IB Diploma Program Mathematics Course Companion Higher Level Option: Statistics Worked Solutions [1 ed.] 0198304854, 9780198304852

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O X

F O R

D

I B

D

I P L O

M

A

P

R

O

G R

A

M

M

E

WORKED SOLUTIONS M AT H E M AT I C S

HI GH E R

LE V E L

:

S TAT I S T I C S C O U R S E

C O M PA N I O N

Josip Harcet

Lorraine Heinrichs

Palmira Mariz Seiler

Marlene Torres Skoumal

WORKED SOLUTIONS

Exploring further probability distributions

1 Skills check 1 a

4

V

¦x

Var( X )

Mode (X) = −1, 1 because those two values share the highest probability of 0.3.

2 i

pi  P 2

i 1

12 u 0.4  22 u 0.3  32 u 0.2  4 2 u 0.1  22

Median, m = 1 since P(X ≤ 0) = 0.4 and P(X ≤ 1) = 0.7

54

1

6

P

¦x p

E( X )

i

1 u 0.3  0 u 0.1  1 u 0.3

i

1.1

1.2

Skills check

1.3

B

C

D

MinX Q1X MedianX... Q3X MaxX

=OneVar(a 1. 1. 2. 3. 4.

i 1

 2 u 0.1  3 u 0.05  4 u 0.15

0.95

8 9 10 11 12 D10 =2.

6

V

¦x

Var( X )

2 i

pi  P 2

i 1

6

¦x

2 i i

( 1)2 u 0.3  0 2 u 0.1  12 u 0.3

P

i 1

1.1

 22 u 0.1  32 u 0.05  4 2 u 0.15 3.85

V

3.85  0.9025 1.1

B

1.72

8 9 10 11 12 D10 =1.

C

D

MinX Q1X MedianX... Q2X MaxX

=OneVar(a –1. –1. 1. 2. 4.

Skills check

1.3

B

C

1.3

2 a

1.4

D

=OneVar(a 0.3 x 2. 0.2 Σx 2. 0.1 Σx2 5. sx : = sn–... #UNDER... 1. σx : = σn...

2 2 3 3 4 4 5 6 D6 =1.

Skills check

1.2

1.2

A

Skills check

1.5

4

{

3 3 2 – ·x , 0 2 3

1 1.1

Skills check

1.2

A

B

C

D

=OneVar(a 0.95 2 0 0.1 x 0.95 3 1 0.3 Σx 3.85 4 2 0.1 Σx2 5 3 0.05 sx : = sn–... #UNDEF... 6 4 0.15 σx : = σnx.. 1.71683 D6 =1.7168284713389

–3 –2 –1 0 –1

1

2

3

4

5

Mode (X) = 0, the part of parabola is opened § ©

downwards and it has its vertex at ¨ 0,

3· ¸. 4¹

m

b

xi

1

2

3

4

pi

0.4

0.3

0.2

0.1

Mode (X) = 1 because it has the highest probability (0.4)

3 1 §3 ·  x 2 ¸ dx Ÿ 4 16 2 ¹ 0© 3 1 3 1 m m Ÿ m 0.695 4 16 2 2 3 3 P E( X ) ³ §¨ x  x 3 ·¸ dx 4 16 ¹ 0©

Median, ³ ¨

2

3 4º ª3 2 «¬ 8 x  64 x »¼ 0

Median, m = 2 since P(X ≤ 1) = 0.4 and P(X ≤ 2) = 0.7 4

P

E( X )

¦x p i

i

1 u 0 .4  2 u 0 .3

i 1

 3 u 0 .2  4 u 0 . 1

2

V

Var( X )

2

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

3 3  2 4

3 4

3 4· §3 2 §3· ³0 ¨© 4 x  16 x ¸¹ dx  ¨© 4 ¸¹

2

Statistics: Chapter 1

1

WORKED SOLUTIONS 2

ª x3 3x 5 º 9  « »  80 ¼ 0 16 ¬4 19 80 1.3

1.4

nSolve

Skills check

1.7

π 4

∫

0.11685

2 –π (x ·cos(2·x))dx 4

√ 0.11685027506808

Skills check

1.5

1.6

4

0.487

( 34 ·m– 161 ·m = 12 ,m )

0.341834

|

0.694593

3

6/99

2

∫( ∫(

1.5

6 9  5 16

2

0.75

x1·f1(x))dx

c

0 2

1.6

0.2375

x2·f1(x))dx–(0.75)2

1.7

⎧ 6,36

1.5

0

1

3/99

Skils check

1.8

y 2

0.5 1.3

1.4

Skills check

1.5

0

2

∫( ∫(

1

0.75

x·f1(x))dx

2

3

4

5

6

7

x

8

0

2

0.2375

x2·f1(x))dx–(0.75)2

Mode (X) = 3 because it is a decreasing function on the interval [3, 6].

0

0.48734

√0.2375

m

|

0

1.4

1.5

{

0 –1

1

V

2

f (x )  P 2

S  4

1.7

1.8

nSolve

(∫

∫( 3

1.9

Skils check

6 m

1 6 dx= ,m 2 x2

)

4.

|4.15888



S 4

Var( X )

3/4

∫( ) ∫( ) 1.7



1.8

1.9

Skils check

x2

3

0

18.

6

6 x 2 dx x 2•

S 4

³x

)

6 x• 2 dx x

0

S 4

³ xf ( x )dx

0.839

3

5

1 Ÿm 2

Median, ³ f ( x )dx

V

³x

Var( X )

6

m

E( X )

4.16

6

π 4 π 4

Mode (X ) = 0, it has the maximum point at (0, 1).

P

³ xf ( x ) dx

E( X )

3

cos(2·x), –π < x < 4 f1(x)= –π >x> 0, 4 1 (0,1)

–1

P

Skills check

1.6

4

6

4/99

b

1 Ÿm 2

Median, ³ f ( x ) dx

2

f (x )  P 2

3

0.342

√ 18–(4.1588830833593)2

S 4

0.838863

3/4 1.5

1.6

nSolve

∫

(∫

Skills check

1.7 m –π 4

1 cos(2·x) dx = ,m 2

π 4 –π 4

)

0.

3 a

1 − 0.5 + 0.25 − 0.125 + ... ⇒ u1 = 1, r = −0.5 Sum

1 1  ( 0.5)

1 1 .5

2 3

0.

(x·cos(2·x))dx

b

2 1

4/5

Sum

2 1  ! Ÿ u1 2 2

2 1 1 2

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

2 2 1

u

2, r 21 21

1 2

22 2

Statistics: Chapter 1

2

WORKED SOLUTIONS 4 a

1 ,x ≠2 2−x −1 × − 1 1 Ÿ f ′( x ) = = ,x ≠2 2 (2 − x ) (2 − x )2

f (x ) =

dx

Ÿ ³ f ( x ) dx b

 ln(2  x )  c , x  2

³ f ( x )dx = ³ e

3 x +1

dx =

b

3 π − 2x f ( x ) = sin 2 3

3 π − 2x § 2 · π − 2x × ¨ − ¸ = − cos cos 2 3 3 © 3¹ 3 S  2x Ÿ ³ f ( x ) dx ³ sin dx 2 3 3 1 9 S  2x S  2x u cos c cos c 2 2 3 4 3  3

2( x 2  2 ) u 2 x

Ÿ ³ f ( x )dx

³ (x

4 a

4

³ (x

2

4 x ( x 2  2)

 2)2 dx

 4 x  4 )dx

k 1 k 2k   12 12 12

a

1 Ÿ 3  3k

12 Ÿ k

0

1

2

F (x)

3 12

7 12

12 12

b

3

1 1  2 6

P( X

2)

F (2)  F (1)

1

a2 a4 a6 a8    20 20 20 20

5

a

20 Ÿ a

1)

F (1)

P( X

3)

F (3)  F (1)

P( X

5)

F (5)  F (3)

P( X

7)

F (7)  F (5)

P( X

9)

F (9)  F (7)

x)

­x ° , x 1, 3, 5, 7, 9 ® 25 °¯ 0, otherwise

3 25

³ 2 sin bx dx

§ bS · ¸ © 3 ¹

1



b 2

b Ÿb 2

1

( ( ) =1– 2b ,b)

0.

( ( b•3π ) =1– 2b ,b,0.5)

1.

b•π nSolve cos 3 nSolve cos

4

F (x)

8 20

14 20

18 20

20 20

7 10

1

Exercise A

1.1

3

14 20

16 1 ! 25 2 S

Ÿ cos ¨

2

F (2)

9 25

ª 1 º3 1 Ÿ 2 «  cos bx » ¬ b ¼0

§ bS · ¸  cos 0 © 3 ¹

1

P( X d 2)

5 25 7 25

9 1  and F (7) 25 2

nSolve

(∫

π 3

1.

)

(2•sin(b•x))dx=1,b

0

1/3 x

c

1 . 2

Median, m = 7 since

10

0, x  1 ­ ° 2 °9x  x , x 1, 2, 3, 4 ® ° 20 °¯ 1, x ! 4

2)

1 25

P( X

Ÿ cos ¨

1

1 2

The modal value of the variable X is 2 since it

0

X=x

⇒ F (x )

1 2

2 , 6

­ x 1 , x 0, 1, 2 ° ® 6 °¯ 0, otherwise

x)

S 3

0, x  0 ­ ° 2 ° x  7x  6 , x 0, 1, 2 ® 24 ° 1, x ! 2 °¯

Ÿ 4a  20 b

F (1)  F (0)

F (5)

X=x

⇒ F (x )

2

1)

P( X

x5 4x3   4x  c 5 3

2

Exercise 1A

b

P( X

( x 2  2)2

f (x )

Ÿ f c( x )

1 a

F (0 )

has the highest probability, P( X

Ÿ f ′( x ) =

d

0)

3 Ÿ P( X 6

1 3 x +1 e +c 3

1 , 6

P( X

f ( x ) = e3 x +1 Ÿ f ′( x ) = 3e3 x +1 Ÿ

c

³ 2x

3 a

b

³

F (x )

f

x

f (t )dt Ÿ F ( x )

0

[2 cos t ]0x

Ÿ F (x )

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

³ 2 sin t dt

 2 cos x  2

­ 0, x  0 ° ° S ®2  2 cos x , 0 d x d 3 ° ° S 1, x ! ¯ 3 Statistics: Chapter 1

3

WORKED SOLUTIONS c

6

a

§ ©

S· ¸ 6¹

§ © §S · 1  2  2 cos ¨ ¸ ©6¹

P¨ X t

1 P¨ X d

S· ¸ 6¹

§S ·

2 a

1 F¨ ¸ ©6¹

3 1

b

ª1

xº

lim « arcsin » b o2 S 2¼ ¬ 2

1ª xº arcsin » S «¬ 2 ¼ 2

S· 1 §S ¨  ¸ S©2 2¹

= 1 – P(X ≤ 6) = 0.000729 c

X ~ Geo(0.3) Ÿ P(5 d X d 7 ) = P(X ≤ 7) – P(X ≤ 4) = 0.158

b

d

X ~ Geo(0.991) Ÿ P(1  X d 7 )

b

= P(X ≤ 7) – P(X ≤ 1) = 0.009

1 (arcsin 1  arcsin(1)) S

1.1

1 +∞

The function is well defined if

³

CH1 Exercise B

1.2

geomCdf(0.25,1,4)

0.683594

1–geomCdf(0.7,1,6)

0.000729

geomCdf(0.3,1,7)–geomCdf(0.3,1,4)

f ( x )dx = 1

0.157746

−∞

b

geomCdf(0.991,1,7)–geomCdf(0.991,1,1)

The modal value is the value of the random variable X where probability reaches its maximum. 3  1 § 1· 2 2 ( 4 x ) (2 x )   ¨ ¸ S © 2¹

f c( x ) x

3 2 2

f c( x )

,

0Ÿx

0

0.009

4/99

3

X ~ Geo(0.73) Ÿ P( X d 3)

4

X ~ Geo(0.92)

S (4  x ) 1.1

Exercise A

1.2

0.684

X ~ Geo(0.7 ) Ÿ P( X ! 6)

0.732

Since the function is not defined for the values of −2 and 2 we need to use limits. §b · dx lim ¨¨ ³ ¸ 2 ¸ b o2 © b S 4  x ¹

X ~ Geo(0.25) Ÿ P( X d 4 )

y 9

5

a

P( X

5)

b

P( X t 4 )

0.980

0.0000377 1  P ( X d 3)

0.000512

X ~ Geo(0.15)

6 1.1

3 –12 –9 –6 –3 0 –3 –6

3 f1(x)=

–9

6

9

12

x

1 π·√4 – x2

0.980317

geomPdf(0.92,5)

0.000038

1–geomCdf(0.92,1,3)

0.000512

geomPdf(0.15,4)

0.092119

6

a

P( X

4)

b

P( X t 7 )

1 4)

= 0.53 × 0.5 = 0.0625 d X ~ Geo(0.88) Ÿ P( X

0.37715

0.0921 1  P( X d 6 )

X ~ Geo( p ) Ÿ P( X ! k ) f

= 0.862 × 0.14 = 0.104 c X ~ Geo(0.5) Ÿ P( X

¦q

0.377 1  P( X d k )

k

n 1

p

¦q

1 p

n 1

n1 

n k 1

inf. geom. seq.

5)

= 0.124 × 0.88 = 0.000182 1.1

0.24

geomCdf(0.73,1,3)

5/99

Exercise 1B X ~ Geo(0.6) Ÿ P( X 2) 0.4 u 0.6 b X ~ Geo(0.14 ) Ÿ P( X 3)

CH1 Exercise B

1.3

1–geomCdf(0.15,1,6)

By looking at the graph of the probability function we notice that the graph has only one extreme point and that is the minimum point at 0, therefore the modal value doesn’t exist. 1 a

1.2

1 p

1  qk 1 q

1 p

1  qk p

1  (1  q k )

qk

CH1 Exercise B

geomPdf(0.6,2) geomPdf(0.14,3) geomPdf(0.5,4) geomPdf(0.88,5)

OR

0.24 0.103544 0.0625 0.000182

X ~ Geo( p ) Ÿ P( X ! k ) f

¦

k

q n 1 p

n k 1

qk p u

¦q

n 1

n1 

qk p

1 1 q

qk p

1 p

qk

inf. geom. seq.

4/99 © Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 1

4

WORKED SOLUTIONS

Investigation 1 X ~ Geo (p) a

p

P( X ! 2 ) b

p

1  P ( X d 5) 1  P ( X d 3)

1  P( X d 2 )

0.12 0.882

Var( X )

1  P( X d 6 ) 1  P( X d 2 )

1  P( X d 4 )

P( X ! 7 ) 1.2

1.3

1  P( X d 7 ) 1.4

0.75 0.252

Var( X ) 0.409

0 .3 0 .7 2

Var( X )

1–geomCdf(0.4,2)

0.36

1–geomCdf(0.7,6) 1–geomCdf(0.7,2)

0.0081

1–geomCdf(0.7,4)

0.0081

4,

12

X ~ Geo(0.7) Ÿ E( X )

b

CH1 Exercise B

0.36

1 0.25

0.155

X ~ Geo(0.25) Ÿ E( X )

a

0.409

1–geomCdf(0.4,5) 1–geomCdf(0.4,3)

1.14,

For 1B, Question 2:

0.12 Ÿ P( X ! 12| X ! 5) 1  P( X d 12) 1  P ( X d 5)

1 0.88

0.0081

0.0081

P(( X ! 12) ˆ ( X ! 5)) P ( X ! 5)

0 .7 0 .3 2

Var( X )

1.43,

1 0 .3

3.33,

0.612

X ~ Geo(0.3) Ÿ E( X )

c

1 0 .7

7.78

|

1 0.991

X ~ Geo(0.991) Ÿ E( X )

d 4/99 CH1 Exercise B

1.2 1.3 1.4 1–geomCdf(0.4,2)

0.36

1–geomCdf(0.7,6) 1–geomCdf(0.7,2)

0.0081

1–geomCdf(0.7,4)

0.009 0.9912

Var( X ) 2

0.0081 0.408676

a

E( X )

1–geomCdf(0.12,7)

0.408676

b

V

1 0.73

x max

X ~ Geo( p ); m, n  ]  ; m  n

P( X ! n ) P( X ! m )

0.27 0.732

Var( X )

P(( X ! n ) ˆ ( X ! m )) P( X ! m )

q n m

P( X ! n  m )

X ~ Geo(0.6) Ÿ E( X ) Var( X )

b

Var( X )

1.36986

1 0 .6

0.27 (0.73)2 3.50527

√

0.27 1 +3∙ 0.73 (0.73)2

1.67,

16/99

1.11 3

X ~ Geo(0.14 ) Ÿ E( X ) 0.86 0.14 2

3.51

0.711802

√ 0 .4 0 .6 2

0.27 0.732

CH1 Exercise C

1 0.73

For 1B, Question 1: a

1 3u 0.73

E( X )  3V

1.1

Exercise 1C 1

0.712

Mario must make four shots to destroy the balloon.

n

q qm

0.00916

1.37

| 6/99

1.01,

X ~ Geo(0.73)

1–geomCdf(0.12,12) 1–geomCdf(0.12,5)

Ÿ P( X ! n | X ! m )

2,

2

X ~ Geo(0.88) Ÿ E( X )

d

0.7 Ÿ P( X ! 6| X ! 2)

P( X ! 4 ) p

0.36

0.36

P(( X ! 6) ˆ ( X ! 2)) P( X ! 2 ) c

0 .5 0 .5 2

Var( X )

0.4 Ÿ P( X ! 5| X ! 3) P(( X ! 5) ˆ ( X ! 3)) P ( X ! 3)

1 0 .5

X ~ Geo(0.5) Ÿ E( X )

c

1 0.14

X ~ Geo(0.54 )

E( X )

1 0.54

V

0.46 0.54 2

1.26

7.14, Var( X )

43.9 x max

E( X )  3V

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

1 3u 0.54

1.85

0.46 0.54 2

5.62

Statistics: Chapter 1

5

WORKED SOLUTIONS Therefore, we must select 6 students at random to ensure that one of them will be familiar with the election procedure.

b

X a NB(0.5, 2) Ÿ P( X ! 6) 6

1

¦ P( X

k)

k 2

0.109

CH1 Exercise C

1.1

1 0.54

1.1

1.85185

CH1 Exercise D

1.2

k=3

1.25599

√

7 64

0.46 (0.54)2

0.050781

6 5.61981

√

0.46 1 +3∙ 0.54 (0.54)2

(nCr(k–1, 2–1)∙(0.5)k–2∙(0.5)2)

1– k=2

0.109375 |

19/99

2/99

Exercise 1D 1 a

X a NB(1, 0.2) Ÿ P( X

§ 2  1· ¨ ¸ 0 .8 u 0 . 2 ©1  1¹ b X a NB(3, 0.5) Ÿ P( X

= 1.1

0.109375

7

(nCr(k–1, 4–1)∙(0.18)k–4∙(0.82)4)

9) k=5

0.524751

0.235 |

3/99

32)

§ 32  1 · 9 32 ¸ 0.23 u 0.77 ¨ © 23  1¹

0.00847

d

X a NB(0.43, 5) Ÿ P(8  X d 11) =

CH1 Exercise D

1.1

0.1875

nCr(8,6)∙(0.2)2∙(0.8)7

0.234881

nCr(31,22)∙(0.23)9∙(0.77)32

0.008467

11

¦ P( X

= k ) = 0.326

k =9

0.16

nCr(3,2)∙0.5∙(0.5)3

CH1 Exercise D

1.2

k=2

X a NB(23, 0.77 ) Ÿ P( X

nCr(1,0)∙0.8∙0.2

= k ) = 0.525

0.188

X a NB(7, 0.8) Ÿ P( X

1.1

7

¦ P( X k =5

4)

§ 9  1· 2 7 ¨ ¸ 0 .2 u 0 . 8 7  1 © ¹ d

X a NB(0.82, 4 ) Ÿ P(5 d X d 7 )

0.16

§ 4  1· 3 ¨ ¸ 0 .5 u 0 . 5 3  1 © ¹ c

c

2)

CH1 Exercise D

1.2

k=5

0.524751

11

(nCr(k–1, 5–1)∙(0.57)k–5∙(0.43)5) k=9 0.325959

4/99 |

2 a

X a NB(0.25, 3) Ÿ P( X d 4 ) P( X 1.1

1.2

3)  P( X

4)

4/99

13 256

k=3

3 a

P (X

3)

CH1 Exercise D

4

(

0.508

)

nCr(k–1, 3–1)∙(0.75)k–3∙(0.25)3

0.050781

|

§3  1· 24 2 Ÿ¨ ¸ (1  p ) u p 2  1 125 © ¹

­ p1 = − 0.274 ° 24 12 2 ° = Ÿ p 2 − p3 = Ÿ ® p2 = 125 125 5 ° °¯ p3 = 0.874 2 Thus p = = 0.4 5

1/99

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 1

6

WORKED SOLUTIONS b

§ ©

We can say that it is almost certain that he will not need to interview more than a dozen students.

2·

X a NB ¨ 2, ¸ Ÿ P(3 d X d 5) 5 ¹

5

¦ P( X

1572 3125

k)

k 3

1.2

1.1

(

(

–p3+p2–

12 ,p 125

(nCr(k–1, 5–1)∙(0.08) – ∙(0.92) )

)

k 5

k –2

2 ∙ 5

k=5 0.999999

2

4/99

X a NB(0.85, 3) Ÿ P( X

7 a

k=3 2/99

§

1  P( X d 6)

1·

§ x  1· § 3 · ¸¨ ¸ © 2  1¹ © 4 ¹ x 2

3 4x

3 32

x)

x 2

§1· © ¹

2

1.4

5

(x–1)·3x–2 3 – =0,x 32 4x

(

CH1 Exercise D

3 32

3 Ÿx 32

3

(nCr(k–1, 3–1)∙(0.15) – ∙(0.85) ) k 3

1–

0.005885 | 6/99

Exercise 1E 4x

x

1

)

k 2

P( X

3. 47 128

2

3

k=3

§4· § 1· § 1· ¨ ¸¨ ¸ ¨ ¸ 2 ¹ © 2 ¹ 

© x ¹ ©N

x)

heads

(34) – ∙(14) )

nCr(k–1, 2–1)∙

0.082907

6

CH1 Exercise D

(

nSolve

1.5

0.999999

§ 1· c X a NB ¨ 2, ¸ Ÿ P ( X d 5) © 4¹ 5 47 P (X k ) ¦ 128 k 2 1.3

0.0829

0.589

nCr(4,2)∙(0.15)2∙(0.85)3

u¨ ¸ 4

٬

1.2

1.4

1.3

§ 1· b X a NB ¨ 2, ¸ Ÿ P( X © 4¹

5)

X a NB(0.85, 3) Ÿ P( X t 7 )

b

X a NB ¨ 2, ¸ © 4¹

Ÿ ( x  1)

5

1572 3125

( ) ( ))

3 nCr(k–1,2–1)∙ 5

0.263633

12

CH1 Exercise D

|

4 a

CH1 Exercise D

0.047339

{–0.274456, 25 ,0.874456} 5

1.5

nCr(5,4)∙0.08∙(0.92)5

1.3

polyRoots

1.4

1.3

0.503

0

1

2

3

4

pi

1 16

4 16

6 16

4 16

1 16

1 1 3 1 1  t  t2  t3  t4 16 4 8 4 16

|

2

2/99

xi

1

2

3

k

...

...

k 1

5 a b

X a NB(0.85, 4 ) Ÿ P( X

5)

0.313

X a NB(0.85, 4 ) Ÿ P( X t 7 ) 1  P( X d 6) 1.4

1.3

1.5

G (t )

0.0473

b

5 6k

...

not one

5k 1 k t ! 6k

k

1 f §5 · t ¦ ¨ t¸ 6 k  © 6¹

0  inf. geom. sequence

1 1 tu 5 6 1 t 6

(nCr(k–1, 4–1)∙(0.15) – ∙(0.85) ) k 4

4

0.047339

X a NB(0.92, 5) Ÿ P( X

6)

X a NB(0.92, 5) Ÿ P( X d 12)

1 6 tu 6 6  5t

t 6  5t

5 5 6 6 t  1 Ÿ 1  t  1 Ÿ   t  6 6 5 5

2/99

6 a

1 §5· ¨ ¸ u 6 ©6 ¹ one

...

k 1

1 5 25 3 t  t2  t ! 6 36 216

0.313204

6 k=4

5 25 36 216



CH1 Exercise D

nCr(4,3)∙0.15∙(0.85)4 1–

1 6

pi

4

tails

xi

G (t )

k=2

§4· § 1· ¨ ¸¨ ¸ ©x¹©2¹

0.264

3 a

2 3t

G (t )

0.999999 | 1 

2 3 1 1 t 3

2 2 2 2  t  t 3 9 27

2 3 2 t  !  k 1 t k  ! Ÿ P( X 81 3

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

0)

2 3

Statistics: Chapter 1

7

WORKED SOLUTIONS b c

P( X d 1)

0 )  P( X

P( X

P( X t 3)

8 9

1  P( X d 2 ) 2 · §2 2   ¸ © 3 9 27 ¹

1¨ d

2 2  3 9

1)

P( X t k )

1

26 27

1 27

1  P ( X d k  1) 2· §2 2 ! k ¸ =1 ¨  3 ¹ ©3 9 2§ 1· ¨1  k ¸ 3 3 ¹ 1 © 1 1 3

4 a

1 11 k 3

1 3k

X  B(1, p ) xi pi

0 1–p

1 p

5

For question 1: G (t )

1 1 3 1 1  t  t2  t3  t4, 16 4 8 4 16

G c(t )

1 3 3 1  t  t 2  t 3 Ÿ G c(1) 4 4 4 4

G cc(t )

3 3 3  t  t 2 Ÿ G cc(1) 4 2 4

E( X )

G c(1)

2

3

2; Var( X ) = G ′′(1)

+ G ′(1)(1 − G ′(1)) = 3 + 2(1 − 2) = 3 − 2 = 1 1.1

CH1 Exercise E

1.2

f1(1)

1

d dx (f1(x)) x=1

2

d2 (f1(x)) x=1 dx2

3

G(t) = 1 – p + pt G c(t )

p, G cc(t )

E( X )

G c(1)

Var( X )

3/99

0 For question 2:

p,

0  p (1  p ) b

§ pt · r¨ ¸ © 1  qt ¹ §

r 1

pt · ¸ 1  qt ¹ ©

r 1

rp ¨

G ′(1) = rp r p − ( r +1)

r

1 (1  qt )2

+ G ′(1)(1 − G ′(1)) = 60 + 6(1 − 6) = 60 − 30 = 30 CH1 Exercise E

(f1(x1.2)) x=1

dx2

 rp r t r 1 u ( (r  1))(1  qt ) ( r  2 ) u ( q )

f2(1)

1

d dx (f2(x)) x=1

6

d2 (f2(x)) x=1 dx2

60

rp r t r  2 (1  qt ) ( r  2 ) ((r  1)(1  qt )  (r  1)qt )

G cc(1)



Var( X )

G c(1)



6/99

For question 3:

rp r u p  r  2 ((r  1) p  (r  1)q )

r  r  p  rp  rq  q p2

E( X )

r ( p  r  q ) p2

G (t )

2 , G c(t ) 3t

 rp  r

 rq  rp  r p2

2

4(3  t )3 Ÿ G cc(1)

E( X )

G c(1)

Var( X ) rq p2

1 2

G cc(t )

G cc(1)  G c(1)(1  G c(1)) r r§ r· ( p  r  q )  ¨1  ¸ p© p¹ p2

2(3  t )2

2(3  1)2

Ÿ G c(1)

r , p

2

60

E( X ) = G ′(1) = 6; Var( X ) = G ′′(1)

1.1

rp r (r  1)t r  2 (1  qt ) ( r 1)

G cc(t )

60(6  5t )2 Ÿ G cc(1)

rp r t r 1 (1  qt ) ( r 1)

r = p

6

 12(6  5t )3 u ( 5)

G cc(t )

p (1  qt )  pt u ( q ) (1  qt )2

6  5t  t u ( 5) (6  5t )2

t , G c(t ) 6  5t

6(6  5t )2 Ÿ G c(1)

p (1  p ) [ pq ]

§ pt · ¨ ¸ © 1  qt ¹

X a NB(r , p ), G (t ) G c(t )

G (t )

G cc(1)  G c(1)(1  G c(1))

4(3  1)3

1 2

1 , 2

G cc(1)  G c(1)(1  G c(1)) 1 1§ 1·  ¨1  ¸ 2 2© 2¹

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

1 1  2 4

3 4

Statistics: Chapter 1

8

WORKED SOLUTIONS CH1 Exercise E

1.1

(f2(x1.2)) x=1

CH1 Exercise E

1.1

√3(x1.2)) x=1

dx2

2

dx2

f3(1)

1

f4(1)

1

d dx (f3(x)) x=1

1 2 1 2

d dx (f4(x)) x=1

5 3 19 9

d2 (f3(x)) x=1 dx2 |

d2 (f4(x)) x=1 dx2 |

9/99

12/99

4 3 1 3 2 ; P( X 4 ) u u u 7 7 3 7 3 4 3 2 3 1 4 t + × t2 + × × t3 b G (t ) = 7 7 3 7 3 7

6 a

P( X

4 § 1 1 4 · = t ¨1 + t 2 + t +!¸ 7 © 7 49 ¹ 2 § 1 1 4 · + t 2 ¨1 + t 2 + t + !¸ 7 © 7 49 ¹

42 7 1

4 t  2t 2 , G c(t ) 7  t2

c G (t )

G c(t ) G cc(t )

4 t  2t 2 7  t2

1

2

2

1 u 2(2  7t  3t 2 )(7  6t ) 144 1 Ÿ E( X  Y ) GXc Y (1) u 12 u 13 72

G cX Y (t )

( 4  4 t )(7  t 2 )  2t ( 4 t  2t 2 ) (7  t 2 )2 2 a G X Y (t )

4 (7  7t  t 2 ) (7  t 2 )2

5 3

3

E( X  Y )

c

Var( X  Y )

15

G cc(1)  G c(1)(1  G c(1))

234  15 u ( 14 )

4 (7  2t )(7  t )

2

 2(7  t ) u ( 2t ) u 4 (7  7t  t )

(( ((

234  210

24

CH1 Exercise F

1.1

2 2

3

3

G c(1)

b

13 6

§ t · § t · ¨ 2  t ¸ u ¨ 3  2t ¸ © ¹ © ¹

G X (t ) u GY (t )

§ · t2 ¨ 2 ¸ © 6  7 t  2t ¹

4 (7  7t  t 2 )(7  t 2 )2 ,

2 3

§ 1  3t · ¨ ¸ © 4 ¹ 2 § 2  7t  3t 2 · ¨ ¸ © 12 ¹

GX (t ) u GY (t )

2

The expected number of shots before the game is over is 2. d

4.67 .

§ 2  7t  3t 2 · 4 28  t ! b G X Y (t ) ¨ ¸ 12 144 144 © ¹ 4 28 32 2 Ÿ P( X  Y d 1)  144 144 144 9 c

4 (7  7  1) (7  1)2

G c(1)

Exercise 1F §2  t · ¸ © 3 ¹

4 (7  7  t 2  t 3  2t 2  t 3 ) (7  t 2 )2

E( X )

19 10  1 9 9 5 14 3u 1 3 3

Therefore, the maximum number of shots to be made before the game is over is 5.

u¨

4 t  2t 2 7 u 7 7  t2 6 6

E( X )  3V

1 a G X Y (t )

2 2·§ 1 2 1 4 §4 · ¨ t  t ¸ ¨ 1  t  t  !¸ 7 ¹© 7 49 ©7 ¹

Verifying: G(1)

G cc(1)  G c(1)(1  G c(1))

19 5 § 5·  ¨1  ¸ 9 3© 3¹

x max

3 1 3 2 3 1 3 1 4 + × × × t4 + × × × × t3 7 3 7 3 7 3 7 3 7 3 1 3 1 3 2  u u u u u t4 ! 7 3 7 3 7 3

4 t  2t 2 1 u 1 7 1  t2 7

Var( X )

2 49

1)

) ( )) t t 2–t ) •( 3–2•t) ) t=1

d t 3 t 3 dt 2–t • 3–2•t t=1

15

d2 dt2

234

3

3

4 ((7  2t )(7  t 2 )  4 t (7  7t  t 2 )) (7  t 2 )3 4 ( 49  14 t  7t 2  2t 3  28t  28t 2  4 t 3 ) (7  t 2 )3 4(49 + 42t + 21t 2 + 2t 3 ) = (7 − t 2 )3

Ÿ G ′′(1) =

4(49 + 42 + 21 + 2) 19 = (7 − 1)3 9

2/99

3 a b

G X (t ) = e0.25( t −1) , GY (t ) = e0.15( t −1) , G Z (t ) = e0.05( t −1)

G X Y  Z (t )

G X (t ) u GY (t ) u G Z (t )

= e0.25( t −1) × e0.15( t −1) × e0.05( t −1) = e0.45( t −1) G X +Y + Z (t ) = e0.45( t −1) Ÿ X + Y + Z ∼ Po(0.45)

P( X  Y  Z d 2) © Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

0.989 Statistics: Chapter 1

9

WORKED SOLUTIONS 1.1

Review exercise

CH1 Exercise F

1.2

poissCdf(0.45,0,2)

0.989121

1

|

a

4t 5t

G (t )

4 1 tu 1 5 1 t 5

4 1 1 2 1 3 § · t u ¨1  t  t  t  "¸ 5 5 25 125 © ¹ 4 4 4 3 4 4 t  t2  t  t " 5 25 125 625

1/99

4

pt 1  qt

X a Geo( p ), GX (t )

Y ∼ NB(r , p ), Y =  X + X  +!+ X  Ÿ GY (t )

b P( X t 2 )

r terms(sucesses)

G X  X !  X ( t ) 

G )  u ! u G X (t ) X ( t ) u G X (t 

r terms

r factors

§ pt · ¨ ¸ © 1  qt ¹

pt pt pt u u !u 1  qt 1  qt  1  qt

c

r

=

5 a

X 2 a B(n2 , p ), G X 2 (t ) G X1 ( t ) u G X 2 ( t ) (q  pt )

b

(q  pt )n1 ½° ¾ Ÿ G X1  X 2 ( t ) (q  pt )n2 ¿°

n1  n2

d

(q  pt )n1 u (q  pt )n2

1  P( X d 1)

4t , G c(t ) 5t

G (t )

r factors

X1 a B(n1 , p ), G X1 (t )

20(5  t )2 ,

G cc(t )

 40(5  t )3 u ( 1)

Let’s assume that for n = k this statement is true: k

i

i 1

· § k a B ¨ ¦ ni , p ¸ ¹ ©i 1

d 4•t It=1 dt 5–t

5 4

d2 4•t It=1 dt2 5–t

5 8

5 5 5 + • 1– 8 4 4

5 16 3/99

G X (t ) = e0.6( t −1) , GY (t ) = e0.12( t −1) , GZ (t ) = e0.28( t −1) b G X Y  Z (t ) G X (t ) u GY (t ) u G Z (t )

2 a

3rd Step k 1

For n

40(5  t )3

5§ 5· 5 5 5 = ¨1 − ¸ = − 4© 4 ¹ 8 16 16

( ) ( ) ( )

2nd Assumption

1 25

CH1 Rewiev ...ise

1.1

X1  X 2 a B(n1  n2 , p )

24 25

G cc(1)  G c(1) (1  G c(1))

= 40 × 4 −3 +

For n = 1, the statement is trivial: X1 a B(n1 , p ) For n = 2, in part a we proved that

1

4 (5  t )  4 t (5  t ) 2

G c(t )

Ÿ Var( X )

Ÿ X1  X 2 a B(n1  n2 , p )

4 · §4  ¸ © 5 25 ¹

1¨

624 625

20 20 5 Ÿ E ( X ) = G ′(1) = = 2 (5 − t ) 16 4

1st Base

¦X

4 4 4 4    5 25 125 625

Ÿ P 1 d X d 4

k  1, ¦ X i i 1

k

¦X

i

 X k 1 ,

= e0.6( t −1) × e0.12( t −1) × e0.28( t −1) = et −1 G X Y  Z (t ) et 1 Ÿ E( X  Y  Z )

i 1

by the assumption the first term has a binomial distribution and by the given proposition in the question X k 1 ~ B(nk 1 , p ). Now we use the fact from part a that the sum of two binomial variables with the same probability p satisfies the following:

G X Y  Z (1)

c

k 1

§ k 1 X ~ B ¦ i ¨ ¦ ni , i 1 ©i 1

1

G X +Y + Z (t ) = et −1 Ÿ X + Y + Z ~ Po(1)

P ( X + Y + Z ≥ 2 ) = 1 − P ( X + Y + Z ≤ 1) = 0.264

·½ § k k X ~ B ¦ i ¨ ¦ ni , p ¸ ° Ÿ X  X ~ i 1 ¹¾ ©i 1 ¦ i k 1 i 1 X k 1 ~ B(nk 1 , p ) °¿ · p¸ Ÿ ¹

e0

The expected number of animals at the meadow at the given moment is 1.

k

§ k B ¨ ¦ ni  nk 1 , ©i 1

e11

1.1

1.2

CH1 Rewiev ...ise

1–poissCdf(1,0,1)

0.264241

· p¸ ¹

4th Conclusion

k § k Now we can conclude that Y = ¦ X i ~ B ¨ ¦ ni , i 1 ©i 1 for all the positive integer values of k.

· p¸ ¹

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

1/99

Statistics: Chapter 1

10

WORKED SOLUTIONS X ~ Geo(0.75) Ÿ P(X = 4) = 0.0117

3 a i

5 a

G (1) = 1 Ÿ

Y ~ NB(2, 0.75) Ÿ P(Y = 4)

ii

§ 4  1· 2 2 ¨ ¸ 0.25 u 0.75 © 2  1¹ iii

P ( X t 3)

iv

Z ~ NB(6, 0.75)

27 256

1  P( X d 2 )

Ÿ P  Z d 10

0.105

b

§ 6  1· 6 ¨ ¸ 0.75 © 6  1¹

§ 7  1· 6 ¨ ¸ 0.75 u 0.25  ! © 6  1¹

1.2

1.3

nCr(3,1)·(0.25)

 a  bt a

(a − b )

=

2

2

a =a 12

G cc(t )

a u (2)(a  bt )3 u ( b )

2ab (a  bt )3

G cc(1)  G c(1) 1  G c(1) 2a (a  1)  a(1  a ) 13

0.922

2a 2  2a  a  a 2 6 a

a2  a

X1 ~ Po(m ), G X1 (t ) = em ( t −1) ½° ¾ Ÿ G X1 + X 2 ( t ) X 2 ~ Po(m ), G X 2 (t ) = em ( t −1) °¿ = G X1 (t ) × G X 2 (t ) = em ( t −1) × em ( t −1) = em ( t −1) + m ( t −1)

e2 m ( t 1) Ÿ X1  X 2 ~ Po(2m ) 0.011719

2·(0.75)2

a 1

a

a (a  bt )2 ,

CH1 Rewiev ...ise

geomPdf(0.75,4)

1Ÿ b

G c(t )

Var( X )

students needed to be selected if we need six involved in the programme is eight. 1.1

1Ÿ a  b

Ÿ E( X ) = G ′(1) = c

6 = 8, the expected number of b E( Z ) = 0.75

ab

a  bt  t u ( b ) a  bt 2

G c(t )

0.0625

§ 10  1· 6 4 ¨ ¸ 0.75 u 0.25 ©6  1 ¹

1

b

approxFraction(5.I 27 256

1–geomCdf(0.75,2)

1st Base For n =1, the statement is trivial: X 1 ~ Po(m ) For n = 2, in part a we proved that X1 + X 2 ~ Po(2m )

0.0625

2nd Assumption Let’s assume that for n = k this statement is true:

3/99

k

CH1 Rewiev ...ise 1.1 1.2 1.3 1–geomPdf(0.75,4)

¦X

0.0029

(nCr(5+k,5)·(0.25) )

k 1

k

k=0

8.

F (2) a

b

1Ÿ

22 a2

1 Ÿ a 2 Ÿ 22 Ÿ a

2 or

 2 since a must be positive.

F c( x )

f (x ) Ÿ f (x )

­ 2x ° 2,0d x d2 ®2 °¯ 0, otherwise

­x ° ,0d x d2 ®2 °¯ 0, otherwise c

¦X

i 1

5/99

4 a

k

k  1, ¦ X1

For n

0.921873

6 0.75 |

~ Po(km )

3rd Step

4 (0.75)6·

1

i 1

Since f (x) is increasing on the whole interval [0, 2] the modal value of the variable X is 2.

i

 X k 1, by the

i 1

assumption the first term has a Poisson distribution with the coefficient km and by the given proposition in the question X k 1 ~ Po(m ). Now we use the fact that the coefficient of the sum of two Poisson variables is the sum of the coefficients of those two variables. k

½ ~ Po(km ) ° ¾Ÿ i 1 X k 1 ~ Po(m ) °¿

¦X

i

k 1

Ÿ

¦X

i

k

¦X

i

 X k 1 ~ Po(km  m )

i 1

~ Po  (k  1)m

i 1

4th Conclusion n

Now we can conclude that Y = ¦ X i ~ B(nm ) k =1

for all the positive integer values of n.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 1

11

WORKED SOLUTIONS

Expectation algebra and Central Limit Theorem

2 Skills check np 1

Exercise 2A

2½

3 Ÿq 2

° 3 ¾ Ÿ 2q 2 °¿

npq

1 ,n 4

p

§8· §1· §3· ¨ ¸¨ ¸ ¨ ¸ ©2¹ ©4¹ ©4¹

2)

6

0.311 k

3

§8·§1· §3· ¨ ¸ ©4¹

k 1

1 binomPdf 8, 4 ,2

0.311462

1 binomCdf 8, 4,1,3

0.786072

( ) ( )

3 u 1 .2

10.8

b

E( X  3)

5 .3  3

8.3, Var( X  3)

c

E( 4 X  1)

3

3

2

6a  9 a  2 Ÿ 6a  7 a  9 a  2

Ÿ (2a  1)(3a  1)(a  2)

e

4 u 1 .2

19.6

2 u 5 .3  5

5.6,

2 2 u 1 .2

E(kX  p )

4 .8

k u 5.3  p,

Var(kX  p ) 2·

§ ©

0

3

10 u

2 3 u 5 5

a

E(3 X  2)

b

Var (3 X  2)

1 (–0.5,0) (–0.333,0)

§2· © ¹

1

–3

1 3 2 §2· ¨ ¸ ©3¹

–3

1 2

 , a2

1 3

2

We see that all three values of a give positive values of the parameter m: 7 , m2 4

7 , m3 9

3 4

1 2 3

3u4  2 9u

3 2

1.5,

2u

3 1 2

4

22 u Var (Y )

Y ~ Po(2) Ÿ E(Y ) a

E(3  2Y )

b

Var (3  2Y )

14 108 5

12 5

21.6

0.75

2 u E(Y )  1

Var (2Y  1)  , a3

2 .4

3

E(2Y  1)

4,

32 u Var ( X )

Y ~ Geo ¨ ¸ Ÿ E(Y ) 3

(2,0) x

12 5

2 5

3 u E( X )  2

CH2 Skills check

1.2

10 u

¹

Var( X )

0

k 2 u 1.2 = 1.2k 2

X ~ B ¨ 10, ¸ Ÿ E( X ) 5

Var(Y )

m1

1 .2

22.2,

2

E(2 X  5)

3 y f1(x)=6·x3–7·x2–9·x–2

Ÿ a1

4 u 5.3  1

Var(2 X  5)

Var ( X )

X ~ Po(m ) Ÿ E( X )

1.1

Var(3 X )

d

0.786

2/99

7a

15.9,

2

8k

2

2

3 u 5 .3

E(3 X )

1 .2

CH2 Skills check

1.1

2

5.3, Var ( X )

Var( 4 X  1)

¦ ¨© k ¸¹ ¨© 4 ¸¹

P(1 d X d 3)

b

E( X ) a

§ 1· 8 Ÿ X ~ B ¨ 8, ¸ © 4¹ 2

P( X

a

1

3 , 4

4u

2, Var (Y )

3  2 u E(Y ) 22 u Var (Y )

3 4

2, 3

2

32u2 4u2

1 8

28

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 2

12

WORKED SOLUTIONS 5

1· 3¹

§ ©

8 1 3

X ~ NB ¨ 8, ¸ Ÿ E( X ) 2 3 2 §1· ¨ ¸ ©3¹

Exercise 2B

24,

1 a

a b 6

b

22 u Var ( X )

Var (2 X  11)

X ~ B(15, p ) Ÿ E( X )

18 5 u 5 2

E(X )

³ 0

1 6 6 9 6

Var( X )

³ 0

1 3

3 5

6u

45 192

6Ÿ p 18 5

c

d

2 5

E(2Y  Z )

e

3.6

6

³x

2

1 ª x3 º « » 3 ¬ 3 ¼0

dx

0

2 6 3

E(2 Z  7 X )

6

³

x 3 dx 

0

6

1 ªx4 º 8 « »  3 ¬ 4 ¼0 3

8 3

8 3 3

3

1 3

3  ( 5)  12

E( X  Y  Z )

∫(

1.63299 1

4

)

·x dx–(1.6329931618555)2

CH2 Exercise A

1.1

∫

0 .5  1 . 4  2 . 8

4 .7

3 u 12  2 u 3  4 u ( 5)

10 ,

3 2 u 2 .8  2 2 u 0 . 5

5 Ÿ Y ~ Po(5) Ÿ Var (Y )

2 and 5

a

E(3 X  5Y )

3u2  5u5

b

Var(11Y  7 X )

31

112 u 5  7 2 u 2

703

E( X )

9 Ÿ n1 p

9 Ÿ Var ( X )

9(1  p )

E(Y )

4 Ÿ n2 p

4 Ÿ Var ( X )

4 (1  p )

22 u 9(1  p )  32 u 4 (1 – p) 72  72 p

1/3

x2·

 14,

2 Ÿ X ~ Po(2) Ÿ E( X )

Var ( X )

)

0 2 ·√6 3

(

3  ( 5)  12

4 .7

49.6

1 x· ·x dx 3

2 ·√6 3 √6

20,

0 .5  1 . 4  2 . 8

E(3Z  2 X  4Y )

35.7

1.63299

√6

x2·

3, 2

E( X  Y  Z )

Var(2 X  3Y )

CH2 Exercise A

1.1

√6

2

2

1 8 u 62  12 3

∫ (

2 u 12  7 u 3

2 u 2 .8  7 u 0 .5

E(Y )

2

8 .4

Var(2 Z  7 X )

 4 2 u 1 .4 2

1 §2 · x u x dx  ¨ 6 ¸ 3 3 © ¹

2 u 1 .4  2 .8

Var(3Z  2 X  4Y )

6

 22,

2

Var( X  Y  Z ) f

1 .9

2 u ( 5)  12

Var( X  Y  Z )

90

1 3

1 x u x dx 3

4 u 48

np Ÿ 15 p

npq Ÿ Var ( X )

6

2 u 24  3

 2,

0 .5  1 . 4

Var(2Y  Z ) 2 u E( X )  3

Var(5X  3)

7

48

E ( 2 X  3)

Var( X )

3  ( 5)

Var( X  Y )

8u

Var(Y )

E( X  Y )

1.63299

)

0 0.333333| 1/3

Var (3X  2)

3u

2 62 3

1 32 u 3

3

8Ÿ

8 12 8  12 p p

Var( X  Y )

1 ·x dx–(1.6329931618555)2 3

E(3X  2)

r1 8Ÿ p 1 p 8 Var( X ) 8 u 8 p p r E(Y ) 12 Ÿ 2 12 Ÿ p 1  p 12 Var(Y ) 12 u  12 p p

E( X )

2 6  2,

5

20  20 p

Given the n independent variables and their corresponding parameters X n , E( X n ), Var ( X n ), n  '  1st Base n=1 X1 , E( X1 ), Var ( X1 ) Ÿ E(aX1 ) Var (aX1 )

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

aE ( X1 ),

2

a Var ( X1 ) Statistics: Chapter 2

13

WORKED SOLUTIONS 2nd Assumption

b

Since we have 6 independent flips of the same coin we add six instances of the variable X.

c

Find the expected value and variance of the random variable Y.

Let’s assume that for n = k the statement is true so k

¦a X

the parameters of

i

i

are calculated as follows:

i 1

k

k

· § E ¨ ¦ ai X i ¸ ¹ ©i 1

¦ a E( X ) and i

E(Y )

i

i 1

1 2

· § Var ¨ ¦ ai X i ¸ ¹ ©i 1

¦ a Var ( X ) 2 i

d

i

E(Y ) r 3 u Var (Y )

1 4

6u

3, Var (Y )

k

k

3 2

3 2

3r3u

i 1

3 r 3.67 Ÿ y  [0.67, 6.67]

3rd Step k 1

For n

6u

k

k  1, ¦ X i

¦X

i 1

· § k 1 E ¨ ¦ ai X i ¸ ¹ ©i 1

We notice that all the possible outcomes {0, 1, 2, 3, 4, 5, 6} are covered by the 99.7% empirical rule.

 X k 1 ,

i

i 1

2

· § k E ¨ ¦ ai X i  ak 1 X k 1 ¸ ¹ ©i 1

Let’s roll an unbiased die 4 times. Record the number of multiples of 3 obtained. a

· § k E ¨ ¦ ai X i ¸  E(ak 1 X k 1 ) ¹ ©i 1 k

¦a EX  a i

k 1

i

E  X k 1

i 1

· § k 1 Var ¨ ¦ ai X i ¸ ¹ ©i 1

k 1

i

¦ ai2 Var  X i

1

P{X = x}

2 3

1 3

0u

i 1

ak21 Var  X k 1

2 1  1u 3 3

02 u

Var( X )

· § k Var ¨ ¦ ai X i  ak 1 X k 1 ¸ ¹ ©i 1 k

0

E( X )

¦a EX i

X=x

1 , 3

2 1 §1·  12 u  ¨ ¸ 3 3 ©3¹

2

1 1  3 9

2 9

b

Since we have 4 independent rolls of the same die we add four instances of the variable X.

c

E(Y )

i 1

· § k 1 = Var ¨ ¦ ai X i ¸ ¹ ©i 1

3 a

4th Conclusion b

i

i

c

i 1

2 i

i

0

1

P{X = x}

1 2

1 2

Var( X )

3u3

2

3 u Var ( X )

3u3

9,

3u4

12

9, 9u4

36

3 u Var ( X ) 12 u 4

12 Var ( X )

62 Var ( X )

8 9

36 u 4

48

144 Ÿ

Var ( X  X  X  3 X ) z Var (6 X )

X=x

E( X )

2 9

3 u Var ( X )

Var ( X  X  X  3 X )

Var (6 X )

i 1

4u

3 u E( X )

3 u E( X )

 3 u Var ( X )

n

¦ a Var ( X ) 4 a

Let’s independently flip 6 unbiased coins and record the number of heads obtained. a

E(3 X )

2

Exercise 2C 1

4 , Var (Y ) 3

E( X  X  X )

Var (3 X )

n

¦ a E( X ),

· § n Var ¨ ¦ ai X i ¸ ¹ ©i 1

1 3

Var ( X  X  X )

The parameters of a linear combination of n random variables can be calculated by: · § n E ¨ ¦ ai X i ¸ ¹ ©i 1

4u

0u

1 1  1u 2 2

02 u

E( X  X  X  X  X ) Var ( X  X  X  X  X )

b

E(Y  Y  Y ) Var (Y  Y  Y )

c

1 , 2

1 1 §1·  12 u  ¨ ¸ 2 2 ©2¹

3 u E(Y )

5 u E( X )

5u2

5 u Var ( X )

3u5

3 u Var (Y )

10

5u1

5

15

3u3

9

E( X  X  X  X  X  Y  Y  Y ) E( X  X  X  X  X )  E(Y  Y  Y )

2

1 1  2 4

1 4

10  15

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

25

Statistics: Chapter 2

14

WORKED SOLUTIONS Var(X + X + X + X + X + Y + Y + Y )

5 a

6

E( XY ) Ÿ P u ( P  2)

E( X ) u E(Y )

Var( X  X  X  X  X )  Var (Y  Y  Y )

Ÿ P 3  P 2  2P

59

Ÿ P ( P  1)( P  2)

14

X = xi

1

2

3

4

P{X = xi}

1 4

1 4

1 4

1 4

1 u (1  2  3  4 ) 4

E( X )

Ÿ P

0 or P

2

3

P(Y = yi)

1 2

1 3

1 6

1 1 1 u1 u 2  u3 2 3 6

E(Y ) b

1 1 u 4 2

1: P ^1, 1`

5 3

1 1 u 4 3

8 : P ^4, 2`

1 1 u 4 6

9 : P ^3, 3`

5 24 b

1 6

1 1 1 1 u  u 4 3 4 2 1 1 1 1 u  u 4 6 4 3

5 24 1 8

0.16  0.25  1.21 1.62

P

0  1  ( 2)  3

0 .5 2,

1  0.16  0.25  1.21

Ÿ P( X  Y  Z  W ! 0 ) c

P

3 u 0  1  ( 2)  3

V2

1 24

2.62

0.892

0,

32 u 1  0.16  0.25  1.21 10.62

3 X  Y  Z  W ~ N (0, 10.62) Ÿ P(3X  Y ! Z  W )

1 24

P(3 X  Y  Z  W ! 0 )

3

4

6

8

9

12

P(Z = zi)

1 8

5 24

1 6

5 24

1 8

1 12

1 24

1 24

d

P

1  22 u 0.16  32 u 0.25  1.21

5 .1

Ÿ P( X  3Z d 2Y  W )

5 1 1  6u  8u 24 8 12 1 1 25 9u  12 u 24 24 6

E( Z )

 1,

X  2Y  3Z  W ~ N (1, 5.1)

4u

25 6

0 .5

0  2 u 1  3 u ( 2)  3

V2

1 5 1 2u 3u 8 24 6

5 5 u 2 3

0, V 2

X  Y  Z  W ~ N (2, 2.62)

2

E( X ) u E(Y )

1  ( 2)  3

V2

1

1u

P

Ÿ P(Y  Z  W  0 )

Z = zi

E( Z )

5

Y  Z  W ~ N (0, 1.62)

1 12

1 1 u 4 6

12 : P ^4, 3`

1

Exercise 2D 1 a

1 1 1 1 3 : (1, 3) or (3, 1) o u  u 4 6 4 2

6 : P ^2, 3` or ^3, 2`

(2,0)

–3

1 1 1 1 u  u 4 3 4 2

4 : P ^2, 2` or ^4, 1`

1 or P = 2.

f1(x)=x3–x2–2·x 1 (0,0) (–1,0)

5 2

1 8

2 : P ^1, 2` or ^2, 1`

0

3 y

–5

1

0

CH2 Exercise C

1.1

1 u 10 4

Y = yi

P3

P( X  2Y  3Z  W d 0 ) e

E( XY )

P

0.329

0  4 u (2)  3 u 0  2( 2)

4,

σ 2 = 1 + 4 2 × 0.25 + 32 × 1 + 22 × 0.25 = 15 X  =  X  2 Z  N(4, 15) Ÿ P( X   = d X  2 Z ) P( X  4 Z  3 X  2 Z d 0 )

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

0.151

Statistics: Chapter 2

15

WORKED SOLUTIONS f

P

3 u 1 2 u 3

σ2

 9,

c Y a N (60, 22 ) Ÿ 4Y a N(4 u 60, 4 2 u 22 )

1.21  0.16  22 u 0.16  32 u 1.21 12.9

= N(240, 82 )

W − Y − 2Y − 3W  N( −9, 12.9)

X  4Y a N(225  240, 122  82 )

Ÿ P(W  Y d 2Y  3W ) = P(W  Y  2Y  3W d 0 ) = 0.994



= N 15,  4 13



2

CH2 Exercise D

1.5

normCdf(–1000,0,0,√1.62)

Ÿ P  X  4Y ! 0

0.5

normCdf(0,1000,2,√2.62)

0.891697

normCdf(–1000,0,–1,√5.1)

0.671047

normCdf(–1000,0,2,√14.25)

0.298121

normCdf(–1000,0,–9,√17.21)

0.984976

d

0.149

Y ~ N  60, 22 Ÿ Y  Y  Y  Y ~ N  4 u 60, 4 u 22 N(240, 4 2 )

5/99

2 a

X  Y  Y  Y  Y a N  225  240, 122  4 2

X a N (240, 20 2 ), Y a N (730, 50 2 )



N 15,  4 10

Ÿ Y  3 X a N(10, 50 2  32 u 20 2 ) Ÿ P(Y  3 X t 0 )

Ÿ P X  Y  Y  Y  Y ! 0

2

2

X a N (240, 20 ), Y a N (730, 50 )

Ÿ P(2Y  4 X t 2500 )

0.162

L a N (35, 52 ) Ÿ P( L  30 )

1.3

0.118

CH2 Exercise D

1.4

normCdf(250,9.E999,225,12) normCdf(1000,9.E999,900,24) 1.5463523972415E-5 normCdf(0,9.E999,–15,4·√13) normCdf(0,9.E999,–15,4·√10)

Ÿ 2Y  4 X a N(2420, 2 u 50 2 u 4 u 20 2 )

3 a



0.551 1.2

b

2

0.01861 0.000015 0.000015 0.149155 0.11784

0.159 5/99

b

2

V a N( 45, 8 ) Ÿ  V  V  V  V a N(5 u 45, 5 u 82 )

W



N 225,  8 5 P (W ! 240 ) c

2



Exercise 2E 1 a

0.201

N 5,  4 61 1.1

1.2

1.3

2

§ 2 · ¸ © 10 ¹

©

b

4 L  3V a N( 4 u 35  3 u 45, 4 2 u 52  32 u 82 )



§

Ÿ P( 4 L  3 L ! 0 )

c 0.158655

normCdf(240,9.E999,225,8·√5)

0.200868

normCdf(0,9.E999,5,4·√61)

0.563578

§ 3 · ¸ © 7¹

P  2 d X d 8

CH2 Exercise D

normCdf(–9.E999,30,35,5)

§

X a N ¨¨ 5, ¨ ©

0.564

2

X a N ¨¨ 1.5, ¨

§

2

· ¸¸ Ÿ P X  5 d 3 ¹



§ 4 · ¸ © 22 ¹

©

b

X a N (225, 12 ) Ÿ P( X ! 250 )

( ( (

)

3 normCdf 2,8,5, √7

X  X  X  X a N ( 4 u 225, 4 u 122 )

Ÿ P( X  X  X  X ! 1000 )

· ¸¸ Ÿ P  X t 0.8 ¹

0.639

0.776549

)

4 normCdf –0.8,0.8,–0.2, √22

0.0186



CH2 Exercise E

2 normCdf 1,3,1.5, √10

4 a

2

1  P  0.8 d X d 0.8

3/99

0.777

0.992

X a N ¨¨ 0.2, ¨

1.1

2

· ¸¸ Ÿ P 1 d X d 3 ¹

0.991849

)

0.63867

N (900, 24 2 )

3/99

0.0000155 2

§

§ 1.2 · ¸ © 15 ¹

X  N  5.5, 1.22 Ÿ X  N ¨¨ 5.5, ¨ ©

Ÿ P  X  5 © Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

2

· ¸¸ ¹

0.0533 Statistics: Chapter 2

16

WORKED SOLUTIONS 3

§

§ 6 · ¸ © 5¹

X  N 35, 62 Ÿ X  N ¨¨ 35, ¨ ©

a b

2

§

· ¸¸ ¹

e



§

( ( (

· ¸ 120 © ¹

©

2

· 1 ¸¸ Ÿ P §¨ X  2 t ·¸ 5¹ © ¹

P 1.8  X  2.2

CH2 Exercise E

1.2

0.0983

6

§

X  N ¨¨ 1.9, ¨

f

⇒ P (X + X + X + X + X > 180) = 0.355 1.1

· ¸ Ÿ P §¨ X  1 ·¸ ¸ 2¹ © ¹

1· § 1 X  ¸ 2¹ © 2

X + X + X + X + X ~ N(5 × 35, 5 × 6 ) 2

2

P¨ 

2



6 · ¸¸ © 40 ¹

¨ ©

P( X t 37) = 0.228

N 175,  5 u 6

§

X  N ¨ 1, ¨¨

) √ )

1.2 normCdf 5,9.E999,5.5,√15

0.946708

6 normCdf 37,9.E999,35, 5

0.228028

3 normCdf 1.5,2.5,2 √30

0.354694

0.2 normCdf 1.25,1.35,1.3, √50

CH2 Exercise F

1.1

normCdf 180,9.E999,175,√5 ∙6

)

0.280

( (

)

0.63869

)

0.9229

normCdf(–0.48,9.E999,–0.5,0.1)

0.42074

|

3/99

3/99 2 § § 18.5 · · 4 a X  N(62.5, 18.5 ) Ÿ X  N ¨¨ 62.5, ¨ ¸ ¸¸ © 12 ¹ ¹ ©

2

P  X d 70 b

( ( (

X  !  X  N(12 u 62.5, 12 u 18.5 ) X



N 750,  4107

1.1

2

0.035295 0.098299

√ )

–1 1 6 normCdf 2 , 2 ,1, 40

12 terms

⇒ P(T

√ )

234 normCdf –9.E999,397,400, 85

0.920 2

T

CH2 Exercise F

1.1



0.280493

)

6 normCdf 1.8,2.2,1.9,√120

≤ 800) = 0.782

1.2

1.3

6/99

CH2 Exercise E

( ( (

)

2

)

§ 8000 · ¸ © 15 ¹

X  N ¨¨ 30 000, ¨ ©

)

normCdf –9.E999,800,12∙62.5,√12.18.5 0.782364 normCdf –9.E999,800,750,√4107

§

0.919895

18.5 normCdf –9.E999,70,62.5, √12

P( X d 32 000)

2

· ¸¸ ¹

0.834

0.782364 1.1

|

1.2

CH2 Exercise F

(

)

8000 normCdf –9.E999,32000,30000, √15

3/99

0.833539 |

Exercise 2F § § 3 ·2 · ¸ Ÿ P(1.5 d X d 2.5) 1 a X  N ¨ 2, ¨ ¨ © 30 ¸¹ ¸ © ¹ b

§

§ 0.2 · ¸ © 50 ¹

X  N ¨1.3, ¨ ¨ ©

2

0.639

· ¸ Ÿ P(1.25 d X d 1.35) ¸ ¹

0.923

2 § § 1 · · c X  N ¨ 0.5, ¨ ¸ ¸ Ÿ P( X t  0.48) © 10 ¹ ¹ ©

0.421

2 § § 234 · · ¨ d X  N 400, ¨¨ ¸ ¸ Ÿ P  X  397 ¨ 85 ¸¹ ¸ © © ¹

0.0353

1/99

3

§

§ 2 · ¸ © 3¹

X  N(63, 22 ) Ÿ X  N ¨¨ 63, ¨ ©

a

P  X ! 64

0.193

b

P( X  60)

0.00469

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

2

· ¸¸ ¹

Statistics: Chapter 2

17

WORKED SOLUTIONS c

P(61 d X d 65) 1.1

§ § 4 ·2 § 9 ·2 · ¸ ¸¸ ¸ ¨ © 4¹ ¹ © © 5¹

0.917

B  S a N ¨¨ 4, ¨

CH2 Exercise F

1.2

( ( (

2 normCdf 64,9.E999,63, √ 3 normCdf –9.E999,60,63, 2 normCdf 61,65,63, √ 3

2 √3

) )

Ÿ PB ! S

0.193238

P  B  S ! 0

0.796

0.004687 CH2 Rewiew ...ise

1.1

(

0.916736

)

√( ) ( ) )

4 2 9 2 normCdf 0,9.E999,4, √5 + 2

|

0.795603

4/99

§ § 4 ·2 · 4 X  N(6, 4 ) Ÿ X  N ¨¨ 6, ¨ ¸ ¸¸ © © n¹ ¹ 2

§

a

b

2

· ¸¸ Ÿ P(5.4 d X d 7.6) 0.579 ¹ 76· §56 dZ d ¨ 4 4 ¸ 0.95 Ÿ P ¨ ¸¸ ¨ n ¹ © n

§ 4 · ¸ © 10 ¹

X  N ¨¨ 6, ¨ ©

P(5 d X d 7)

§ n· 0.95 Ÿ P ¨ Z d ¸ © 4 ¹

0.975

2 a

Var(2X ) = (E(X ))2 – 5 ⇒ 22Var(X ) = (E(X ))2 – 5 ⇒ 4 m = m2 – 5 ⇒ m2 – 4m – 5 = 0 ⇒ (m – 5)(m + 1) = 0 ⇒ m = 5 or m Variance is always positive.

b

P(X ≥ 6) = 1 – P(X ≤ 5) = 0.384

c

Var (3Y )

18 Ÿ 32 Var (Y )

Ÿ Var (Y )

n Ÿ 4

) (0.975) Ÿ n

Ÿ n

7.839856 Ÿ n

1.1

1/99

1

1.2

1.3

4 u 1.95996

(

2 Ÿ Y  Po(2)

Ÿ X  Y  Po(7 ) Ÿ P( X  Y  5)

61.4633 | 62

P( X  Y d 4 )

)

18

Ÿ X  Y  Po(5  2)

CH2 Exercise F

4 normCdf 5.4,7.6,6, √10

1 .

0.173

0.57942 1.1

|

1.2

CH2 Rewiev ...ise

1–poissCdf(5,0,5)

0.384039

poissCdf(7,0,4)

0.172992

1/99 2/99 1.2

y 1

1.3

1.4

CH2 Exercise F

E(Z ) = E(3X – 4Y) = 3 × 5 – 4 × 2 = 7 Var(Z ) = Var(3X – 4Y ) = 32 × 5 + 42 × 2 = 77 e The random variable Z has no Poisson distribution since 7 = E(Z) ≠ Var(Z) = 77. d

0.5

f1(x)=normCdf(5,7,6,

4 )–0.95 √x

0 10

(61.5,0)

80 x

–0.5

3 a b

Review exercise 1

X  N  400, 20 2 Ÿ P( X ! 450 )

Y  N(350, 152 ) Ÿ Y  Y  N(2 u 350, 2 u 152 ) Ÿ P(Y  Y  670 )

§

§ 4 · ¸ © 5¹

2

B  N(202, 4 2 ) Ÿ B  N ¨¨ 202, ¨ ©

§

§ 9 · ¸ © 4¹

S  N(198, 92 ) Ÿ S  N ¨¨ 198, ¨ ©

2

· ¸¸ ¹

· ¸¸ ¹

c

0.00621

0.0787

Z  N(320, 122 )

Ÿ X  Z  N(400  320, 202  122 )



N 720,  4 34

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

2



Statistics: Chapter 2

18

WORKED SOLUTIONS d



2

X  Z  2Y  N 720  2 u 350,  4 34  22 u 152



c

§ ©

N  20, 382 Ÿ P( X  Z  2Y  0 ) 1.1

1.2

1.3

1.3

y 1

0.00621

normCdf(–9.E999,670,700,15∙√2) 0.07865 normCdf(–9.E999,0,20,38)

2

· ¸¸ ¹

Ÿ P(1050  X  12500)

0.299

CH2 Rewiew ...ise

normCdf(450,9.E999,400,20)

§ 3200 · ¸ © n ¹

X  N ¨¨12000, ¨

0.5

1.4

0.85 Ÿ n

44.3 | 45

CH2 Rewiew...ise

1.5

(

)

3200 f1(x)=normCdf 10500,12500,12000, –0.85 √x

0.299334 0 10

100 x

(44.3,0)

–0.5 3/99

4 a

X  B(n, p ) Ÿ Var ( X ) n

27 Ÿ 27 u pq

Ÿ 9 p2  9 p  2 1 or p 3

p

E(3X  7)

6 a

2 9

E  X  E( X )

2



E X 2  2 X E( X )   E( X )

0

E  X 2  2  E( X )   E( X ) 2

3np  7

1 3 u 27 u  7 3 2 3 u 27 u  7 3

b

X  Geo( p ) and Y  Geo(q ), where p  q

20 Ÿ E( X  Y )

E( X )  E(Y )

47 q  p pq

E( Z )  12

0 Ÿ (m  4 )(m  3)

Ÿ Var ( X  Y )

0

1 pq

Var ( X )  Var (Y ) q 3  p3 p2q 2

1 

 (q  p ) (q 2  pq  p 2 ) p2q 2

X  N (12 000, 3200 2 ) Ÿ P( X > 13 000) = 0.377

2 § § 3200 · · b X  N ¨ 12 000, ¨ ¸ ¸¸ Ÿ P( X  11500) ¨ © 30 ¹ ¹ ©

1

1 1  p q

Var(5 + 2Z) = 22 Var(Z ) = 4 × 4 = 16

q

2

 p  3 pq p2q 2

0.196

1  3 pq p2q 2

E( X  Y )(E( X  Y )  3)

CH2 Rewiew ...ise

normCdf(13000,9.E999,12000,3200) 0.37733 3200 normCdf –9.E999,11500,12000, √30 0.19605

(

2

t0

q p  2 p2 q

1.4

2

E  X 2   E( X ) t E  X 2  

m = 4 or m = −3 Variance is always positive.

1.3



2

Z a Po(m ) Ÿ ( Var ( Z ))2

1.2

2

E  X 2  2E( X ) u E( X )   E( X )

0 Ÿ (3 p  1)(3 p  2)

3E( X )  7

Ÿ m 2  m  12

5 a

6 Ÿ p (1  p )

6

2 3

­ °° E(3X  7 ) Ÿ® °E(3X  7 ) °¯ b

6 Ÿ npq

)

· 1 § 1  3¸ ¨ pq © pq ¹ 1  3 pq p2q 2

1 3  2 pq p q 2

Var( X  Y ) Q.E.D.

2/99

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 2

19

WORKED SOLUTIONS

Exploring statistical analysis methods

3

Y ~ Po (0.4) Ÿ P (Y 0) = 0.670 b Y ~ Po (7) Ÿ P (3 d X < 8) = P ( X d 7)  P ( X d 2) = 0.569

Skills check 1

1.1

CH3 Skils check

1.2

A

3 a

B

C

D

=OneVar(a 1 22 Title One–Var... 1 3 37 x 3.61818 2 5 46 Σx 398. 3 7 5 Σx2 1750. 4 5 sx : = sn–... 1.68633 D2 =3.6181818181818 1.1

C

5 6 7 8 9 E6 =d62

x 2 a

b

D

1.678652 1·

§ ©

poissCdf(3,5)

0.916082

1–poissCdf(4.7,4)

0.505391

poissCdf(7,7)–poissCdf(7,2)

0.569078

0

5

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}Ÿ x 11, s 2 33 b {21, 24, 36, 28, 30, 22, 25, 26, 38, 32, 34, 29, 37, 33, 32, 34, 29, 37, 33, 31, 30}Ÿ x 29.75, s 2 25.3 67, s 2 1518 c {1, 4, 7, 10, !, 133} Ÿ x

1 a

1.1

5)

¹

§1· §1· ¨ ¸ ¨ ¸ ©2¹ ©2¹

Exercise 3A

1 32

0.03125

1·

§ ©

¹

= P(3 ≤ X ≤ 7) =

10 − k

§ 10 · § 4 · ¨ ¸¨ ¸ © k ¹©5¹

7

¦ k =3

0.322 1.3

1.3

CH3 Exercise A

11

varPop(seq(2·x,x,1,10))

33

k

3/6 1.1

1.2

1.2

1.3

CH3 Exercise A

CH3 Skils check

1 binomPdf 5, ,5 2 1 binomCdf 8, ,5 3

( (

(

§1· ¨ ¸ ©5¹

1.2

mean(seq(2·x,x,1,10))

mean({21,24,36,28,30,22,25,26,38,32,34,29 119 4 varPop({21,24,36,28,30,22,25,26,38,32,34,29 405 16

X ~ B ¨ 10, ¸ Ÿ P(3 ≤ X < 8) 5

5

0.67032

4/99

2.82

X ~ B ¨ 5, ¸ Ÿ P( X 2

1.1

CH3 Skils check

1.4

E

=OneVar(a sx : = sn–... 1.68633 σx : = σnx... 1.67865 2.81785 n 110. MinX 1. Q1X 3.

3.62, V 2

§5· =¨ ¸ ©5¹

1.3

CH3 Skils check

1.2

B

1.2

poissPdf(0.4,0)

nCr(8,k).

) )

varPop({21,24,36,28,30,22,25,26,38,32,34,2 405 16

0.03125 0.980338

8–k

k

( 23 ) .( 13 )

)

mean(seq(x,x,1,133,3)) varPop(seq(x,x,1,133,3))

2144 2187

k=0

(

1.3

CH3 Skils check

nCr(8,k).

( 3 ) .( 3 )

3 1–binomCdf 12, ,4 7

(

)

2187

0.640537

)

1 1 binomCdf 10, ,7 –binomCdf 10, ,2 5 5 0.322123

(

)

(

1518 6/99

3/99

1.2

67

|

k=0

1.1

119 4

)

2

x

0.65, s 1.1

B

0.864 CH3 Exercise A

1.2

C

D

E =OneVar(a 1 Title 22 One–Var... 2 x 12 0.65 3 Σx 4 26. 4 Σx2 2 46. 5 sx : = sn–... 0.863802 E5 =0.86380197160799

5/99

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

20

WORKED SOLUTIONS x 33.7 b s = 23.8

b

3 a

1.1

1.2

99% CI [0.104, 0.167] 1.1

B

C

CH3 Exercise B

1.2

zInterval 0.04,{0.1,0.12,0.15,0.18,0.13,0.12 "Title" "z Interval" "CLower" 0.104389 "CUpper" 0.16652 "x" 0.135455 "ME" 0.031066 "sx := sn–1x" 0.035317 11. "n" 0.04 "σ" 3/99

CH3 Exercise A

1.3

D

E =OneVar(a 1 Title 6 One–Var... 2 x 13 33.7143 3 Σx 26 2360. 4 Σx2 17 118550. 5 8 sx : = sn–... 23.7695 E5 =23.769510891752

c

95% CI [320, 325]

Exercise 3B 1.1

1 a

CH3 Exercise B

1.1

zInterval 1,5,10,0.99: stat.results "Title" "z Interval" "CLower" 4.18545 "CUpper" 5.81455 "x" 5. "ME" 0.814549 "n" 10. 1. "σ" 1/99

b

3

x  P d 2 Ÿ P  x d 2 Ÿ 2 d P  x d 2 Ÿx 2dPdx 2

95% CI [−12.8, − 9.20]

V

5.5 Ÿ 1.960 u

zInterval 4.3,–11,22,0.95: stat.results "Title" "z Interval" "CLower" –12.7968 "CUpper" –9.20318 "x" –11. "ME" 1.79682 "n" 22. 4.3 "σ"

Ÿ n

|

1.1

1.2

1.3

1.1

1.2

CH3 Exercise B

29.052

CH3 Exercise B

5 a

x

14.2  17.4 2

b

V

3 Ÿ 1.645 u

3/99

90% CI [3.90, 6.10]

5.39 Ÿ n

zInterval 3.5,{70,75,77,71,68,80,85,72},1,0 "Title" "z Interval" "CLower" 72.3247 "CUpper" 77.1753 "x" 74.75 "ME" 2.42533 "sx := sn–1x" 5.70088 "n" 8. "σ" 3.5 3/99

CH3 Exercise B

1.1

2

4 95% CI for the weight if elementines is [72.3 g, 77.2 g]

90% CI [2819, 2889] zInterval 327,2854,230,0.9: stat.results "Title" "z Interval" "CLower" 2818.53 "CUpper" 2889.47 "x" 2854. "ME" 35.4659 230. "n" 327. "σ"

n

We should take at least 30 elements.

2/99

2 a

5 .5

CH3 Exercise B

1.1

c

CH3 Exercise B

1.2

zInterval 4.5,{321,325,330,324,325,326,317 "Title" "z Interval" "CLower" 320.5 "CUpper" 325.214 "x" 322.857 "ME" 2.3572 "sx := sn–1x" 5.5727 14. "n" 4.5 "σ" 3/99

99% CI [4.19, 5.81]

15.8 3 n

15.8  14.2

Ÿ n 3.08 Ÿ n 9.51 The sample size should be 10.

zInterval 2,{1,2,3,4,5,6,7,8,9},1,0.9: stat.res "Title" "z Interval" "CLower" 3.90343 "CUpper" 6.09657 "x" 5. "ME" 1.09657 "sx := sn–1x" 2.73861 "n" 9. "σ" 2. 3/99

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

21

WORKED SOLUTIONS

Investigation

Exercise 3C

Let’s consider samples of different sizes: all have the mean value x 100 and they come from a population with V 2 64. a i

1 a

1.2

1.3

1.4

CH3 Exercise B

zInterval 8,100,10,0.95: stat.results "Title" "z Interval" "CLower" 95.0416 "CUpper" 104.958 "x" 100. "ME" 4.95836 "n" 10. "σ" 8. |

1/99

b

1.2

1.3

1.4

CH3 Exercise B

zInterval 8,100,25,0.95: stat.results "Title" "z Interval" "CLower" 96.8641 "CUpper" 103.136 "x" 100. "ME" 3.13594 "n" 25. "σ" 8.

2/99

c

|

1.2

1.3

1.4

CH3 Exercise B

3/99

2 a

|

1.1

150 Ÿ 95% CI [98.7, 101.3]

n

1.2

1.3

1.4

CH3 Exercise B

zInterval 8,100,150,0.95: stat.results "Title" "z Interval" "CLower" 98.7198 "CUpper" 101.28 "x" 100. "ME" 1.28024 "n" 150. "σ" 8.

b

4/99

At the same significance level the larger sample size we take the narrower a confidence interval we ª

get. In the formula « x  ¬

º V V u zD , x  u z D » the n n 2 2 ¼

length of the interval is symmetrical with respect to the mean of the sample by the same term that is obtained by division by n therefore the larger n we take the smaller number we obtain.

1.2

CH3 Exercise C

95% CI [0.112, 0.159] 1.1

|

b

99% CI [1.94, 8.06] tInterval {1,2,3,4,5,6,7,8,9},1,0.99: stat.results "Title" "t Interval" "CLower" 1.93696 "CUpper" 8.06304 "x" 5. 3.06304 "ME" 8. "df" "sx := sn–1x" 2.73861 9. "n" 1/99

3/99

iv

CH3 Exercise C

tInterval 3478,429,310,0.95: stat.results "Title" "t Interval" "CLower" 3430.06 "CUpper" 3525.94 "x" 3478. 47.9434 "ME" 309. "df" 429. "sx := sn–1x" 310. "n"

n = 50 Ÿ 95% CI [97.8, 102.2] zInterval 8,100,50,0.95: stat.results "Title" "z Interval" "CLower" 97.7826 "CUpper" 102.217 "x" 100. "ME" 2.21745 "n" 50. "σ" 8.

95% CI [3430, 3526] 1.1

2/99

iii

CH3 Exercise C

tInterval –23,5.8,32,0.99: stat.results "Title" "t Interval" "CLower" –25.8135 "CUpper" –20.1865 "x" –23. "ME" 2.81348 "df" 31. "sx := sn–1x" 5.8 32. "n"

25 Ÿ 95% CI [96.9, 103.1]

n

99% CI [−25.81, −20.19] 1.1

1/99

ii

CH3 Exercise C

1.1

tInterval 15,1.2,15,0.9: stat.results "Title" "t Interval" "CLower" 14.4543 "CUpper" 15.5457 "x" 15. "ME" 0.545722 "df" 14. "sx := sn–1x" 1.2 15. "n"

10 Ÿ 95% CI [95.0, 105.0]

n

90% CI [14.45, 15.55]

1.2

CH3 Exercise C

tInterval {0.1,0.12,0.15,0.18,0.13,0.12,0.09 "Title" "t Interval" "CLower" 0.111728 "CUpper" 0.159181 "x" 0.135455 "ME" 0.023726 "df" 10. "sx := sn–1x" 0.035317 11. "n" 2/99

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

22

WORKED SOLUTIONS c

90% CI [319.6, 324.9] 1.1

c

CH3 Exercise C

1.2

tInterval {321,325,330,324,325,326,317,318 "Title" "t Interval" "CLower" 319.612 "CUpper" 324.921 "x" 322.267 "ME" 2.65433 "df" 14. "sx := sn–1x" 5.83667 15. "n" 3/99

For the same set of data, a higher significance level will mean a wider confidence interval. The result is expected since the confidence interval is symmetrical about the mean value x r

s n

u tc

Exercise 3D 1 a

CH3 Exercise D

1.1

A

B

C

3 99% CI for the weight of mandarins is [67.6 g, 82.5 g] 1.1

1.2

1.3

1 15 2 17 3 23 4 17 5 18 C1 =–3

CH3 Exercise C

tInterval {66,70,75,84,90,77,71,68,80,85,72, "Title" "t Interval" "CLower" 67.6363 "CUpper" 82.5304 "x" 75.0833 "ME" 7.44705 "df" 11. "sx := sn–1x" 8.30617 12. "n" 1/99

4 a

x

b

s

12.345  14.355 2 1 .5

1 .5 Ÿ t c u

8

1.3

1.4

7) = 0.900

CH3 Exercise C

1.89505

99% CI [−2.18, 1.98] b

1.1

CH3 Exercise D

1.2

C

D

=a[]–b[] 8 1 9 –2 10 –1 11 12 F13

0.900069 |

2/99

The confidence level is 90%.

1.1

1.4

1.5

=d[]–e[]

5/99

90% CI [−13.73, −7.44] c

1.1

=d[]–e[] 4 –8 5 –14 6 –7 7 –13 8 –9 I8 =0.07

"t Interval" 462.981 609.019 536. 73.0187 14. 95. 15.

CH3 Exercise D

1.2

F

CH3 Exercise C

"Title" "CLower" "CUpper" "x" "ME" "df" "sx := sn–1x" "n"

"t Interval" –13.7287 –7.43794 –10.5833 3.14539 11. 6.06717 12.

|

99% CI [463.0, 609.0] 1.5

–9 –6 –3 –12 –18

CH3 Exercise D

1.2

CH3 Exercise C

1/99

1.4

F 102 108 109 111 103

"Title" "CLower" "CUpper" "x" "ME" "df" "sx := sn–1x" "n"

tInterval 536,95,15,0.95: stat.results "Title" "t Interval" "CLower" 483.391 "CUpper" 588.609 "x" 536. 52.6092 "ME" 14. "df" 95. "sx := sn–1x" 15. "n"

1.3

E 93 102 106 99 85

95% CI [483.4, 588.6] 1.3

"t Interval" –2.18091 1.98091 –0.1 2.08091 9. 2.02485 10. 2/99

tCdf(–1.895046173579,1.895046173579,7)

b

CH3 Exercise D

1.2

|

1.5

5 a

E

–3 2 0 –2 3

"Title" "CLower" "CUpper" "x" "ME" "df" "sx := sn–1x" "n"

13.35  12.345

Ÿ t c 1.895 Ÿ P( 1.895  t  1.895|v 1.2

1.1

18 15 23 19 15

13.35

(13.35–12.345)·√8

D

=a[]–b[]

G

H

I =g[]–h[]

0.58 0.82 0.77 0.9 1.02

0.69 0.78 0.65 0.8 0.95

–0.11 0.04 0.12 0.1 0.07

|

2/99

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

23

WORKED SOLUTIONS 1.1

b

CH3 Exercise D

1.2

"Title" "CLower" "CUpper" "x" "ME" "df" "sx := sn–1x" "n"

"t Interval" –0.060895 0.085895 0.0125 0.073395 7. 0.08779 8.

H0 : P 1.1

P0 , H1 : P z P0, D CH3 Exercise E

zTest 2,0.3,1.9,25,0: stat.results "Title" "z Test" "Alternate Hyp" "μ ≠ μ0" "z" –1.66667 "PVal" 0.095581 "x" 1.9 "n" 25. "σ" 0.3 |

6/99

2/99

95% CI [−0.069, 0.859]

Since the p-value is 0.095581 > 0.05 we have no sufficient evidence to reject the null hypothesis at the 5% significance level.

2 a

Blood sample

1

2

3

4

5

6

7

8

9

10 11 12

Bob

144 153 170 183 125 95 148 177 160 155 170 135

Rick

141 161 173 174 119 104 135 175 164 158 167 142

Bob– Rick

3 −8 −3 9

b

95% CI [−4.27, 4.60] 1.1

1.2

B

8 177 9 160 10 155 11 170 12 135 C12 =–7 1.1

1.2

1.3

6 −9 13

c

C =a[]–b[] 175 164 158 167 142

D

1.1

2 −4 −3 3 −7

E

2 –4 –3 3 –7

2 a

E F =tInterval( t Interval –4.26715 4.60048 0.166667 4.43381

CH3 Exercise E

H0 : P 1.1

P0 , H1 : P  P0, D

0 .1

CH3 Exercise E

zTest 10,4,9,20,–1: stat.results "Title" "z Test" "Alternate Hyp" "μ < μ0" "z" –1.11803 "PVal" 0.131776 9. "x" 20. "n" 4. "σ" | 4/99

H 0 : μ = μ0 , H1 : μ ≠ μ0, D 1.1

0.01

Since the p-value is 0.010566 > 0.01 we have no sufficient evidence to reject the null hypothesis at the 1% significance level.

Exercise 3E 1 a

P0 , H1 : P z P0 , D

3/99

CH3 Exercise D

C D =a[]–b[] 1 3 Title 2 –8 CLower 3 –3 CUpper 4 9 x 5 6 ME E1 ="t Interval"

H0 : P

zTest –235,12.8,–238,119,0: stat.results "Title" "z Test" "Alternate Hyp" "μ ≠ μ0" "z" –2.55673 "PVal" 0.010566 "x" –238. "n" 119. "σ" 12.8

CH3 Exercise D

1.3

A

0.05

0 .1

Since the p-value is 0.131776 > 0.1 we have no sufficient evidence to reject the null hypothesis at the 10% significance level.

CH3 Exercise E

zTest 10,2,12,20,0: stat.results "Title" "z Test" "Alternate Hyp" "μ ≠ μ0" "z" 4.47214 "PVal" 0.000008 "x" 12. "n" 20. "σ" 2.

b

H0 : P 1.1

1/99

Since the p-value is 0.000008 < 0.1 we reject the null hypothesis at the 10% significance level.

P0 , H1 : P  P0, D

0.01

CH3 Exercise E

zTest 21.4,0.75,21.2,50,–1: stat.results "Title" "z Test" "Alternate Hyp" "μ < μ0" –1.88562 "z" 0.029673 "PVal" 21.2 "x" 50. "n" 0.75 "σ"

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

5/99

Statistics: Chapter 3

24

WORKED SOLUTIONS Since the p-value is 0.029673 > 0.01 we have no sufficient evidence to reject the null hypothesis at the 1% significance level. c

H0 : P 1.1

1.2

P0 , H1 : P  P0, D 1.3

4 a

b

1.1

P0 , H1 : P ! P0, D

0.05

5 a

CH3 Exercise E

b

1.1

0 .1

6 a

b

Since the p-value is 0.109391 > 0.1 we have no sufficient evidence to reject the null hypothesis at the 10% significance level. 1.1

0.01

CH3 Exercise E

zTest –73,3.72,–71.6,92,1: stat.results "Title" "z Test" "Alternate Hyp" "μ > μ0" "z" 3.60977 "PVal" 0.000153 –71.6 "x" 92. "n" 3.72 "σ" |

9/99

CH3 Exercise E

1.3

B

C

D

1.43 1.52 1.35 1.38 1.42

Since the p-value is 0.335687 > 0.05 we have no sufficient evidence to reject the null hypothesis at the 5% significance level and conclude that the company’s claim is correct.

CH3 Exercise E

P0 , H1 : P ! P0, D

1.2

=zTest(1.4 Title z Test Alternate Hyp μ > μ0 z 0.424264 PVal 0.335687 x 1.445 =zTest(1.4,0.3,a [ ], 1,1): CopyVar

1 2 3 4 5 C

8/99

H0 : P

E

Use the z-test. A fat

zTest 27.3,3.6,28.,40,1: stat.results "Title" "z Test" "Alternate Hyp" "μ > μ0" "z" 1.22977 "PVal" 0.109391 "x" 28. 40. "n" 3.6 "σ"

c

D

H0:“The mean level of fat in the drink is 1.4 g.” ( P 1 .4 ) H1:“The mean level of fat in the drink is more than 1.4 g.” ( P ! 1.4 )

1.1

Since the p-value is 0.036819 < 0.05 we reject the null hypothesis at the 5% significance level.

P0 , H1 : P ! P0, D

C

Since the p-value is 0.00001 < 0.01 we reject the null hypothesis at the 1% significance level and conclude that the harvested snails are not from the population.

7/99

H0 : P

B

=zTest(26, 1 22 Title z Test 2 25 Alternate... μ ≠ μ0 3 31 z 4.42627 4 35 PVal 0.00001 5 28 x 30. C4 =9.5958692475566E-6

zTest 10,5,12,20,1: stat.results "Title" "z Test" "Alternate Hyp" "μ > μ0" "z" 1.78885 "PVal" 0.036819 "x" 12. "n" 20. "σ" 5.

b

CH3 Exercise E

1.2

A worms

CH3 Exercise E

Since the p-value is 0.005283 < 0.01 we reject the null hypothesis at the 1% significance level.

1.1

Use the z-test.

0.01

9/99

H0 : P

26)

H1: “The mean weight is not 26 g.” ( P z 26)

zTest –235,12.8,–238,119,–1: stat.results "Title" "z Test" "Alternate Hyp" "μ < μ0" "z" –2.55673 "PVal" 0.005283 "x" –238. "n" 119. "σ" 12.8

3 a

H0: “The mean weight is 26 g.” ( P

H0: “The mean volume of juice in the bottle is 300 ml.” (μ = 300) H1 : “The mean volume of juice in the bottle is less than 300 ml.”(μ < 300) Use the z-test. 1.2

1.3

1.4

A volume B 1 2 3 4 5 C

CH3 Exercise E

C

D

=zTest(30 295 Title z Test 288 Alternate Hyp... μ < μ0 293 z –2.60364 301 PVal 0.004612 302 x 295.75 =zTest(300,7.3,a [ ], 1,–1): CopyVar

Since the p-value is 0.004612 < 0.1 we reject the null hypothesis at the 10% significance level and conclude that the bottles contain less volume than stated.

Since the p-value is 0.000153 < 0.01 we reject the null hypothesis at the 1% significance level.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

25

WORKED SOLUTIONS

Exercise 3F 1 a

b

H0 : μ = μ0, H1 : μ < μ0, α = 0.01

H0 : μ =μ0 , H1 :μ ≠ μ0, α = 0.05 1.1

CH3 Exercise F

1.1

"Title" "t Test" "Alternate Hyp" "μ < μ0" "t" –2.99871 "PVal" 0.007494 "df" 9. "x" 119.8 "sx := sn–1x" 2.32 10. "n"

CH3 Exercise F

"Title" "t Test" "Alternate Hyp" "μ ≠ μ0" "t" –0.486504 "PVal" 0.638237 "df" 9. "x" 4.8 "sx := sn–1x" 1.3 10. "n"

5/99

|

Since the p-value is 0.007494 < 0.01 we reject the null hypothesis at the 1% significance level.

1/99

Since the p-value 0.6382 > 0.05 we have no sufficient evidence to reject the null hypothesis at the 5% significance level. b

c

CH3 Exercise F

1.1

H0 : μ = μ0, H1 : μ ≠ μ0, α = 0.1 1.1

"Title" "t Test" "Alternate Hyp" "μ < μ0" "t" –0.816497 "PVal" 0.225675 "df" 5. "x" 622.8 "sx := sn–1x" 12.6 6. "n"

CH3 Exercise F

"Title" "t Test" "Alternate Hyp" "μ ≠ μ0" "t" –1.66667 "PVal" 0.10858 "df" 24. "x" 1.9 "sx := sn–1x" 0.3 25. "n"

|

6/99

Since the p-value 0.225675 > 0.1 we have no sufficient evidence to reject the null hypothesis at the 10% significance level.

|

2/99

Since the p-value 0.10858 > 0.1 we have no sufficient evidence to reject the null hypothesis at the 10% significance level. c

H0 : μ = μ0, H1 : μ < μ0, α = 0.1

3 a

CH3 Exercise F

1.1

H0 : μ = μ0, H1 : μ ≠ μ0, α = 0.01 1.1

H0 : μ = μ0, H1 : μ > μ0, α = 0.1 "Title" "t Test" "Alternate Hyp" "μ > μ0" "t" –0.667483 0.743755 "PVal" 19. "df" 0.95 "x" 0.335 "sx := sn–1x" 20. "n"

CH3 Exercise F

"Title" "t Test" "Alternate Hyp" "μ ≠ μ0" "t" 3.48125 "PVal" 0.013122 "df" 6. "x" –35.3 "sx := sn–1x" 0.532 7. "n"

7/99

Since the p-value 0.743755 > 0.1 we have no sufficient evidence to reject the null hypothesis at the 10% significance level.

3/99

Since the p-value 0.013122 > 0.01 we have no sufficient evidence to reject the null hypothesis at the 1% significance level. 2 a

b

H0 : μ = μ0, H1 : μ > μ0, α = 0.05 1.1

1.2

H0 : μ = μ0, H1 : μ < μ0, α = 0.05 1.1

1.3

CH3 Exercise F

"Title" "t Test" "Alternate Hyp" "μ > μ0" 3.53553 "t" 0.004763 "PVal" 7. "df" 26.4 "x" 1.12 "sx := sn–1x" 8. "n"

CH3 Exercise F

"Title" "t Test" "Alternate Hyp" "μ < μ0" –1.99172 "t" 0.027947 "PVal" 29. "df" 14.2 "x" 2.2 "sx := sn–1x" 30. "n"

8/99

4/99

Since the p-value is 0.027947 < 0.05 we reject the null hypothesis at the 5% significance level.

Since the p-value 0.004763 < 0.05 we reject the null hypothesis at the 5% significance level.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

26

WORKED SOLUTIONS H0 : μ = μ0, H1 : μ > μ0, α = 0.01

c

1.1

1.2

1.3

Use the t-test on the difference of times on those two different cubes.

CH3 Exercise F

"Title" "t Test" "Alternate Hyp" "μ > μ0" "t" 1.25463 "PVal" 0.115078 "df" 14. "x" 758.6 "sx := sn–1x" 14.2 15. "n"

A cube1 B cube2 2 3 4 5 6 C

9/99

Since the p-value 0.115078 > 0.01 we have no sufficient evidence to reject the null hypothesis at the 1% significance level. 4

CH3 Exercise F

1.2

1 2 3 4 5 C

C

119 Title 123 Alternate Hyp 121 t 120 PVal 118 df =tTest(120,a [ ], 1,0):

D

=tTest(120 t Test μ ≠ μ0 0.283654 0.784882 7. CopyVar Stat

Since the p-value 0.283654 > 0.01 we have no sufficient evidence to reject the null hypothesis at the 1% significance level and conclude that the factory advertised a correct volume of a particular ice-cream product. 5

1.2

A life 1 2 3 4 5 C

1.3

B

2

1.1

B new

1 85 2 92 3 100 4 97 5 89 C1 =–5

D

1.1

1 2 3 4 5 E

1.2

90 95 98 99 93

C dartdiff D =old–new –5 –3 2 –2 –4

CH3 Exercise G

C dartdiff D

E

=old–new –5 –3 2 –2 –4

=tTest(0, c

Title Alternate... t PVal df =tTest(0, c [ ] ,1,0):

F

t Test μ ≠ μ0 –2.13719 0.085622 5. CopyVar Stat.,

Since the p-value 0.085622 < 0.1 we have to reject the null hypothesis at the 10% significance level and conclude that players score a better result by using the new type of dart.

Exercise 3G

H1 : “There is a difference in finishing times.” (μd ≠ 0)

CH3 Exercise G

1.2

A old

=tTest(300 29500 Title t Test 28350 Alternate Hyp... μ < μ0 30300 t –1.51876 30250 PVal 0.094643 29350 df 5. =tTest(30000,a [ ], 1,–1): CopyVar S

H0 : “There is no difference in finishing times.” (μd = 0)

H0 : “There is no difference in the scores.” (μd = 0) Use the t-test on the difference of scores on the two types of dart.

Since the p-value 0.094543 < 0.1 we reject the null hypothesis at the 10% significance level and conclude that the manufacturer claims a longer life expectancy of the LED lamps.

1

CH3 Exercise G

H1 : “There is a difference in the scores.” (μd ≠ 0)

CH3 Exercise F

C

diff:=a[ ]–b[ ]

Since the p-value 0.874063 > 0.05 we have no sufficient evidence to reject the null hypothesis at the 5% significance level and conclude that there is no difference in finishing times on the two Rubik’s Cubes.

H0: “The mean life expectancy is 30 000 hours.” (μ = 30 000) H1 : “The mean life expectancy is less than 30 000 hours.” (μ < 30 000) 1.1

38 40 34 30 44

C diff D =a[]–b[] –3 1 –4 –2 2

D E F ff =tTest(0, c ]–b[] 2 –3 Alternate Hyp μ ≠ μ0 3 0.164399 1 t 4 0.874063 –4 PVal 5 7. –2 df 6 0.25 2 x E =tTest(0, c [ ] ,1,0): CopyVar Stat.,

H1 : “The mean volume is not 120 ml.” (μ ≠ 120) A volume B

35 41 30 28 46

1.1

H0 : “The mean volume is 120 ml.” (μ = 120) 1.1

CH3 Exercise G

1.1

3

H0 : “There is no difference in the weights.” (μd = 0) H1 : “Students who join the programme drop some weight.” (μd > 0)

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

27

WORKED SOLUTIONS Use the t-test on the difference of weights before and after the programme. 1.1

1.2

3 a

H 0 : X  Po(45) Ÿ D

1  P(X d 52) = 0.133

b

H1 : X  Po(40) Ÿ E

P(X d 52) = 0.972

CH3 Exercise G

1.3

A before B after

C wdiff D =before-a

1 55 52 2 82 84 3 63 61 4 69 69 5 65 62 C wdiff: =before–after

1.1

E

3 –2 2 0 3

1.2

1.3

CH3 Exercise H

1–PoissCdf(45,0,52)

0.132809

poissCdf(40,0,52) |

0.971943

2/99 1.1

1.2

CH3 Exercise G

1.3

C wdiff D =before-a

E

F

Review exercise

E

=tTest(0,w

1

1 2 3 Title t Test 2 4 –2 Alternate... μ > μ0 3 1 2t 1.23335 4 9 0 PVal 0.121576 5 2 3 df 11. C =tTest(0,wdiff,1,1): CopyVar Stat.,

Use z-interval since the standard deviation is known and the value is 114 g. a

1/99

b

Exercise 3H

b

H 0 : X  N(5, 0.4 2 ) Ÿ D 1  P(4.2 d X d 5.8)

0.0455

0.773

CH3 Exercise H

1.1

1–normCdf(4.2,5.8,5,0.4)

0.0455

normCdf(4.2,5.8,4.5,0.4) |

0.772796

CH3 Review...ise

1.1

H1 : X  N(4.5, 0.4 ) Ÿ

P(4.2 d X d 5.8)

99% CI [586, 718] zInterval 114,652,20,0.99: stat.results "Title" "z Interval" "CLower" 586.339 "CUpper" 717.661 "x" 652. "ME" 65.6609 "n" 20. 114. "σ" | 2/99

2

E

CH3 Review...ise

1.1

zInterval 114,652,20,0.95: stat.results "Title" "z Interval" "CLower" 602.038 "CUpper" 701.962 "x" 652. "ME" 49.9618 "n" 20. 114. "σ"

Since the p-value is 0.121576 > 0.05 we have no sufficient evidence to reject the null hypothesis at the 5% significance level and conclude that there is no difference in weight before and after the programme.

1 a

95% CI [602, 702]

c

95% < 99% ⇒ [602, 702] ⊂ [586, 718] We notice that a higher significance level means a wider confidence level.

2 a 2/99

b

1  P( X d 35) § ©

H1 : X  B ¨ 50, 1.1

0.00130

4· ¸ŸE 7¹

P(X d 35) = 0.978

CH3 Exercise H

1.2

(

)

1 1–binomCdf 50, 2 ,0,35

(

)

4 binomCdf 50, 7 ,0,35

0.001301

A

Analyzer I

Difference b

2/99

C

D

E

F

G

H

I

J

235 352 410 280 341 325 428 388 272 310

−2

9

−6

8

5

−4 15 −8

7

−5

H0 : “There is no difference in potassium levels.” (μd = 0) H1 : “There is a difference in potassium levels.” (μd ≠ 0) 1.1

1.2

A a1 0.977867

B

Analyzer II 237 343 416 272 336 329 413 396 265 315

1· 1 § 2 a H 0 : X  B ¨ 50, ¸ Ÿ E(X ) = 50 × = 25 2¹ 2 ©

D

Patient

1 235 2 352 3 410 4 280 5 341 C1 =–2

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

CH3 Review...ise

B a2 237 343 416 272 336

C dl D ='a1–'a2 –2 9 –6 8 5

E

Statistics: Chapter 3

28

WORKED SOLUTIONS 1.1

1 2 3 4 5 E

10

CH3 Review...ise

1.2

C dl D F F ='a1–'a2 =tTest(0,d –2 Title t Test 9 Alternate... μ ≠ μ0 –6 t 0.76657 8 PVal 0.46297 5 df 9. =tTest(0,dl[ ],1,0): CopyVar Stat.,

b

¦x

i

i 1

x

30.97

10 10

§ 3097 · x i2  10 u ¨ ¦ ¸ © 100 ¹ i 1 9

s2 1.1

Since the p-value 0.46297 > 0.01 we have no sufficient evidence to reject the null hypothesis and we conclude that there is no difference in measurement of the two types of biochemical analyzers.

1.2

1.3

2

2681 9000

0.298

CH3 Review...ise

{ 31.2,30.8,30.4,30.8,31.3,32.1,30.3,31.4,3 30.97

varSamp({ 31.2,30.8,30.4,30.8,31.3,32.1,30 0.297889

15

3 a

¦x

i

80 15

i 1

x

15 15

¦x

s2

i 1

184 42 b

2 i

§ 16 ·  15 u ¨ ¸ © 3 ¹ 14

92 21

s = s2 = 1.1

1.2

16 3

1.3

2/99

5.33 2

1280 3 14

488 

c

H1 : “The average time is more than 30 s.” (μ > 30) We use the t-test since the standard deviation is unknown.

4.38 92 = 2.09 21

1.2

CH3 Review...ise

√ 21

"Title" "CLower" "CUpper" "x" "ME" "df" "sx := sn–1x"

"t Interval" 3.72456 6.94211 5.33333 1.60877 14. 2.09307

b

In 99% of the cases the mean value of a sample of 15 observations taken from the population will fall within the confidence interval [3.72, 6.94]. 47.2  55.2 2

x

V

2

25 Ÿ V

51.2

2

55.2  51.2 Ÿ z D 2

D 1.960 Ÿ 2

5

4 6 4

)(1.960)

210

X  N (210, 144) Ÿ σ = 12 Ÿ 1.960 ×

1.3

0.975

s = 1000 m, v = 120 km/h =

t =

204  216 2

x

216  210 Ÿ n 3.92 Ÿ n The sample size should be 16.

6

The confidence level of the interval is 95%. 5 a

6 a b

5 Ÿ zD u

CH3 Review...ise

Since the p-value 0.000163 < 0.05 we reject the null hypothesis and can conclude that the average time is more than 30 seconds. Therefore, the company sets up the speedometers to show a higher speed.

99% CI [3.72, 6.94]

4 a

1.4

1/3

1/99

c

1.3

tTest 30,{31.2,30.8,30.4,30.8,31.3,32.1,30. "Title" "t Test" "Alternate Hyp" "μ > μ0" "t" 5.62011 0.000163 "PVal" 9. "df" 30.97 "x" "sx := sn–1x" 0.545792 10. "n"

tInterval 16 , 92 ,15,0.99: stat.results 3

H0 : “The average time is 30 s.” (μ = 30)

100 m / s, 3

1.4

204+326 2

(

1.5

12 n

15.37

CH3 Review...ise

265

)

nSolve invNorm(0.975)·12 =216–210,n √n

15.3658

|

s 1000 Ÿt = = 30 s 100 v 3

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

2/99

Statistics: Chapter 3

29

WORKED SOLUTIONS 7 a

For the speed values we used the midpoints of the intervals. 1.4

1.5

A

CH3 Review...ise

1.6

B

C

1 5 9 2 15 56 3 25 47 4 35 25 5 45 13 D =OneVar(a[ ],b[ 1.5

1.6

Av

D E =OneVar(a Title One–Var... x 23.4667 Σx 3520 Σx2 99150 sx : = sn–... 10.5383 ]): CopyVar Stat

8

H0 = μ = 10, H1 : μ < 10, using the mean of a sample of size 5.

i

B f

i ii

= s = 10.5

b i

C

1.6

CH3 Review...ise

1.8

(

√ 25 ) =0.1,x) 9.18948

(

(

√ 25 ) =0.05,x)

nSolve normCdf –9.E999,x,10,

8.9597

2/99

1.6

1.7

 1.645 Ÿ x

1.7

x −10

b

CH3 Review...ise

tInterval ν,f ,0.9: stat.results "Title" "CLower" "CUpper" "x" "ME" "df" "sx := sn–1x" "n"

10  1.645 u

2 5

2 5

8.96

The critical region is the interval ]−∞, 8.96[

90% CI [22.0, 24.9] 1.6

5% for N (0, 1) is –1.645 so

CH3 Review...ise

1/99

"t Interval" 22.0425 24.8908 23.4667 1.42417 149. 10.5383 150.

2·

§ ©

H1 : μ = 9.3 Ÿ X  N ¨ 9.3, ¸ 5 ¹

i

E = P (X > 9.19) = 0.569

ii

E = P(X > 8.96) = 0.705 1.6

1.7

1.8

CH3 Review...ise

8.9597

( (

normCdf 9.1894753867941,9.E999,9.3,

We notice that [22.0, 24.9] ⊂ [21.8, 25.2], so a 90% confidence interval is a subset of a 95% confidence interval.

√ 25 )

0.569364

normCdf 8.9597033202516,9.E999,9.3,

√ 25 )

0.704731

2/99

c

9.19

23.5

tInterval ν,ƒ,0.95: stat.results "t Interval" "Title" "CLower" 21.7664 "CUpper" 25.1669 23.4667 "x" 1.70026 "ME" 149. "df" "sx := sn–1x" 10.5383 "n" 150.

1.5

1.7

(

ii

ii

2 5

10  1.282 u

nSolve normCdf –9.E999,x,10,

= 95% CI [21.8, 25.2] 1.5

2 5

The critical region is the interval ]−∞, 9.19[

D E =OneVar(v 1 5 9 Title One–Var... 2 15 56 x 23.4667 3 25 47 Σx 3520 4 35 25 Σx2 99150 5 45 13 sx : = sn–... 10.5383 D =OneVar(v,f): CopyVar Stat., Stat6

=x

x −10

10% for N (0, 1) is –1.282 so

 1.282 Ÿ x

CH3 Review...ise

1.7

2· 5¹

§ ©

a H 0 : μ = 10 Ÿ X  N ¨ 10, ¸

| 4/99

c

When the probability of a Type I error decreases from 10% to 5%, the probability of a Type II error increase from 0.569 to 0.705.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 3

30

WORKED SOLUTIONS

4

Statistical modeling

Skills Check

A xcoord

B ycoord

C

D

E

=

1 a

2E(Z ) – 3E(Y ) + 2E(X)

b

4Var(Z ) – 9Var(Y ) + 4Var(X)

c

E(X )E(Y )E(Z)

2 a

x = 43.5, y = –31.75, Var(X ) = 46.25, Var(Y ) = 98.69

b

d B ycoord

C

D

E

=

F

G

H

=TwoVar(’

1 33 2 49

–21 Title –44 –x

Two–va...

3 50

–23 Σx

4 42 5

–39 Σx2 sx := sn-...

7.85281

6

σx := σn...

6.80074

7 8

n –y

9

Σy

–127.

10 11

Σy Sy := Sn-...

4427. 11.471

12

σy := σn...

9.93416

13

Σxy

43.5 7754.

H

1

4

1

2

1

1

3

3

2

4

2

4

5

1

Title –x Σx

=LinRegM

Two–va...

Title

Linear R... =Linear R..

28.

RegEqn m

m*x+b –0.3409.. 2.95455

2.8 96.

b

3

Σx2 sx := sn-...

1.39841

r2

1.32665

r

0.146104 –0.3822..

Resid

{–0.5909..

6

2

4

σx := σn...

7

5

2

4

1

n –y

10.

8 9

4

1

Σy

20.

10

2

1

11

Σy2 Sy := Sn-...

1.24722

12

σy := σn...

1.18322

13

Σxy

2. 54.

50.

Since r = −0.382, there is a weak negative correlation between the two random variables.

4.

2

G

H1 =“Linear Regression (mx+b)”

174.

–31.75

F =TwoVar(’

2

1.3

2.1

Exercise 4A

2.2

52.9 y

–5637.

E1 =“Two–Variable Statistics”

3 a

0.382

b

0.951

c

0.938

d

0.732

1.1

1.2

1.3

CH4 Skills check

A xcoord

tCdf(–0.4,0.8,2)

0.382266

tCdf(–∞,1.83,10)

0.951412

tCdf(–1.75,∞,7)

0.938203

tCdf(–1.14,1.14,20)

0.732245

Exercise 4A 1.2

1.3

2.1

B ycoord

C

D

=

|

1

x 158.4

81.6

Exercise 4A

0.5 y

E

F

G

=TwoVar(’

1

100

33

2

115

38

3

140

40

4

149

51

5

88

6

H =LinRegM

Title –x Σx

Two–Va...

Σx2 sx := sn-...

165059.

b

36

21.1421

r2

132

40

σx := σn...

20.0572

r

0.630724 0.794182

7

152

51

{–0.9961..

144

43

n –y

Resid

8 9

121

32

Σy

10

128

42

11

Σy2 Sy := Sn-...

12

σy := σn...

13

Σxy

126.9 1269.

10.

Title

Linear R...

RegEqn m

m*x+b 0.245495 9.44674

40.6 406. 16868 6.53537 6.2 52509.

B11

Since r = 0.794, there is a strong positive correlation between the two random variables. 3 –0.3 0.2

1.1 68 y

1.2

Exercise 4A

6.06 x

29.4 © Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

x 96.6

Statistics: Chapter 4

31

WORKED SOLUTIONS A xcoord

B ycoord

C

D

E

=

F

G

=TwoVar(’

H

55

58

2

35

65

Title –x

3

66

52

Σx

4

82

35

b

91

36

Σx2 sx := sn-...

47645.

5

18.7264

r2

6

79

42

σx := σn...

17.7654

r

0.939034 –0.9690...

7

48

60

Resid

{2.39513..

8

52

55

n –y

49.1

9

71

50

Σy

491.

10

88

38

Σy2 Sy := Sn-...

25147.

σy := σn...

10.1926

11 12 13

66.7 667.

10.

Title

Linear R...

RegEqn m

m*x+b –0.5559... 86.1833

10.744

Σxy

= E(X2 + 2XY + Y 2) – (μx μy)2 = E(X 2 + 2XY + Y 2) − μx2 − 2 μx μy – P y2 = (E(X 2) − P y2 )+ (E(Y 2) − P y2 ) + 2(E(XY ) − μx μy) = Var(X) + Var(Y) + 2Cov(X, Y) 8

Since r = −0.970, there is a strong negative correlation between the two random variables.

Exercise 4B

2

3

Hence, Cov ( X , Y ) Hence, p = 0.

1

Cov(X, Y) = E(XY ) – μx μy

A xcoord

B ycoord

C

=

D

E

42 2

= Cov(Y, X )

3

26

14 RegEqn

4

22

20 t

5

15

41 Pval

= E(X 2) − P x2

6

44

44 df

10.

7

50

51 a

9.76713

= Var(X )

8

38

45 b

0.73773

9

12

17 s

9.69498

10

28

27 SESlope... 0.181742

11

22

12 13 13

1

28 r2 1 r

Cov(X, X) = E(XX ) – μx μy

Cov(aX, Y) = E(aXY ) – E(aX )E(Y )

= bE(XY ) – bE(X )E(Y )

= E > X1Y  X 2Y @  > E( X1 )  E( X 2 )@ E(Y ) = > E( X1Y )  E( X1 )E(Y )@  > E( X 2Y )  E( X 2 )E(Y )@ = C ov( X1 , Y )  Cov( X 2 , Y )

Cov(X, Y1, + Y2) = E[(X)(Y1 + Y2)] – E(X)E(Y1 + Y2) = E[XY1 + XY2] – E(X)[E(Y1) + E(Y2)] = [E(XY1) – E(X)E(Y1)] + [E(XY2) – E(X)E(Y2)] = Cov(X, Y1) + Cov(X, Y2)

34 Title

18 Alternate... ß & ρ ≠ ...

Resid

a+b*x 4.05922 0.002289

0.622318 0.788871 {–6.7517...

Since 0.00229 < 0.01, we reject the null hypothesis. There is evidence of significant correlation between the two variables at the 1% level.

= bE(XY ) – E(X )E(bY )

E > ( X1  X 2 )Y @  E( X1  X 2 )E(Y )

G

H0 : p = 0; H1 : p ≠ 0

Cov(X, bY) = E(X(bY )) – E(X )E(bY )

5 Cov( X1  X 2 , Y )

F

D1 =“Linear Reg t Test”

= aCov(X, Y)

= bCov(X, Y)

0.

Linear R...

1 2

= b[E(XY ) – ( μx μy)]

Px P y  Px P y

=LinRegtT

= E(XY ) – μy μx

= a[E(XY ) – (μx μy)]

6

E( XY )  P x P y

Exercise 4C

= aE(XY ) – aE(X )E(Y )

4

Since X and Y are independent variables, E(XY ) = E(X )E(Y ) = μx μy.

30995.

A11

1

Var(X + Y) = E(X + Y)2 – [E(X + Y)]2

=LinRegM

1

Two–Va...

7

2

A xcoord

B ycoord

C

D

=

E

F

G

=LinRegtT Linear R...

1

64

69

Title

2

66

64

Alternate... ß & ρ ≠ ...

3

68

67

RegEqn

4

69

64

t

0.931872

5

70

62

Pval

0.394177

6

72

73

df

5.

7

74

71

a

35.6

8

b

0.457143

9

s

10

SESlope...

0.490564

11

r2 r

0.147977

12 13 13

Resid

a+b*x

4.10435

0.384678 {4.14285...

E1 =“Linear Reg t Test”

H0 : p = 0; H1 : p = 0 Since 0.394 > 0.05, there is not enough evidence to reject the null hypothesis. There is not evidence of significant correlation between the two variables at the 5% level.

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

Statistics: Chapter 4

32

WORKED SOLUTIONS 3

A xcoord

B ycoord

C

D

E

=

F

2

G

A body

=LinRegtT

=

B heart

C

D

E

F

1

67

104

Title

Linear R...

1

30

136

Title

Linear R...

2

70

116

Alternate... ß & ρ ≠ ...

2

37

156

RegEqn

3

70

121

RegEqn

3

38

150

m

m*x+b 1.90885

4

70

126

t

5

71

117

Pval

6

72

113

7

72

124

8

72

126

9

73

127

a+b*x 2.62304

4

32

140

b

82.763

0.034257

5

36

155

0.57174

df

7.

6

32

157

r2 r

a

–93.

7

33

143

Resid

b

3.

8

38

160

s

5.78174

9

44

170

10

SESlope...

1.14371

10

38

144

11

r2 r

0.49569

11

0.704052

Resid

{–4.,–0.9...

12 13 13

12 13 13 E1 =“Linear Reg t Test”

0.756135 {–4.0286...

F1

H0 : p = 0; H1 : p = 0 Since 0.0343 < 0.1, we reject the null hypothesis. There is evidence of significant correlation between the two variables at the 10% level.

r = 0.756, which indicates a strong correlation between the two variables. Hence we find the line of regression, y = 1.91x + 82.8, and when x = 35 g, y = 150 mg. 3 a

H0 : p = 0; H1 : p = 0

b

Exercise 4D 1 a

G

=LinRegM

A x

B y

C

D

A score B age C he... D We... E

F

H

G

=LinRegM

=

I

=LinReg

G

Title

Linear R...

=LinRegT

120

80

Title

Linear R...

Alternate... ß & p ≠ ... m*x+b a+b*x 0.62303 RegEqn

J

2

125

90

RegEqn

=LinRegM

3

130

92

m

1

62

10

145 44.9 Title... Linear R... Title... Linear .... Title... Linear R...

4

135

98

b

94

11

150 42.1 Reg... m*x+b

5

140

100

3

81

8

139 30.2 m

0.014779 m

0.1642... m

–0.2780...

6

145

103

r2 r

4

62

9

148 41.4 b

8.48321 b

131.601 b

61.3778

7

150

105

Resid

2

2

2

0.298465

Reg... m*x+b

F

1

2

Reg... m*x+b

E =LinRegM

=

10.8182 t 0.926185 Pval 0.962385 df

10.8182 0.62303

58

8

130 41.8 r

8

155

108

b

6

86

10

150 38.4 r

–0.54632

9

160

110

s

7

59

11

154 54.1 Res... {0.60049... Res... {3.216... Res... {0.75945...

10

165

110

SESlope...

8

72

9

140 38.6

11

9

54

10

153 52.4

10

60

9

120 38.6

12 13 13

0.0426... r

0.187584 r

0.2064 r

11

8.

{–5.5818... a

5

0.035188 r

10.0189 0.000008

r2 r Resid

2.82414 0.062185 0.926185 0.962385 {–5.5818...

G1

12 13 13

r = 0.962; p = 8 × 10−6

E10

Column F shows the regression data for score versus age. Column H shows the regression data for score versus height. Column J shows the regression data for score versus weight. |r| is largest for the random variables score and weight, hence we use W.

There is very strong evidence to indicate a positive association between the random variables X and Y. d Since r is close to 1, we can find the regression line, y = 0.623x + 10.8, and when x = 138 mm, y = 97 mm.

c

4

From the GDC, S = −0.270W + 61.4 c From the GDC, b

A

B

C

D

E

F

G

m

¦ xy  nx y ¦y

x x 5

2

 ny 2

0 .6 ( y  y ) Ÿ x

Linear R...

2 Alternate... ß & ρ ≠ ... 3 RegEqn

a+b*x

4 t

–1.84488

5 PVal

0.102277

6 df

8.

7 a

61.3778

8 b

–0.2780...

9 s

6.1981

10 SESlope... 0.150699 11 r2 12 r 13 13 Resid

0.6

0.6 y  12.4 ; x = 66.4 g

For the gradient of the Y on X line of regression,

= 1 Title...

§ 182 ·§ 200 · 8390  5 ¨ ¸¨ ¸ © 5 ¹© 5 ¹ 2 § 200 · 9850  5 ¨ ¸ © 5 ¹

m=

¦ xy − nx y ¦x

yy

2

− nx 2

§ 30 · § 86 · 580 − 10 ¨ ¸ ¨ ¸ © 10 ¹ © 10 ¹ = 2.48, hence = 2 § 30 · 220 − 10 ¨ ¸ © 10 ¹

2.48( x  x ) Ÿ y

2.48 x  1.17

For the gradient of the X on Y line of regression,

0.298465 -0.54632 {0.75945..

C13

Since p = 0.102 > 0.05 there is insufficient evidence to reject the null hypothesis, i.e. there is not evidence of significant correlation between the score and the weight at the 5% level.

m

¦ xy  nx y ¦ y  ny

x x

2

2

§ 30 · § 86 · 580  10 ¨ ¸ ¨ ¸ © 10 ¹ © 10 ¹ 2 § 86 · 1588  10 ¨ ¸ © 10 ¹

0.380( y  y ) Ÿ x

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

0.380 , hence

0.380 y  29.6 Statistics: Chapter 4

33

WORKED SOLUTIONS

Review exercise 1 a

5 a

r is the unbiased estimate of ρ, and

¦x y i

r

t

¦ x

2 i

 nx y

i

 nx

2

¦ y

2 i

 ny

2



m=

¦ y 2 − ny 2

x x

x 2

§ 440 606 · 37 000 − 8 ¨ × ¸ 8 ¹ © 8 = = 1.088 2 § 606 · 49 278 − 8 ¨ ¸ © 8 ¹

Var ( X ) Var (Y )

, the maximum

3

If p = 0.9 then t = invt (0.9, 18) solving 1.33039

4 a

r

41.2 Title

10.3

11.1 Alternate... ß & ρ ≠ ...

3

20.1

18.7 RegEqn

4

55

5

39.6

6

24.1

41.2 PVal 26.5 df

7

31.2

29.3 a

2.59473

8

9.5

11.2 b

0.904534

9

22.3

25.1 s

10

35.1

33.2 SESlope...

a+b*x

52.3 t

11

r2

12 13 13

r Resid

20.9954 2.78035... 8.

1.9017 0.043083 0.982175 0.991047 {–1.6465...

y = 0.905(19.8) + 2.59 = 20.5

d

p = 0.00620 < 0.01, hence there is a significant correlation between the random variables.

6 a

The gradient of the Y on X line, y = a + bx, is Cov ( X , Y ) , and the gradient of the X on Y line, Var ( X )

1.33039 , and

Var ( X  Y ) Var ( X  Z )

Linear R...

44.5

c

x = c + dy, is

Cov ( X , Y ) . Since r = 0, Var (Y )

Cov(X,Y ) = 0, hence y = a and x = b, and the two lines are perpendicular.

Cov (( X  Y ), ( X  Z ))

Var (U ) Var (V )

G

The p-value suggests a strong relationship between the two random variables, hence it makes sense to find the equation of the regression line: y = 0.905 x + 2.59

18 , gives r = 0.299 1  r2

Cov (U , V )

U

10.2

F

b

Cov(X, Y ) will occur when ρ = 1. Hence, Covmax(X, Y) = 17 u 15

E

=LinRegT

From the GDC, r = 0.991; p = 2.78 × 10−8

32.4 C ov( X ,Y )

D

D5 =2.7803542327572E–8

1.088( y  y ) ⇒ x = 1.088 y – 27.4 ;

Since U

C

2

n2 = 10.7 r 1  r2

¦ xy − nx y

B ycord

1

= 0.975

Hence, since p =.000039 < 0.05, there is evidence of a strong positive relationship between the two random variables. b

A xcord =

b

1 d 1 . d

For y = a + bx, m = b, and for x = c + dy, m = .

Cov ( XX )  Cov ( XZ )  Cov (YX )  Cov (YZ )

r

> Var ( X  Y )@ > Var ( X  Z )@ =

Var( X )  0 2V

2

2V

2

V2 2V 2

U

Cov (V , W ) Var (V ) Var (W )

c

Cov (( X  Y ), ( X  Y ))

Cov ( XX )  Cov ( XY )  Cov (YX )  Cov (YY )

bd

> Var ( X  Y )@ > Var ( X  Y )@ Var ( X )  Var (Y )  Y )@

bd Ÿ bd

1, b

Cov ( X , Y ) and d Var ( X ) ª Cov ( X , Y ) º « Var ( X ) Var (Y ) » ¬ ¼

a  bx and x

c  dy ,

Cov ( X , Y ) . Hence Var (Y ) 2

r 2. Since b and d are

either both positive or both negative, r  bd if both are positive, and r  bd if both are negative.

0

We know that if V and W are independent then ρ = 0, but the converse does not necessarily hold.

For the regression lines y b

Var ( X  Y ) Var ( X  Y )

> Var ( X  Y )@ > Var ( X

1. Since r 2

Hence, the two lines have the same gradient. Since both regression lines must go through ( x , y ), the lines are identical.

1 2

Hence it cannot be said that U and V are independent. b

r1 Ÿ r2

d

Since b and d are both positive, r

 bd

© Oxford University Press 2015: this may be reproduced for class use solely for the purchaser’s institute

0.19 u 0.77

0.382

Statistics: Chapter 4

34