 37.888, 197.79 + 37.888) = (159.90, 235.68)

11.3.21 The test statistic is t = *

1 1

6 82 * * 2

1 1

ˆ ˆ

1 1

( ) ( )i ii i

s

x x x x

β β

= =

−

+− −∑ ∑

, where s = 5.983 13.804

6 8 4

++ −

= 1.407.

Then t = 0.606 1.07

1 11.407

31.33 46

−

+ = −1.42. Since the observed ratio is not less than −t.05,10 = −1.8125

the difference in slopes can be ascribed to chance. These data do not support further investigation.

11.3.22 s = 2 2 2 21 1(3 4 * ) [3(0.9058 ) 4(1.2368 )]

3 4 7s s+ = +

+ = 1.1071.

Then t = 3.4615 2.7373

1 11.1071

26 39.3333

− +

+ = −2.59. Since t = −2.59 < −t.025,7 = −2.3646, reject H0.



11.3.23 The form given in the text is2

2

2

1

1 ( )ˆVar( )

( )n

ii

x xY

nx x

σ

=

⎡ ⎤⎢ ⎥−⎢ ⎥= +⎢ ⎥

−⎢ ⎥⎣ ⎦

∑. Putting the sum in the brackets

over a least common denominator gives

2 22

1

2 2

1 1

( ) ( )1 ( )

( ) ( )

n

ii

n n

i ii i

x x n x xx x

nx x n x x

=

= =

− + −−+ =− −

∑

∑ ∑

=

2 2 2 2

1

2

1

( 2 )

( )

n

ii

n

ii

x nx n x x xx

n x x

=

=

− + + −

−

∑

∑ =

2 2 2 2

1 1 1

2 2

1 1

2 2

( ) ( )

n n n

i i ii i i

n n

i ii i

x nx nxx x nx x x

n x x n x x

= = =

= =

+ − + −=

− −

∑ ∑ ∑

∑ ∑

=

2

1

2

1

( )

( )

n

ii

n

ii

x x

n x x

=

=

−

−

∑

∑. Thus

2 2

1

2

1

( )ˆVar( )

( )

n

iin

ii

x x

Y

n x x

σ=

=

−=

−

∑

∑.

11.3.24 2 20 1

1 1

ˆ ˆˆ( ) ( )n n

i ii i

Y Y x Yβ β= =

− = + −∑ ∑ =

2 2 2 21 1 1 1 1

1 1 1

ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )n n n

i i ii i i

Y x x Y x x x xβ β β β β= = =

− + − = − = −∑ ∑ ∑ .

An application of Equation 11.3.3 completes the proof. Section 11.4: Covariance and Correlation

11.4.1 E(XY) = 1 2(1) 2 2(1) 1 2(3) 2 2(3)

1 2 3 622 22 22 22

+ + + ++ + + = 80/22 = 40/11

E(X) = 10 12

1 222 22

+ = 34/22=17/11

E(X 2) = 10 12

1 422 22

+ = 58/22=29/11

E(Y) = 7 15

1 322 22

+ = 52/22=26/11

E(Y 2) = 7 15

1 922 22

+ = 142/22 = 71/11

Cov(XY) = 40/11 − (17/11)(26/11) = −2/121 Var(X) = 29/11 − (17/11)2 = 30/121 Var(Y) = 71/11 − (26/11)2 = 105/121

ρ(X,Y) = 2 /121 2 2

30 /121 105/121 3150 15 14

− − −= = = -0.036

Section 11.4: Covariance and Correlation 175


11.4.2 E(XY) =

12 3 21 1 1

0 0 00

( )2 3 6 6

x x x xxy x y dydx dx

⎛ ⎞+ = + = +⎜ ⎟⎝ ⎠∫ ∫ ∫ = 1/3.

fX(x) = x + 1

2, so E(X) =

1

0

1

2x x dx⎛ ⎞+⎜ ⎟⎝ ⎠∫ = 7/12

E(X 2) = 1

2

0

1

2x x dx⎛ ⎞+⎜ ⎟⎝ ⎠∫ = 5/12. Var(X) = 5/12 − (7/12)2 = 11/144.

By symmetry Y has the same moments, so Cov(X, Y) = 1/3 − (7/12)(7/12) = −1/144. Then

ρ = ( )2

1/144

11 /12

− = −1/11.

11.4.3 3

0( ) 8 4

x

Xf x xy dy x= =∫ . 1

3

0( ) (4 ) 4 / 5E X x x dx= =∫ .

12 2 3

0( ) (4 ) 2 / 3E X x x dx= =∫ .

Var(X) = 2/3 − (4/5)2 = 2/75

1

3( ) 8 4( )Yy

f y xy dx y y= = −∫ . 1

4

0( ) 4 ( ) 8/15E Y y y dy= − =∫ .

12 3 5

0( ) 4 ( ) 1/ 3E Y y y dy= − =∫ .

Var(Y) = 1/3 − (8/15)2 = 11/225.

1

2 2

0 0( ) 8 4 / 9

xE XY x y dy dx= =∫ ∫ .

4 4 8 8Cov( , )

9 5 15 450X Y = − ⋅ =

ρ = 8 / 450

2 / 75 11/ 225 = 0.492

11.4.4 E(XY) = 1 1 1 1

2 3 82 8 4 8

⎛ ⎞+ + +⎜ ⎟⎝ ⎠ = 3

E(X) = 3 1

1 24 4+ = 5/4

E(X 2) = 3 1

1 44 4+ = 7/4

E(Y) = 1 1 1 1

1 2 3 48 2 4 8+ + + = 19/8

E(Y 2) = 1 1 1 1

1 4 9 168 2 4 8+ + + = 51/8

Cov(XY) = 3 − (5/4)(19/8) = 1/32 Var(X) = 7/4 − (5/4)2 = 3/16 Var(Y) = 51/8 − (19/8)2 = 47/64

ρ(X, Y) = 1/ 32 1

3/16 47 / 64 3 47= = 0.0842

11.4.5 ρ(a + bX, c + dY) = Cov( , )

Var( )Var( )

a bX c dY

a bX c dY

+ ++ +

= 2 2

Cov( , )

Var( ) Var( )

bd X Y

b X d Y, the equality in the

numerator stemming from Question 3.9.14. Since b > 0, d > 0, this last expression is Cov( , ) Cov( , )

( , )X Y X Y

bd X Y X YX Y

bdρ

σ σ σ σ= = .



11.4.6 To find ρ(X, Y), we need the first four moments of X.

E(X) = 1 1

1 1 1 ( 1) 1

2 2

n n

k k

n n nk k

n n n= =

+ +⎛ ⎞ = = =⎜ ⎟⎝ ⎠∑ ∑

E(X 2) = 2 2

1 1

1 1 1 ( 1)(2 1) ( 1)(2 1)

6 6

n n

k k

n n n n nk k

n n n= =

+ + + +⎛ ⎞ = = =⎜ ⎟⎝ ⎠∑ ∑

E(X 3) = 2 2 2

3

1

1 1 ( 1) ( 1)

4 4

n

k

n n n nk

n n=

+ += =∑

E(X 4) = 2 2

4

1

1 1 ( 1)(2 1)(3 3 1) ( 1)(2 1)(3 3 1)

30 30

n

k

n n n n n n n n nk

n n=

+ + + − + + + −= =∑

Note: We have already encountered the sums of the integers to the first and second powers. The

sums for the third and fourth powers can be found in such books of mathematical tables as the CRC Standard Mathematical Tables and Formulae.

Cov(X, Y) = E(XY) − E(X)E(Y) = E(X 3) − E(X)E(X 2)

= 2 2( 1) ( 1) ( 1)(2 1) ( 1) ( 1)

4 2 6 12

n n n n n n n+ + + + + −− =

Var(X) = E(X 2) − E(X)2 = 2( 1)(2 1) ( 1) ( 1)( 1)

6 4 12

n n n n n+ + + + −− =

Var(Y) = E(Y 2) − E(Y)2 = E(X 4) − E(X 2)2 = 2 2 2( 1)(2 1)(3 3 1) ( 1) (2 1)

30 36

n n n n n n+ + + − + +−

= 2( 1)(2 1)(8 3 11)

180

n n n n+ + + − =

( 1)(2 1)( 1)(8 11)

180

n n n n+ + − +

ρ(X, Y) =

2( 1) ( 1)15( 1)12

( 1)( 1) ( 1)(2 1)( 1)(8 11) (2 1)(8 11)12 180

n nn

n n n n n n n n

+ −+

=+ − + + − + + +

115 1

15( 1) 15lim ( , ) lim lim

4(2 1)(8 11) 1 112 8

n n n

n nX Y

n n

n n

ρ→∞ →∞ →∞

⎛ ⎞+⎜ ⎟⎝ ⎠+= = =

+ + ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

11.4.7 (a) Cov(X + Y, X − Y) = E[(X + Y)(X − Y)] − E(X + Y)E(X − Y) = E[X 2 − Y 2] − (µX + µY)(µX − µY)

= 2 2 2 2( ) ( )X YE X E Yµ µ− − −

(b) ρ(X + Y) = Cov( , )

Var( )Var( )

X Y X Y

X Y X Y

+ −+ −

.

By part (a) Cov(X + Y, X − Y) = Var(X) − Var(Y). Var(X + Y) = Var(X) + Var(Y) + 2Cov (X, Y) = Var(X) + Var(Y) + 0. Similarly, Var(X − Y) = Var(X) + Var(Y). Then

Var( ) Var( ) Var( ) Var( )

( )Var( ) Var( )(Var( ) Var( ))(Var( ) Var( ))

X Y X YX Y

X YX Y X Yρ − −+ = =

++ +

Section 11.4: Covariance and Correlation 177


11.4.8 Multiply the numerator and denominator of Equation 11.4.1 by n2 to obtain

R = 1 1 1 1 1 1

2 22 2 2 2

1 1 1 1 1 1

( ) ( )

n n n n n n

i i i i i i i ii i i i i i

n n n n n n

i i i i i ii i i i i i

n X Y X Y n X Y X Y

n X X n Y Y n X X n Y Y

= = = = = =

= = = = = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=⎛ ⎞ ⎛ ⎞− − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑

11.4.9 By Equation 11.4.2 r = 1 1 1

2 2

2 2

1 1 1 1

n n n

i i i ii i i

n n n n

i i i ii i i i

n x y x y

n x x n y y

= = =

= = = =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ ∑ ∑

=

2

2

1 1 1 1 12 2 2

2 2 2

1 1 1 1 1 1

n n n n n

i i i i i ii i i i i

n n n n n n

i i i i i ii i i i i i

n x y x y n x x

n x x n x x n y y

= = = = =

= = = = = =

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⋅⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∑ ∑ ∑ ∑ ∑

∑ ∑ ∑ ∑ ∑ ∑

=

2

2

1 11 2

2

1 1

ˆ

n n

i ii i

n n

i ii i

n x x

n y y

β = =

= =

⎛ ⎞− ⎜ ⎟⎝ ⎠

⎛ ⎞− ⎜ ⎟⎝ ⎠

∑ ∑

∑ ∑

11.4.10 1 1 1

2 2

2 2

1 1 1 1

n n n

i i i ii i i

n n n n

i i i ii i i i

n x y x y

r

n x x n y y

= = =

= = = =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=⎛ ⎞ ⎛ ⎞

− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ ∑ ∑

= ( )2 2

21(7,319,602) (45,110)(3042.2)

21(109,957,100) 45,110 21(529,321.58) (3042.2)

−

− − = 0.730

Since 2 2(0.730) 0.5329r = = , we can say that 53.3% of the variability is explained by cigarette consumption.

11.4.11 r = 1 1 1

2 2

2 2

1 1 1 1

n n n

i i i ii i i

n n n n

i i i ii i i i

n x y x y

n x x n y y

= = =

= = = =

⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑ ∑

∑ ∑ ∑ ∑

= 2 2

12(480,565) (4936)(1175)

12(3,071,116) (4936) 12(123,349) (1175)

−

− − = −0.030.

The data do not suggest that altitude affects home run hitting.



11.4.12 r = 2 2

10(325.08) (123.1)(25.80)

10(1529.63) (123.1) 10(74.00) (25.80)

−

− − = 0.726

11.4.13 r = 2 2

17(4,759,470) (7,973)(8,517)

17(4,611,291) (7,973) 17(5,421,917) (8,517)

−

− − = 0.762. The amount of variation

attributed to the linear regression is r2 = (0.762)2 = 0.581, or 58.1%.

11.4.14 r = 2 2

18(221.37) ( 0.9)(160.2)

18(92.63) ( 0.9) 18(6437.68) (160.2)

− −

− − − = 0.337. The amount of variation attributed to

the linear regression is r2 = (0.337)2 = 0.114, or 11.4%. Section 11.5: The Bivariate Normal Distribution 11.5.1 Y is a normal random variable with E(Y) = 6 and Var(Y) = 10. Then P(5 < Y < 6.5) =

5 6 6.5 6

10 10P Z⎛ ⎞− −

< <⎜ ⎟⎝ ⎠ = P(−0.32 < Z < 0.16) = 0.5636 − 0.3745 = 0.1891. By Theorem 11.5.1,

Y|2 is normal with E(Y|2) = µY +

110

2(2 ) 6 (2 3)2

YX

X

ρσ µσ

− = + − = 5.209

Var(Y|2) = 2 2(1 ) Yρ σ− = (1 − 0.25)10 = 7.5, so the standard deviation of Y is 7.5 = 2.739.

P(5 < Y|2 < 6.5) = 5 5.209 6.5 5.209

2.739 2.739P Z

− −⎛ ⎞< <⎜ ⎟⎝ ⎠= P(−0.08 < Z < 0.47) = 0.6808 − 0.4681

= 0.2127

11.5.2 (a) The lemma on page 424 can be used to show that X and Y Xρ− are bivariate normal. Thus it

suffices to show that Cov(X, Y − ρX) = 0.

Cov(X, Y − ρX) = E[X(Y − ρX)] − E(X)E(Y − ρX) = E(XY) − ρE(X 2) − E(X)E(Y) + ρE(X)2

= Cov(X, Y ) − ρVar(X) = Cov(X, Y) − Cov( , )

Var( )Var( ) Var( )

X YX

X Y

= Cov(X, Y) − Cov(X, Y) = 0, since Var(X) = Var(Y) . (b) The lemma on page 424 can be used to show that X Y+ and X Y− are bivariate normal. By Question 11.4.7, Cov(X + Y, X − Y) = Var( ) Var( )X Y− . Since the variances are equal,

Cov(X + Y, X − Y) = 0, and the two variables are independent.

11.5.3 (a) fX+Y(t) = 2 222

1 1 1exp ( ) 2 ( )

2 12 1t y t y y y dyρ

ρπ ρ

∞

−∞

⎧ ⎫⎛ ⎞⎪ ⎪⎡ ⎤− − − − +⎨ ⎬⎜ ⎟ ⎣ ⎦⎝ ⎠−⎪ ⎪− ⎩ ⎭∫

The expression in the brackets can be expanded and rewritten as t2 + 2(1 + ρ)y2 − 2t(1 + ρ)y = t2 + 2(1 + ρ)[y2 − ty]

Section 11.5: The Bivariate Normal Distribution 179


= t2 + 2(1 + ρ)2

2 21(1 )

4 2

ty ty tρ⎡ ⎤

− + − +⎢ ⎥⎣ ⎦

= 2 212(1 )( / 2)

2t y t

ρ ρ− + + − . Placing this

expression into the exponent gives

fX+Y(t) = 2 2

2 2

1 1 1 1 12(1 )( / 2)

2 2 21 1

2

1

2 1

t y t

e e dy

ρ ρρ ρ

π ρ

⎛ ⎞ ⎛ ⎞−− − + −∞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− −

−∞− ∫

= fX+Y(t) =

2 21 1 ( / 2)

2 2(1 ) 2 (1 ) / 2

2

1

2 1

t y t

e e dyρ ρ

π ρ

⎛ ⎞ ⎛ ⎞−− −⎜ ⎟ ⎜ ⎟∞+ +⎝ ⎠ ⎝ ⎠

−∞− ∫ .

The integral is that of a normal pdf with mean t/2 and σ2 = (1 + ρ)/2.

Thus, the integral equals 2 (1 ) / 2 (1 )π ρ π ρ+ = + .

Putting this into the expression for fX+Y gives

fX+Y(t) =

21

2 2(1 )1

2 2(1 )

t

eρ

π ρ

⎛ ⎞− ⎜ ⎟+⎝ ⎠

+, which is the pdf of a normal variable with

µ = 0 and σ2 = 2(1 + ρ).

(b) E(X + Y) = cµX + dµY; Var(X + Y) = 2 2 2 2 2 ( , )X Y X Yc d cd X Yσ σ σ σ ρ+ +

11.5.4 E(Y|55) = µY + 0.6 2.6

(55 ) 11 (55 56)1.2

YX

X

ρσ µσ

− = + − = 10.117

Var(Y|55) = 2 2 2(1 ) (1 0.6 )2.6Yρ σ− = − = 1.664, so the standard deviation of Y is 1.664 = 1.290.

P(10 ≤ Y ≤ 10.5|x = 55) = 10 10.117 10.5 10.117

1.290 1.290P Z

− −⎛ ⎞≤ ≤⎜ ⎟⎝ ⎠ = P(−0.09 ≤ Z ≤ 0.30)

= 0.6179 − 0.4641 = 0.1538

The mean of Y also = 10.117. However, the standard deviation is 1.290 / 4 = 0.645. Then

(10.5 11 55)P Y x≤ ≤ = = 10.5 10.117 11 10.117

0.645 0.645P Z

− −⎛ ⎞≤ ≤⎜ ⎟⎝ ⎠ = P(0.59 ≤ Z ≤ 1.37)

= 0.9147 − 0.7224 = 0.1923

11.5.5 E(X) = E(Y) = 0; Var(X) = 4; Var(Y) = 1; ρ(X, Y) = 1/2; k = 1/(2 3)π

11.5.6 −(ax2 − 2uxy + by2) = 2 2

2 2 2

1 12

2 1 X YX Y

x x y yρσ σρ σ σ

⎛ ⎞⎛ ⎞− − +⎜ ⎟⎜ ⎟⎝ ⎠− ⎝ ⎠

so we get the following equations:

a = 2 2 2 2

1 1 1 1 1 1;

2 21 1X Y

bρ σ ρ σ

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− −

; and u = 2

1 1

2 1 X Y

ρσ σρ

⎛ ⎞⎜ ⎟⎝ ⎠−

. From the first two

equations we obtain 212 (1 )

X

a ρσ

= − and 212 (1 )

Y

b ρσ

= − . Substituting these values in the

expression for u gives u = 2 22

1 12 (1 ) 2 (1 )

2 1a b a bρ ρ ρ ρ

ρ⎛ ⎞

− − =⎜ ⎟⎝ ⎠−, or

u

a bρ = .



From this equation, we get the only conditions on the parameters a, b, and u:

1u

a bρ= ≤ or |u| ≤ a b

Then we can solve for 2Xσ and 2

Yσ in terms of a, b, and ρ : 22

1 1 1

2 1X aσ

ρ⎛ ⎞

= ⎜ ⎟⎝ ⎠−;

22

1 1 1

2 1Y bσ

ρ⎛ ⎞

= ⎜ ⎟⎝ ⎠−

11.5.7 r = −0.453. T18 = 2 2

2 18( 0.453)

1 1 ( 0.453)

n r

r

− −=

− − − = −2.16

Since −t.005,18 = −2.8784 < T18 = −2.16 < 2.8784 = t.005,18, accept H0.

11.5.8 r = 2 2

14(710,499) (2458)(4097)

14(444,118) (2458) 14(1,262,559) (4097)

−

− − = −0.312

T12 = 2 2

2 12( 0.312)

1 1 ( 0.312)

n r

r

− −=

− − − = −1.14

Since −t.025,12 = −2.1788 < T12 = −1.14 < 2.1788 = t.025,12, accept H0. The data sets appear to be

independent.

11.5.9 From Question 11.4.11, r = −0.030. T10 = 2

10( 0.030)

1 ( 0.030)

−

− − = −0.09.

Since −t.025,10 = −2.2281 < T10 = −0.09 < 2.2281 = t.025,10, accept H0.

11.5.10 From Question 11.4.13, r = 0.762. T15 = 2

15(0.762)

1 (0.762)− = 4.56.

Since T15 = 4.56 > 2.9467 = t.005,15 reject H0.

11.5.11 r = 2 2

10(1349.66) (18.33)(738)

10(34.1267) (18.33) 10(54756) (738)

−

− − = −0.249

Since the correlation coefficient is negative, there is no need to test 1 : 0H ρ > .

To see if there is any effect at all, one could test against 1 : 0H ρ ≠ . In that case the test statistic

is 2

8( 0.249)0.727

1 ( 0.249)

−=

− −. Since the test statistic lies between .05,8 1.8595t− = − and

.05,8 1.8595t = , do not reject 0H

chapter 11: regression - pepperdine university

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