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MT427 Problem Set 8

Chapter 9

Section 9.

9.2.1 t =

65.2 75.

p 1/^ 1/^ 13.9 1/ 9^ 1/

x y

s n m

!!

"

#

= !1.

Since! t (^) .05,19 = !1.7291 < t = !1.68, accept H 0.

9.2.2 sp =

2 2 2 2 ( 1) ( 1) 3(267 ) 3(224 )

n s X m sY

n m

! #!

"

#! #! 2

= 246.

t =

1133.0 1013.

p 1/^ 1/^ 246.44 1/ 4^ 1/ 4

x y

s n m

!!

"

#

= 0.

Since! t (^) .025,6 = !2.4469 < t = 0.69 < t .025,6 = 2.4469, accept H 0.

9.2.3 sp =

2 2 ( 2) ( 1) 5(167.568 11(52.072)

n s X m sY

n m

! #!

"

#! #!

= 9.39.

t =

28.6 12.

p 1/^ 1/^ 9.39 1/ 6^ 1/

x y

s n m

!!

"

#

= 3.

Since t = 3.37 > 2.9208 = t .005,16, reject H (^) 0.

9.2.4 sp =

2 2 5(15.1 ) 8(8.1 )

#!

= 11.

t =

70.83 79.

11.317 1/ 6 1/ 9

!

= !1.

Since! t (^) .005,13 = !3.0123 < t = !1.43 < t (^) .005,13 = 3.0123, accept H 0.

9.2.5 sp =

7(7.169) 7(10.304)

#!

= 2.

t =

11.2 9.

2.956 1/ 8 1/ 8

!

= 0.

Since t = 0.96 < t (^) .05,14 = 1.7613, accept H 0.

9.2.6 sp =

2 2 30(1.469 ) 56(1.350 )

#!

= 1.

t =

3.10 2.

1.393 1/ 31 1/ 57

!

= 2.

Since t = 2.16 > t (^) .025,86=1.9880, reject H 0.

Chapter 9 137

9.2.1 t =

s (^) p 1/ n 1/ m 13.9 1/ 9 1/

"

#

= !1.

Since! t (^) .05,19 = !1.7291 < t = !1.68, accept H 0.

9.2.2 sp =

2 2 2 2 ( 1) ( 1) 3(267 ) 3(224 )

n s X m sY

n m

! #!

"

#! #! 2

= 246.

t =

1133.0 1013.

p 1/^ 1/^ 246.44 1/ 4^ 1/ 4

x y

s n m

!!

"

#

= 0.

Since! t (^) .025,6 = !2.4469 < t = 0.69 < t .025,6 = 2.4469, accept H 0.

9.2.3 sp =

2 2 ( 2) ( 1) 5(167.568 11(52.072)

n s X m sY

n m

! #!

"

#! #!

= 9.39.

t =

28.6 12.

p 1/^ 1/^ 9.39 1/ 6^ 1/

x y

s n m

!!

"

#

= 3.

Since t = 3.37 > 2.9208 = t .005,16, reject H (^) 0.

9.2.4 sp =

2 2 5(15.1 ) 8(8.1 )

#!

= 11.

t =

70.83 79.

11.317 1/ 6 1/ 9

!

= !1.

Since! t (^) .005,13 = !3.0123 < t = !1.43 < t (^) .005,13 = 3.0123, accept H 0.

9.2.5 sp =

7(7.169) 7(10.304)

#!

= 2.

t =

11.2 9.

2.956 1/ 8 1/ 8

!

= 0.

Since t = 0.96 < t (^) .05,14 = 1.7613, accept H 0.

9.2.6 sp =

2 2 30(1.469 ) 56(1.350 )

#!

= 1.

t =

3.10 2.

1.393 1/ 31 1/ 57

!

= 2.

Since t = 2.16 > t (^) .025,86=1.9880, reject H 0.

Chapter 9 137

p n # m! 2 4 # 4! 2

t =

1133.0 1013.

p 1/^ 1/^ 246.44 1/ 4^ 1/ 4

x y

s n m

!!

"

#

= 0.

Since! t (^) .025,6 = !2.4469 < t = 0.69 < t .025,6 = 2.4469, accept H 0.

9.2.3 sp =

2 2 ( 2) ( 1) 5(167.568 11(52.072)

n s (^) X m sY

n m

! #!

"

#! #!

= 9.39.

t =

28.6 12.

p 1/^ 1/^ 9.39 1/ 6^ 1/

x y

s n m

!!

"

#

= 3.

Since t = 3.37 > 2.9208 = t .005,16, reject H (^) 0.

9.2.4 sp =

2 2 5(15.1 ) 8(8.1 )

#!

= 11.

t =

70.83 79.

11.317 1/ 6 1/ 9

!

= !1.

Since! t (^) .005,13 = !3.0123 < t = !1.43 < t (^) .005,13 = 3.0123, accept H 0.

9.2.5 sp =

7(7.169) 7(10.304)

#!

= 2.

t =

11.2 9.

2.956 1/ 8 1/ 8

!

= 0.

Since t = 0.96 < t (^) .05,14 = 1.7613, accept H 0.

9.2.6 sp =

2 2 30(1.469 ) 56(1.350 )

#!

= 1.

t =

3.10 2.

1.393 1/ 31 1/ 57

!

= 2.

Since t = 2.16 > t (^) .025,86=1.9880, reject H 0.

Chapter 9 137

9.2.15 For the data given, x = 545.45, sX = 428, and y = 241.82, sY = 183. Then

t = 2 2 2 2

545.45 241.

X /^ Y /^ 428 /11^ 183 /

x y

s n s m

!!

"

#

= 2.

The degrees of freedom associated with this statistic is the greatest integer in

$ %

$ % $ %

(^2 2 2 2 2 )

2 2 2 2 2 2 2 2

/ / (428 /11 183 /11)

/ ( 1) / /( 1) (428 /11) /10^ (183 /11) /

X Y

X Y

s n s m

s n n s m m

#

"

! #!

= 13.5.

The greatest integer is 13. Since t = 2.16 > t (^) .05,13 = 1.7709, reject H 0.

9.2.16 Decreasing the degrees of freedom also decreases the power of the test.

9.2.17 a) The sample standard deviation for the first data set is approximately 3.15; for the

second, 3.29. These seem close enough to permit the use of Theorem 9.2.2.

b) Intuitively, the states with the comprehensive law should have fewer deaths. However,

the average for these data is 8.1, which is larger than the average of 7.0 for the states

with a more limited law.

Section 9.

9.3.1 a) The critical values are F .025,25,4 and F (^) .975,25,4. These values are not tabulated, but in this

case, we can approximate them by F .025,24,4 = 0.296 and F .975,24,4 = 8.51. The observed

F = 86.9/73.6 = 1.181. Since 0.296 < 1.181 < 8.51, we can accept H 0 that the

variances are equal.

b) Yes, we can use Theorem 9.2.2, since we have no reason to doubt that the variances are

equal.

9.3.2 The observed F = 0.

2 /0.

2 = 2.935. Since F .025,9,9 = 0.248 < 2.935 < 4.03 = F .975,9,9, we

can accept H 0 that the variances are equal.

9.3.3 a) The critical values are F (^) .025,19,19 and F .975,19,19. These values are not tabulated, but in this

case, we can approximate them by F .025,20,20 = 0.406 and F .975,20,20 = 2.46. The observed

F = 2.41/3.52 = 0.685. Since 0.406 < 0.685 < 2.46, we can accept H 0 that the variances

are equal.

b) Since t = 2.662 > t (^) .025,38 = 2.0244, reject H 0.

9.3.4 The observed F = 3.

2 /5.

2 = 0.315. Since F .025,9,9 = 0.248 < 0.315 < 4.03 = F .975,9,9, we can

accept H 0 that the variances are equal.

9.3.5 F = 0.

2 /0.

2 = 0.292. Since F .025,9,9 = 0.248 < 0.292 < 4.03 = F .975,.9,9, accept H 0.

9.3.6 The observed F = 398.75/274.52 = 1.453. Let & = 0.05. The critical values are F .025,13,11 and

F (^) .975,13,11. These values are not in Table A.4, so approximate them by F .025,12,11 = 0.301 and

F .975,12,11 = 3.47. Since 0.301 < 1.453 < 3.47, accept H 0 that the variances are equal. Theorem

9.2.2 is appropriate.

140 Chapter 9

9.2.15 For the data given, x = 545.45, sX = 428, and y = 241.82, sY = 183. Then

t = 2 2 2 2

545.45 241.

X /^ Y /^ 428 /11^ 183 /

x y

s n s m

!!

"

#

= 2.

The degrees of freedom associated with this statistic is the greatest integer in

$ %

$ % $ %

(^2 2 2 2 2 )

2 2 2 2 2 2 2 2

/ / (428 /11 183 /11)

/ ( 1) / /( 1) (428 /11) /10^ (183 /11) /

X Y

X Y

s n s m

s n n s m m

#

"

! #!

= 13.5.

The greatest integer is 13. Since t = 2.16 > t (^) .05,13 = 1.7709, reject H 0.

9.2.16 Decreasing the degrees of freedom also decreases the power of the test.

9.2.17 a) The sample standard deviation for the first data set is approximately 3.15; for the

second, 3.29. These seem close enough to permit the use of Theorem 9.2.2.

b) Intuitively, the states with the comprehensive law should have fewer deaths. However,

the average for these data is 8.1, which is larger than the average of 7.0 for the states

with a more limited law.

Section 9.

9.3.1 a) The critical values are F .025,25,4 and F (^) .975,25,4. These values are not tabulated, but in this

case, we can approximate them by F .025,24,4 = 0.296 and F .975,24,4 = 8.51. The observed

F = 86.9/73.6 = 1.181. Since 0.296 < 1.181 < 8.51, we can accept H 0 that the

variances are equal.

b) Yes, we can use Theorem 9.2.2, since we have no reason to doubt that the variances are

equal.

9.3.2 The observed F = 0.

2 /0.

2 = 2.935. Since F .025,9,9 = 0.248 < 2.935 < 4.03 = F .975,9,9, we

can accept H 0 that the variances are equal.

9.3.3 a) The critical values are F (^) .025,19,19 and F .975,19,19. These values are not tabulated, but in this

case, we can approximate them by F .025,20,20 = 0.406 and F .975,20,20 = 2.46. The observed

F = 2.41/3.52 = 0.685. Since 0.406 < 0.685 < 2.46, we can accept H 0 that the variances

are equal.

b) Since t = 2.662 > t (^) .025,38 = 2.0244, reject H 0.

9.3.4 The observed F = 3.

2 /5.

2 = 0.315. Since F .025,9,9 = 0.248 < 0.315 < 4.03 = F .975,9,9, we can

accept H 0 that the variances are equal.

9.3.5 F = 0.

2 /0.

2 = 0.292. Since F .025,9,9 = 0.248 < 0.292 < 4.03 = F .975,.9,9, accept H 0.

9.3.6 The observed F = 398.75/274.52 = 1.453. Let & = 0.05. The critical values are F .025,13,11 and

F (^) .975,13,11. These values are not in Table A.4, so approximate them by F .025,12,11 = 0.301 and

F .975,12,11 = 3.47. Since 0.301 < 1.453 < 3.47, accept H 0 that the variances are equal. Theorem

9.2.2 is appropriate.

140 Chapter 9

9.3.7 Let! = 0.05. F = 65.25/227.77 = 0.286. Since 0.208 = F .025,8,5 < 0.286 < 6.76 = F .975,8,5,

accept H 0. Thus, Theorem 9.2.2 is appropriate.

9.3.8 For these data,

2 s (^) X = 56.86 and

2 s Y (^) = 66.5. The observed F = 66.5/56.86 = 1.170. Since

F .025,8,8 = 0.226 < 1.170 < 4.43 = F .975, 8, 8, we can accept H 0 that the variances are equal.

Thus, Theorem 9.2.2 can be used, as it has the hypothesis that the variances are equal.

9.3.9 If

2 2 X Y

2

" # " # " , the maximum likelihood estimator for "

2 is

2 2 2

1 1

ˆ ( ) ( )

n m

i i i i

x x y y n m

"

$ %

# ( & ' & )

' * +

, ,. Then

2 2 2 1 1

( ) / 2 1 ( ) ( ) 2 ˆ 2

( ˆ)

2 ˆ

n m i i i i

n m (^) x x y y

L e

"

-

."

' $^ % & ( & ' & ) ( )

, ,

$ %

# ( )

* +

=

( ) / 2 ( ) / 2 2

2 ˆ

n m n m e

."

' $ % & ' ( )

If

2 2

" X / " Y the maximum likelihood estimators for

2 2

" X and" Y are

2 2

1

ˆ (

n

X i i

x x ) n

"

(^) , & , and

2 2

1

ˆ ( )

m

Y i i

y y m

"

(^) , &.

Then

2 2 2 2 1 1

/ 2 1 / 2^1 ( ) ( ) ˆ^1 2 ˆ^12 ˆ ( )

n m i i X i Y i

n (^) x x m y y

L e e

" # " #

$ % $ & ( & ) & ( & ( ) (

$ % ,^ $ % ,

0

% ) )

9.4.

ˆ

x y p n m

!!

" "

!!

= 0.

z = ˆ (1 ˆ ) ˆ (1 ˆ)

x y

n m

p p p p

n m

#

!

=

0.188(0.812) 0.188(0.812)

!

= #2.

For this experiment, H 0 : pX = p (^) Y and H 1 : pX < p (^) Y. Since z = #2.38 < #1.64 = # z .05, reject H 0.

9.4.3 Let $ = 0.05.

ˆ

p

!

"

!

= 0.

z =

0.836(0.164) 0.836(0.164)

!

= #0.

For this experiment, H 0 : pX = p (^) Y and H 1 : pX % p (^) Y. Since #1.96 < z = #0.17 < 1.96 = z .025,

accept H 0 at the 0.05 level of significance.

9.4.

ˆ

p

!

"

!

= 0.

z =

0.627(0.373) 0.627(0.373)

!

= #0.

Since #2.58 < z = #0.92 < 2.58 = z .005, accept H 0 at the 0.01 level of significance.

9.4.

ˆ

p

!

"

!

= 0.

z =

0.54(0.46) 0.54(0.46)

!

= 1.

The P value is P ( Z & #1.70) + P ( Z ' 1.70) = 2(1 # 0.9554) = 0.0892.

9.4.

ˆ

p

!

"

!

= 0.

z =

0.697(0.303) 0.697(0.303)

!

= 1.

Since #1.96 < z = 1.50 < 1.96 = z .025, accept H 0 at the 0.05 level of significance.

142 Chapter 9

9.4.

ˆ

p

!

"

!

= 0.

z =

0.358(0.642) 0.358(0.642)

!

= #7.93. Since z = #7.93 < #1.96 = # z .025, reject H 0.

9.4.

ˆ

p

!

"

!

= 0.

z =

0.256(0.744) 0.256(0.744)

!

= 0.25. In this situation, H 1 is p (^) X > p (^) Y.

Since z = 0.25 < 1.64 = z .05, accept H 0. The player is right.

9.4.9 From Equation 9.4.1,

$ =

(55 60) (160 192 55 60)

55 105 60 132

[(55 60) /(160 192)] [1 (55 60) /(160 192)]

(55 /160) [1 (55 /160)] (60 /192) [1 (60 /192)]

!! !! #!!

#

=

115 237 160 192

352 55 105 60 132

115 (237 )(160 )(192 )

352 (55 )(105 )(60 )(132 )

. We calculate ln $, which is #0.1935. Then #2ln $ =

0.387. Since #2ln $ = 0.387 < 6.635 =

2

% .99,1, accept H 0.

Section 9.

9.5.1 The center of the confidence interval is x # y = 1007.9 # 831.9 = 176.0. The radius is

/ 2, 2

2.0739(411)

t (^) n m sp n m

&! #!^ "^!^ = 359.4. The confidence interval is

(176.0 # 359.4, 176.0 + 359.4) = (#183.4, 535.4)

9.5.2 The center of the confidence interval is x # y = 6.7 # 5.6 = 1.1. sp =

2 2 8(0.54 ) 6(0.36 )

!

=

0.47. The radius is (^) / 2, 2

1.7613(0.47)

t (^) n m sp n m

&! #!^ "^!^ = 0.42. The confidence

interval is (1.1 # 0.42, 1.1 + 0.42) = (0.68, 1.52). Since 0 is not in the interval, we can reject

the null hypothesis that ' X = ' Y.

9.5.3 The center of the confidence interval is x # y = 83.96 # 84.84 = #0.88. The radius is

/ 2, 2

2.2281(11.2)

t (^) n m sp n m

&! #!^ "^!^ = 14.61. The confidence interval is (#0.88^ #

14.61, #0.88 + 14.61) = (#15.49, 13.73). Since the confidence interval contains 0, the data do

not suggest that the dome makes a difference.

Chapter 9 143

9.4.

ˆ

p

!

"

!

= 0.

z =

0.256(0.744) 0.256(0.744)

!

= 0.25. In this situation, H 1 is p (^) X > p (^) Y.

Since z = 0.25 < 1.64 = z .05, accept H 0. The player is right.

9.4.9 From Equation 9.4.1,

$ =

(55 60) (160 192 55 60)

55 105 60 132

[(55 60) /(160 192)] [1 (55 60) /(160 192)]

(55 /160) [1 (55 /160)] (60 /192) [1 (60 /192)]

!! !! #!!

#

=

115 237 160 192

352 55 105 60 132

115 (237 )(160 )(192 )

352 (55 )(105 )(60 )(132 )

. We calculate ln $, which is #0.1935. Then #2ln $ =

0.387. Since #2ln $ = 0.387 < 6.635 =

2

% .99,1, accept H 0.

Section 9.

9.5.1 The center of the confidence interval is x # y = 1007.9 # 831.9 = 176.0. The radius is

/ 2, 2

2.0739(411)

t (^) n m s (^) p n m

&! #!^ "^!^ = 359.4. The confidence interval is

(176.0 # 359.4, 176.0 + 359.4) = (#183.4, 535.4)

9.5.2 The center of the confidence interval is x # y = 6.7 # 5.6 = 1.1. sp =

2 2 8(0.54 ) 6(0.36 )

!

=

0.47. The radius is (^) / 2, 2

1.7613(0.47)

t (^) n m s (^) p n m

&! #!^ "^!^ = 0.42. The confidence

interval is (1.1 # 0.42, 1.1 + 0.42) = (0.68, 1.52). Since 0 is not in the interval, we can reject

the null hypothesis that ' X = ' Y.

9.5.3 The center of the confidence interval is x # y = 83.96 # 84.84 = #0.88. The radius is

/ 2, 2

2.2281(11.2)

t (^) n m s (^) p n m

&! #!^ "^!^ = 14.61. The confidence interval is (#0.88^ #

14.61, #0.88 + 14.61) = (#15.49, 13.73). Since the confidence interval contains 0, the data do

not suggest that the dome makes a difference.

Chapter 9 143

9.5.5 Equation (9.5.1) is (^) / 2, 2 / 2, 2

( )

X Y n m n m

p

X Y

P t t

S

n m

$ $

#

"! "!

% &

' (

!!!

' ! ) ) (

' (

' " (

* +

= 1! $

which implies

/ 2, 2 / 2, 2

P t (^) n m S (^) p X Y ( (^) X Y ) t (^) n m Sp n m n m

$ "! #^ # $ "!

% &

'!^ "^ )^!^!^!^ ) (

' (

* +

" = 1! $, or

/ 2, 2 / 2, 2

P ( X Y ) t (^) n m S (^) p ( (^) X Y ) ( X Y ) t (^) n m Sp n m n m

$ "! #^ # $ "!

% &

'!^!^!^ "^ )!^!^ )!^!^ "^ " (

' (

* +

= 1! $

Multiplying the inequality above by !1 gives the inequality of the confidence interval of

Theorem 9.5.1.

9.5.6 Begin with the statistic X! Y , which has E ( X! Y )= # X! # Y and Var( X! Y )=

. Then

2 2

, X / n " , Y / m / 2 / 2

2 2

( )

/ /

X Y

X Y

X Y

P z z

n m

$

#

, ,

% &

!!!

' ! ) ) (

' " (

* +

$ = 1^!^ $, which implies

-.

2 2 2 2

P! z $ / 2 , X / n " , Y / m ) X! Y! ( # X! # Y ) ) z $/ 2 , X / n " , Y / m = 1! $

Solving the inequality for # X! # Y gives

-.

2 2 2 2

P X! Y! z $ / 2 , X / n " , Y / m ) # X! # Y ) X! Y " z $/ 2 , X / n " , Y / m = 1! $.

Thus the confidence interval is -.

2 2 2 2

x! y! z $ / 2 , X / n " , Y / m x ,! y " z $ / 2 , X / n ", Y / m

9.5.7 Approximate the needed F .025,25,4 and F (^) .975,25,4 by F .025,24,4 = 0.296 and F .975,24,4 = 8.51. The

confidence interval is approximately

2 2

2 0.025,24,4^ , 2 0.975,24,

X X

Y Y

s s F F s s

% &

' (

* +

=

73.6 73.

(0.296), (8.51)

86.9 86.

% &

' (

* +

= (0.251, 7.21). Because the confidence interval contains 1, it supports the conclusion of

Question 9.3.1 to accept H 0 that the variances are equal.

9.5.8 The confidence interval is

2 2

2 .025,5,7^ , 2 .975,5,

X X

Y Y

s s F F s s

% &

' (

* +

=

137.4 137.

(0.146), (5.29)

340.3 340.

% &

' (

* +

= (0.06, 2.14)

Since the confidence interval contains 1, we can accept H 0 that the variances are equal, and

Theorem 9.2.1 applies.

144 Chapter 9

9.5.9 Since

2 2

2 2

/

/

Y Y

X X

S

S

!

!

has an F distribution with m " 1 and n " 1 degrees of freedom,

2 2

/ 2, 1, 1 2 2 1 / 2, 1, 1

/

/

Y Y m n m n X X

S

P F F

S

!

!

" " " " "

$ %

' &^ & (

) *

= 1 " # or

2 2 2

2 / 2, 1, 1 2 2 1 / 2, 1, 1 1

X X X m n m n Y Y Y

S S

P F F

S S

!

!

" " " " "

$ %

' &^ &^ (+^ "

) *

The inequality provides the confidence interval of Theorem 9.5.2.

9.5.10 The center of the confidence interval is

x y

n m

" + " (^) = 0.031. The radius is

.

x x y y

n n m m z n m

$ %$ % $ %$ %

' (' "^ ( ' (' " (

) *) * ) *) *

, =

$ %$ % $ %$ %

' (' "^ ( ' (' " (

) *) * ) *) *

, = 0.064.

The 80% confidence interval is (0.031 " 0.064, 0.031 + 0.064) = ("0.033, 0.095)

9.5.11 The approximate normal distribution implies that

( )

( / )(1 / ) ( / )(1 / )

X Y

X Y

p p n m P z z X n X n Y m Y m

n m

# #^

$ %

' "^ "^ " (

' " & & (+ "

' " " (

' , (

) *

or

( / )(1 / ) ( / )(1 / )

( X Y )

X n X n Y m Y m X Y P z p p n m n m

$ " "

'"^ ,^ &^ "^ "

'

)

"

( / )(1 / ) ( / )(1 / )

X n X n Y m Y m z n m

#^

" "^ %

& , (

(

*

+ "

which implies that

( / )(1 / ) ( / )(1 / )

( X Y )

X Y X n X n Y m Y m P z p n m n m

$ $ % " "

' "^ ' "^ ("^ ,^ & "^ "

'

) )^ *

p

( / )(1 / ) ( / )(1 / )

X Y X n X n Y m Y m z n m n m

#^

$ % "^ "^ %

& " ' " ( , , (+ "

(

) * *

Multiplying the inequality by "1 yields the confidence interval.

9.5.12 The center of the confidence interval is

x y

n m

" + " = "0.083. The radius is

.

x x y y

n n m m z n m

$ %$ % $ %$ % $ %$ % $ %$ %

' (' "^ ( ' (' "^ ( ' (' "^ ( ' (' " (

) *) * ) *) * ) *) * ) *) *

, + , = 0.

The 95% confidence interval is ("0.083 " 0.058, "0.083 + 0.058) = ("0.141, "0.025)

Since the confidence interval lies to the left of 0, there is statistical evidence that the suicide

rate among women members of the American Chemical Society is higher.

Chapter 9 2 145