ct3-pxs-11

CT3 – P XS – 11

Series X Solutions

ActEd Study Materials: 2011 Examinations

Subject CT3

Contents

Series X Solutions

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CT3: Assignment X1 Solutions Page 1

Assignment X1 Solutions Markers: This document sets out one approach to solving each of the questions. Please give credit for other valid approaches. Solution X1.1

By definition, the 22c distribution is the same as the distribution. [1] (1, ½)Gamma

The exponential distribution with mean ½ has parameter 2. This is a distribution, and so is not equivalent to the other two. [1]

(1, 2)Gamma

Therefore the student is wrong. [Total 2] Be careful to distinguish between the parameter l and the mean 1 l for the exponential distribution. Solution X1.2

Let and be the events “a patient has high levels of the first, second, third and fourth chemical”.

1 2 3, ,C C C 4C

Let be the event “a patient has the disease”. D We have:

[1] 1 2 3 4

1 2 3

( ) 0.4 ( ) 0.2 ( ) 0.1 ( ) 0.3( | ) 0.2 ( | ) 0.4 ( | ) 0.8 ( | ) 0.1

P C P C P C P CP D C P D C P D C P D C

= = = == = = 4 =

) We want : 3( |P C D

3 33 4

1

( | ) ( )( | )

( | ) ( )

0.8 0.1(0.2 0.4) (0.4 0.2) (0.8 0.1) (0.1 0.3)

i ii

P D C P CP C D

P D C P C=

=

¥=¥ + ¥ + ¥ + ¥

Â

[1]

[1] 0.296= [Total 3]


CT3: Assignment X1 Solutions

Solution X1.3

This question is based on Subject C1, September 1996, Question 10. (i) PDF of Y Consider the distribution function (DF) of Y:

( ) ( )( ) ( ) (6 ) (1 ) 66

61

YXF y P Y y P y P X y X P X y y

X

yP Xy

Ê ˆ= £ = £ = £ - = + £Á ˜Ë ¯-Ê ˆ

= £Á ˜+Ë ¯ [1]

Since (6 )X- and (1 )y+ are both positive the sign remains unchanged during multiplication.

Integrating the PDF (from 1 to 61

yy+ ) or using the DF of X from page 13 of the Tables:

6 16 6 1 61( ) 1

1 1 6 1 5 1Y X

yy y yyF y P X Fy y

-Ê ˆ Ê ˆ È+= £ = = = -

y˘

Í ˙Á ˜ Á ˜+ + - +Ë ¯ Ë ¯ Î ˚ [1]

To obtain the PDF of Y , we need to differentiate the DF of Y :

21 6 1 6 6( ) ( ) 15 1 5 (1 ) 5(1 )

Ê ˆÊ ˆ= = - = =¢ Á ˜Á ˜+Ë ¯ + +Ë ¯Y Y

d yf y F ydy y 2y y

[1]

Alternatively, we could use the formula from Chapter 3, Section 4 with ( )y u x= :

( )( ) [ ( )]Y Xdw yf y f w y

dy= , where 1( ) ( )w y u y-=

1 6( ) ( )6 1

x yy u x x u yx y

-= = fi = =- + 2

6(1 )

dxdy y

fi =+

2 2 26 6 6 1( )

5(1 ) (1 ) (1 ) 5(1 )Y Xf y f 26

y y y

Ê ˆfi = ¥ = ¥ =Á ˜+ + +Ë ¯ y+

(ii) Range

This is valid for the range 1 61 66 1 6 6 5

x y< < fi < < fi >- -

1y . [1]

[Total 4]



Solution X1.4

The mean of the gamma distribution is 6 25.al= .

The variance is 15 6252 .al

= , so the standard deviation is 15 625 3 95285. .= . [1]

The mode will be the value of x, which gives the greatest value of the PDF. We can find the mode by differentiating the PDF or (equivalently) the log of the PDF:

1( )

( )log ( ) ( 1) log log log ( )

xf x x e

f x x + x

aa ll

aa a l l

- -=G

fi = - - - aG

1log ( )d f xdx x

a l-fi = - [1]

Setting this equal to zero to obtain the maximum turning point gives:

1 3 75 x .al-fi = = for the mode. [1]

Alternatively, differentiating the original function gives:

2 1

2

( ) ( 1)( )

[( 1) ]( )

x x

x

d f x x e x edx

x e x

aa l a l

aa l

l a la

l a la

- - - -

- -

È ˘= - -Î ˚G

= - -G

Setting this equal to zero gives the same value as before. So:

6 25 3 75 = 0 6323 95285. .

.k -= . [1]

[Total 4]



Solution X1.5

(i) Total probability The total probability is given by:

11 1 3 4

2

00 0

( 0) ( ) (1 ) 3 4 12x xP X f x dx x dx + +x

ba b a b aÈ ˘

= + = + - = - =Í ˙Î ˚

Ú Ú [1]

(ii) Mean

1

011 4 5

2

00

( ) 0 ( 0) ( )

0 (1 ) 4 5 20

E X P X xf x dx

x x x x x dx ba b b

= ¥ = +

È ˘= ¥ + - = - =Í ˙

Î ˚

Ú

Ú [1]

(iii) Standard deviation We are told that the mean, determined in part (ii), equals 0.4:

0.4 820b b= fi = [½]

The total probability given in part (i) must equal 1:

81 112 12 3

+ ba a= fi + = fi = 1a [½]

Now:

12 2 2

011 5 6

2 4

0 0

[ ] 0 ( 0) ( )

1 8 80 8 (1 ) 3 5

E X P X x f x dx

x x x x dx

= ¥ = +

È ˘= ¥ + - = - =Í ˙

Í ˙Î ˚

Ú

Ú4

6 15 [1]

Hence, the standard deviation is given by:

2 2 24var( ) ( ) ( ) 0.4 0 1067 0 32715

X E X E X . .= - = - = = [1]

[Total 5]



Solution X1.6

(i) Distribution function

22 2

3 20 0 0

2 5 5 5( ) ( ) ( ) 15(5 ) (5 )

xx xF x P X x f t dt dt

xt t

È ˘¥ Ê ˆ= £ = = = - = -Í ˙ Á ˜Ë ¯++ +Í ˙Î ˚Ú Ú [1]

Alternatively, we could just recognise that it’s a Pareto distribution with and

and quote the DF from page 14 of the Tables. 2a =

5l = (ii) Conditional probability

( 5 and 3) ( 5( 5 | 3)( 3) ( 3

P X X P XP X XP X P X> >> > = =

> >))

> [1]

Calculating the probabilities using the DF from part (i):

25( 5) 1 (5)

10 4P X F Ê ˆ> = - = =Á ˜Ë ¯

1 [1]

25 2( 3) 1 (3)

8 6P X F Ê ˆ> = - = =Á ˜Ë ¯

54

[½]

So:

16( 5 | 3) 0.625

P X X> > = = 4 [½]

[Total 3] (iii) Simulation We need to rearrange to get 1( )x F u-= :

25 51

5 1u x

x uÊ ˆ= - fi = -Á ˜Ë ¯+ -

5 [1]



Substituting in our random numbers, we get:

5 5 3.521 0.656

x = - =-

[1]

5 5 0.9131 0.285

x = - =-

[1]

[Total 3] Solution X1.7

(i) Derive a recursive relationship for the binomial distribution

1 1 1

!( ) (1 ) (1 )( )! !

!( 1) (1 ) (1 )1 ( 1)!( 1)!

x n x x n x

1x n x x n x

n nP X x p p p px n x x

n nP X x p p p px n x x

- -

- - + -

Ê ˆ= = - = -Á ˜ -Ë ¯

Ê ˆ= - = - = -Á ˜- - + -Ë ¯

- +

( ) 1( 1) 1

1( ) (1

P X x n x pP X x x p

n x pP X x P X xx p

= - +fi = ¥= - -

- +fi = = ¥ = --

1) [Total 2]

(ii)(a) Probability of no deaths during the year Let X be the random variable “number of deaths during the next year”: [1] 1,000~ (1000,0.01) ( 0) 0.99 0.0000432X Bin P Xfi = = = (ii)(b) Probability of more than two deaths during the year Using the recursive relationship gives:

0.01( 1) 1,000 0.0000432 0.0004360.99

999 0.01( 2) 0.000436 0.002202 0.99

P X

P X

= = ¥ ¥ =

= = ¥ ¥ =

( 2) 1 ( 2) 1 [ ( 0) ( 1) ( 2)] 0.997321P X P X P X P X P Xfi > = - £ = - = + = + = = [1]



(ii)(c) Probability of exactly twenty deaths during the year

[1] 20 9801,000( 20) 0.01 0.99 0.00179

20P X

Ê ˆ= = ¥ ¥ =Á ˜Ë ¯

[Total 3] Or using a Poisson approximation . ~ (1,000 0.01) (10)X Poi Poi¥ =

10 2010( 20) 0.0018720!

eP X-

fi = = =

Solution X1.8

Results for male benefit packages:

[1] 1 3

Median 2221 24.5 (24.75)Q Q

== =

Results for female benefit packages:

[1] 1 3

Median 2221 (20.5) 25.5 (26.25)Q Q

== =

The results in brackets correspond to the alternative quartile formulae in the Core

Reading 1 3( 1),4 4

n n+ +ÊÁË ¯

ˆ˜ . Either method is acceptable as long as students are

consistent.

15 20 25 30 35

female

male

amount (000's)

[2] Looking at the boxplots, we see that the median of both distributions is £22,000. This suggests that the benefit packages for male and female students have the same average. [1]



However, the overall spread of the figures for the male students appears to be greater than the corresponding spread for female students (although they have a smaller IQR). This suggests that the variability in the benefit packages for male students is greater than the corresponding variability for female students (although conclusions drawn from such small sample sizes should be treated with caution). The value of 32 for the males could be an outlier. [1] [Total 6] Solution X1.9

(i)(a) Probability of 30 claims in a 3-month period for the whole portfolio The number of claims for the whole portfolio in a 3-month period has a

distribution, using . [1] (50 0.5) (25)Poi Poi¥ = ( ) ( ) ~ ( )Poi Poi Poil m l+ m+

30

2525(30 claims) 0.045430!

P e-= = [1]

(i)(b) Probability of a wait of more than ½ month before a claim is made The number of claims for the whole portfolio in a one-month period has a

( ) ( )25 13 8Poi Poi= 3 distribution. So the waiting time in months, T , between events is:

( )1

3~ 8T Exp [1]

So the probability of waiting more than ½ month is:

1 1 13 3 68 8 41

3½½

( ½) 8 0.0155t tP T e dt e e• •

- - -È ˘> = = - = =Í ˙Î ˚Ú [1]

[Total 4] Alternatively, students could simply use the distribution function:

164( ½) 1 (½) 0.0155P T F e-> = - = =

Students could also work in a 3-month period, using . In which case, they

should calculate

~ (25T Exp )16(P T > ) instead.



(ii) Simulation of claims made in a 3-month period by a single policy We require the distribution function of . From the Tables: ~ (0.5X Poi )

F

, etc [1]

(0) ( 0) 0.60653(1) ( 1) 0.90980(2) ( 2) 0.98561

F P XF P XF P X

= = == £ == £ =

Since , our simulated value is 2. [1] (1) 0.975 (2)F < < [Total 2]



Solution X1.10

(i) Overall mean and standard deviation First, we need the sum of the cash payments and the sum of the squares of the cash payments. To do this we use the definitions of the sample mean and sample variance. Calling the cash payments to clerk A, Ax , and the cash payments to clerk B, Bx , we get: 1

1000 250 250,000A A Ax x x= = fi =Â Â

11,500 240 360,000B B Ax x x= = fi =Â Â

So the overall mean can be found by combining these values:

250,000 360,000Overall mean £2441,000 1,500

+=+

= [1]

Similarly for the variance:

{ }2 2 2 21999

2 2 2

1,000 250 30

30 999 1,000 250 63,399,100

A A

A

s x

x

= - ¥ =

fi = ¥ + ¥ =

ÂÂ

{ }2 2 2 211,499

2 2 2

1,500 240 25

25 1,499 1,500 240 87,336,875

B B

B

x x

x

= - ¥ =

fi = ¥ + ¥ =

ÂÂ [1]

Combining these values, the overall variance is given by:

263,399,100 87,336,875 2500 244 758.69

2499+ - ¥ = [1]

The overall standard deviation is 758.69 £27.54= . [1] [Total 4] (ii) Standard deviation for the adjusted marks Adding 10 onto the marks does not affect the standard deviation (although it would increase the mean by 10). However multiplying the marks by 1.3 increases the standard deviation by the same factor. [1]



Hence the standard deviation of the adjusted marks is 1.3 . [1] 6 7.8¥ = [Total 2] Solution X1.11

(i) Mean Using the formula from page 13 the Tables:

1( )1 4 5

E X aa b

= = =+ +

1 [1]

(ii) Median The median, M , is the value that is halfway through the distribution: ( ) [1] 0.5P X M£ = A distribution has PDF (1, 4)Beta 3( ) 4(1 )f x x= - , so:

[1] 3 40

0

( ) 4(1 ) (1 ) 1 (1 )M M

P X M x dx x MÈ ˘£ = - = - - = - -Î ˚Ú 4

Hence: [1] 41 (1 ) 0.5 0.159M M- - = fi = [Total 3] (iii) Distribution shape Using the result from part (i), the mean is 0.2. Since the mean is to the right of the median this suggests that the distribution is positively skewed. [1]

0

1

2

4

0 0.2 0.4 0.6 0.8 1

x

ME(X) 3

(x)

f



Solution X1.12

(i) Expression for μ Using the formulae for the mean and variance of the lognormal distribution:

2 2 2½ 2and ( 1)e m e em s m s s+ += - 2s=

Eliminating m , we obtain:

( )2 2

22 2 2( 1) ln 1 s

mm e s +s s- = fi = [2]

and, from the equation for : m

( )2

2ln ½ ln 1 sm

m +m = - [1]

[Total 3] We can also express this in an alternative form:

( ) ( ) ( )2 2 2

2 22ln ½ ln 1 ½ ln ½ ln ½ lns m s

m mm mm +

+= - + = - =

4

2 2m

m s

(ii) Probability that the next claim exceeds £20,000 We have and , using the equations above gives: 15,000m = 8,000s =

[1] 29.4906 0.25033m s= = So we have : ~ log (9.4906,0.25033) ln ~ (9.4906,0.25033)X N X Nfi

( )ln 20,000 9.49060.25033

( 20,000) (ln ln 20,000)

( 0.825)

P X P X P Z

P Z

-> = > = >

= > [1] [1] 1 ( 0.825) 1 0.79531 0.20469P Z= - < = - = [Total 3]



(iii) Mean number of claims We are counting the number of failures before the first success. Therefore this is a Type 2 geometric distribution, with . [1] 0.20469p =

Hence, the mean is 0.795310.20469 3.9q

p = = claims. [1]

[Total 2] Alternatively, we could find the mean of a Type 1 geometric distribution and subtract 1.



Solution X1.13

(i) CGF of a gamma distribution From the definition of a moment generating function:

( ) ( )

0 0

( )( ) ( )

tX tx x t xXM t E e e x e dx x e dx

a aa l a ll l

a a

• •-1 - -1 - -= = =

G GÚ Ú [1]

We almost have a PDF, so putting in the appropriate constants: ( , )Ga ta l -

1 ( )

0

( )( )( )( )

t xX

tM t x et

a aa l

al l

al

•- - --=

G- Ú dx [1]

provided tt

al ll

Ê ˆ= Á ˜Ë ¯-< [½]

This follows since 1 ( )

0

( )( )

t xt x ea

a lla

•- - --

GÚ dx is the integral of a PDF over

its whole range, so its value is 1. Dividing the numerator and denominator by

( , )Ga ta l -

l gives:

1( ) 11X

tM t tt

a al

l l

-Ê ˆ Ê ˆ= = -Á ˜Á ˜ Ë ¯-Ë ¯< [½]

Hence, the cumulant generating function is:

( ) ln ( ) ln 1X XtC t M t ta ll

Ê ˆ= = - -Á ˜Ë ¯< [1]

[Total 4]



Alternatively we could use a substitution of into the initial integral obtained above:

( )y tl= - x

11 ( )

0 0

1

0

1( )( ) ( )

1( )

1 ( ) by definition of ( )( )

t x yX

y

yM t x e dx et t

y e dyt

t

aa aa l

aaa

aa

l la a l

la l

l a aa l

• • -- - - -

•- -

Ê ˆ= = Á ˜Ë ¯G G -

Ê ˆ= Á ˜Ë ¯G -

Ê ˆ= G GÁ ˜Ë ¯G -

Ú Ú

Ú

dyl -

This gives:

1( ) 11X

tM t tt

a al

l l

-Ê ˆ Ê ˆ= = -Á ˜Á ˜ Ë ¯-Ë ¯<

( ) ln ( ) ln 1X XtC t M t ta ll

Ê ˆ= = - -Á ˜Ë ¯<

(ii) Coefficient of skewness The definition of the coefficient of skewness is:

1.5skew( )

[var( )]X

X

So using the fact that and , we get: var( ) (0)XX C= ¢¢ skew( ) (0)XX C= ¢¢¢

1

( ) 1XtC t a

l l

-Ê ˆ= -¢ Á ˜Ë ¯ [1]

2

2( ) 1 var( ) (0)XtC t X Call l

-Ê ˆ= - fi = =¢¢ ¢¢Á ˜Ë ¯ 2Xa [1]

3

32( ) 1 skew( ) (0)X

tC t X Call l

-Ê ˆ= - fi = =¢¢¢ ¢¢¢Á ˜Ë ¯ 32

Xa [1]

( )3

1.5 1.52

2 2Coefficient of skewness a l aaaa l

fi = = 2= [1]

[Total 4]



Solution X1.14

(i)(a) Proof of mean result From the definition of the MGF: ( ) [ ]tX

XM t E e= Differentiating with respect to t gives: ( ) [ ]tX

XM t E Xe=¢ Putting gives 0t = 0(0) [ ] [ ]XM E Xe E X= =¢ . [1] (i)(b) Proof of variance result Differentiating again gives: 2( ) [ ]tX

XM t E X e=¢¢ Putting gives 0t = 2(0) [ ]XM E X=¢¢ . [1]

[ ]22 2var( ) [ ] [ ] (0) (0)X XX E X E X M Mfi = - = -¢¢ ¢ [1] [Total 3] Alternatively, parts (i)(a) and (i)(b) can be proved using the expansion:

( )

2 2 3 3

2 32 3

( ) 12! 3!

1 ( ) ( ) ( )2! 3!

tXX

t X t XM t E e E tX

t ttE X E X E X

Ê ˆ= = + + + +Á ˜Ë ¯

= + + + +

( )2

2 3( ) ( ) ( ) (0) ( )2!X XtM t E X tE X E X M E Xfi = + + + fi =¢ ¢

( )2 3( ) ( ) (0) ( )X X2M t E X tE X M E Xfi = + + fi =¢¢ ¢¢



(ii)(a) Obtain MGF of X The PDF can be considered in two parts; 1

2( ) xf x e-= for and 0x > 12( ) xf x e= for

. So the MGF is: 0x <

01 12 2

00

( 1) ( 1)1 12 2

0

( ) [ ]tX tx x tx xX

t x t x

M t E e e e dx e e dx

e dx e

•-

-••

+ -

-•

= = +

= +

Ú Ú

Ú Ú dx [1]

0( 1) ( 1)

0

1 12 1 2 1

t x t xe et t

•+ -

-•

È ˘ È ˘= +Í ˙ Í+ -Í ˙ Í ˙Î ˚ Î ˚

˙ [1]

2

1 1 1 12 1 2 1

11

t t

t

Ê ˆ Ê ˆ= -Á ˜ Á ˜Ë ¯ Ë ¯+ -

=-

[1]

Note that we require for the first integral to be finite, and for the second. So . [1]

1 0t + >1

1 0t - <1 t- < <

(ii)(b) Standard deviation Using the expressions derived in part (i) gives:

[1]

2 1

2 2

2 2 2 2 3

2

( ) (1 )

( ) 2 (1 ) ( ) (0) 0

( ) 2(1 ) 8 (1 )

( ) (0) 2

X

X X

X

X

M t t

M t t t E X M

M t t t t

E X M

-

-

- -

= -

= - fi = =¢ ¢

= - + -¢¢

fi = =¢¢

var( ) 2 ( ) 2X sd Xfi = fi = [1] Alternatively, we could expand this expression as a power series and compare it with

221 [ ] [ ]

2ttE X E X+ + + to give and as before. ( ) 0E X = 2( )E X = 2




(ii)(c) E(X6) The general expansion for an MGF is:

( )

2 2 3 3

2 32 3

( ) 12! 3!

1 ( ) ( ) ( )2! 3!

tXX

t X t XM t E e E tX

t ttE X E X E X

Ê ˆ= = + + + +Á ˜Ë ¯

= + + + +

Now expanding our expression from part (ii)(a) as a power series using the formula given on page 2 of the Tables:

2 1

2 2 2 2

2 4 6

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

2! 3!1

XM t t

t t t

t t t

-= -- - - - -= + - - + - + - +

= + + + +

3

[1]

Alternatively, we could spot that the MGF is of the form 1

ar-

with and , so

it is the sum to infinity of the geometric series:

1a = 2r t=

2 3 2 4 61a ar ar ar t t t+ + + + = + + + + Comparing the coefficients of gives: 6t

6 61 ( ) 1 ( ) 6! 7206!

E X E X= fi = = [1]

[Total 8] Another alternative would be to differentiate the MGF six times and substitute in . 0t =



(i) 9,24(F < 3.256)P From page 174 of the Tables, we see that . Hence: 9,24( 3.256) 0.01P F > =

9,24 9,24( 3.256) 1 ( 3.256) 1 0.01 0.99P F P F< = − > = − = [1] (ii) 3,5(F < 0.18836)P

Since 0.18836 is less than 1, we need to use the ,

1

m nF result:

3,53,5

5,3

1 1( 0.18836)0.18836

( 5.309)

0.10

P F PF

P F

⎛ ⎞< = >⎜ ⎟⎜ ⎟

⎝ ⎠= >

= [1] (iii) Find the value of a Since 95% of the distribution is greater than a, this tells us that it must be a lower

critical point. Hence we need to use the ,

1

m nF result again:

8,6 6,88,6

6,8

1 1 1( ) 0.95

1 0.05

1 3.581

0.279

P F a P P FF a a

P Fa

aa

⎛ ⎞ ⎛ ⎞> = < = < =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎛ ⎞⇒ > =⎜ ⎟⎝ ⎠

⇒ =

⇒ = [1] [Total 3]



Solution X2.2

Starting with the MGF of the geometric distribution: 1( ) , 1, 2, 3,xP X x pq x−= = = …

[1] ( ) 1 2 2 3

1( ) tX tx x t t t

Xx

M t E e e pq pe pqe pq e∞

−

=⇒ = = = + + +∑

This is an infinite geometric series with and . So using ta pe= tr qe=1

aSr∞ =

−:

( )1

t

X tpeM t

qe=

− [½]

Now the negative binomial is the sum of k independent geometric distributions:

( ) ( ) ( )( ) ( )

1 1

1

1

(( )

by independence

( ) ( )

k k

k

k

t X X tXtXtYY

tXtX

X X

M t E e E e E e e

E e E e

M t M t

+ += = =

=

=

…

…

… [1]

Since the question states “from first principles” the proof of the above result should not be assumed.

So:

( )1

kt

Y tpeM t

qe

⎛ ⎞= ⎜⎜ −⎝ ⎠

⎟⎟ [½]

[Total 3] Alternatively, using series:

( ) ( )

( ) ( ) ( ) ( )2 3

1( )

1

11

( 1) ( 1)( 2)12! 3!

tx k x kX

x k

k x kt t

x kkt t t t

xM t e p q

k

xpe qe

k

k k k k kpe k qe qe qe

∞−

=∞ −

=

−⎛ ⎞= ⎜ ⎟−⎝ ⎠

−⎛ ⎞= ⎜ ⎟−⎝ ⎠

+ + +⎡ ⎤= + + + +⎢ ⎥⎣ ⎦

∑

∑



Using the binomial expansion on page 2 of the Tables:

( ) ( )( ) 11

ktk kt tX t

peM t pe qeqe

− ⎛ ⎞= − = ⎜ ⎟⎜ ⎟−⎝ ⎠

Solution X2.3

(i) Variance of the sample mean is 2? X has a distribution, so it is true. [1] (40 2)N , (ii) Mean of the sample variance is 20?

2

29Ss

has a 29c distribution, so

2292

9 [ ] 9È ˘

=Í ˙Í ˙

=Î ˚

SE E cs

. So , so it is

true. [1]

2 2[ ] 20= =E S s

Alternatively, using the fact that is an unbiased estimator of for any distribution: .

2S 2s2 2[ ] 20= =E S s

(iii) Probability sample variance exceeds 30 is less than 5%?

2920S is 2

9c . So 22 29

9( 30) 13.5 ( 13.5) 0.141320È ˘= = =Í ˙Î ˚

P > P > P >S S c using page

165 of the Tables. So (c) is false. [1] [Total 3]



Solution X2.4

This is based on Subject 101, September 2000, Q4.

Using 1~XnS n

tm-- from page 22 of the Tables, we have:

83~

9X t

SSfi 8~X t [1]

The required probability can then be calculated as:

83( ) 1 3 (X XP X S P P P t

S SÊ ˆ Ê ˆ

> = > = > = >Á ˜ Á ˜Ë ¯ Ë ¯3) [1]

From the Tables on page 163, the required probability to 2 decimal places is 0.01. [1] [Total 3] Interpolation produces a value of 0.0089, but given to 2 decimal places this is 0.01. Solution X2.5

Using subscript X for the first sample and subscript Y for the second:

2

2/ 20/ 40

X

Y

SS

has an distribution, ie 4,19F2

22 X

Y

SS

is [1] 4,19F

So, using tables of the distribution: F

2

2 24,192

2 [ 2.25 ] 4.5 ( 4.5) 0.01X

X YY

SP S S P P F

S

È ˘> = > = > ªÍ ˙

Í ˙Î ˚ [1]

[Total 2]



Solution X2.6

Let be the number of policies that make one or more claims. Then: N [½] ~ (100, 0.2) ( ) 20 var( ) 16fi = =N Bin E N N Let iX be the number of claims arising on the ith policy making a claim. Then: 2

3~ (1,0.6) ( ) 1 var( ) 1fi = =iX NBin E X X 19 [½]

Let be the total number of claims for this portfolio, such that . S1=

= ÂN

ii

S X

23( ) ( ) ( ) 20 1 33= = ¥ =E S E N E X 1

3 [1]

2

21 29 3

23

var( ) ( ) var( ) ( ) var( )

20 1 (1 ) 16

66

= +

= ¥ + ¥

=

S E N X E X N

[1]

Hence:

23standard deviation 66 8.165= = [1]

[Total 4]



Solution X2.7

Let X be the time for the first review and Y be the time for the second. Then: 2 2~ (8,2 ) and ~ (8,2 )X N Y N We require: ( 3- <P X Y ) So we need the distribution of -X Y , which is: [1] 2 2~ (8 8, 2 2 ) ~ (0, 8)- - +X Y N N Therefore: ( 3) ( 3- < = - < - <P X Y P X Y 3) [1]

3 38 8

( 1.061) ( 1.061)

( 1.061) [1 ( 1.061)]

Ê ˆ= - < <Á ˜Ë ¯

= < - < -

= < - - <

P Z

P Z P Z

P Z P Z [1]

[1]

2 ( 1.061) 1

2 0.85566 1

0.71132

= <

= ¥ -

=

P Z -

[Total 4]



Solution X2.8

(i) Proof var( ) cov( , )X Y X Y X Y+ = + + cov( , ) cov( , )X X Y Y X Y= + + + [1]

cov( , ) cov( , ) cov( , ) cov( , )

var( ) 2cov( , ) var( )X X X Y Y X YX X Y Y

= + + += + +

Y

2

)]

[1] [Total 2] Alternatively:

2

2

2 2

var( ) [{( ) ( )} ]

[{( ( )) ( ( ))} ]

[( ( )) ] [( ( )) ] 2 [( ( ))( ( ))]var( ) var( ) 2cov( , )

X Y E X Y E X Y

E X E X Y E Y

E X E X E Y E Y E X E X Y E YX Y X Y

+ = + − +

= − + −

= − + − + − −= + +

or:

2 2

2 2 2

2 2 2

2 2 2 2

var( ) [( ) ] [ ]

[ 2 ] [ ( ) ( )]

( ) 2 ( ) ( ) ( ) 2 ( ) ( ) ( )

[ ( ) ( )] [ ( ) ( )] 2[ ( ) ( ) (var( ) var( ) 2cov( , )

X Y E X Y E X Y

E X XY Y E X E Y

E X E XY E Y E X E X E Y E Y

E X E X E Y E Y E XY E X E YX Y X Y

+ = + − +

= + + − +

= + + − − −

= − + − + −= + +

(ii) Standard deviation Using the definition of the correlation coefficient gives:

2 2

cov( , )37 5 7

corr( , ) cov( , ) 15X YX Y X Y¥

= - = fi = - [1]

Using the result from part (i): var(3 2 5) var(3 2 )X Y X- + = - Y using var( ) var( )aX b aX+ = var(3 ) var( 2 ) 2cov(3 , 2 )X Y X= + - + - Y [1] 9var( ) 4var( ) 12cov( , )X Y X= + - Y

[½] 2 29 5 4 7 12 15= ¥ + ¥ - ¥ - 601=



Hence the standard deviation is 601 24.5= . [½] [Total 3] Solution X2.9

Now X has a Poisson distribution with mean 6. Hence: ( ) 6 and var( ) 6E X X= = To get the unconditional variance of, we use var( ) [var( | )] var[ ( | )]Y E Y X E Y X= + from page 16 of the Tables:

2

2

var[ ] [var( | )] var[ ( | )]

var[3 5]4

1 [ ] 9var[ ]4

Y E Y X E Y X

XE X

E X X

= +

È ˘= + -Í ˙

Í ˙Î ˚

= + [1]

Now using the fact that 2 2( ) var( ) ( )E X X E= + X :

( )2

2

1var( ) var[ ] [ ] 9var[ ]4

1 (6 6 ) 9 64

64.5

Y X E X= + +

= + + ¥

=

X

[1]

( ) 64.5 8.031sd Yfi = = [1] [Total 3]



Solution X2.10

(i) Discrete conditional distribution of X given Y = y If we draw up a table of possible values for X and Y , we have:

X 0 1 2

0 0 c 2c 1 2c 3c 4c Y 2 4c 5c 6c

Since , we have ( , )= = =ÂÂx y

P X x Y y 1 127

=c . [1]

Now ( ,( | )( )= == =

=P X x Y yP X Y y

P Y y) .

2

0

( ) ( , )

1 ( 2 )27

1 [2 (1 2 ) (2 2 )]27

1 3[3 6 ] [1 2 ]27 27

=

= = = =

= +

= + + + +

= + = +

Â

Â

x

x

P Y y P X x Y y

x y

y y

y

y

y [1]

1 ( 2 ) 227( | ) 3 3(1 2 )(1 2 )27

+ += = =++

x y x yP X Y yyy

[1]

[Total 3] Note: Students do not actually need to calculate the value of c to get the answer and full marks.



Alternatively, we could calculate the three conditional distributions:

1 2( 0 | 0) 0 ( 1 | 0) ( 2 | 0)3 3

= = = = = = = = =P X Y P X Y P X Y

2 3( 0 | 1) ( 1 | 1) ( 2 | 1)9 9

= = = = = = = = =P X Y P X Y P X Y 49

4 5( 0 | 2) ( 1 | 2) ( 2 | 2)15 15 15

= = = = = = = = =P X Y P X Y P X Y 6

or we could give the answer as:

( | 0)3

2( | 1)9

4( | 2)15

xP X x Y

xP X x Y

xP X x Y

= = =

+= = =

+= = =

for . 1, 2, 3x = (ii) Continuous conditional distribution of X given Y = y Now using gives: ( , ) 1=Ú Ú

x y

f x y dxdy

2 2 2 222

00 0 0 0

( 2 ) (2 4)y

x y x x

c x y dxdy c xy y dx c x dx=

= = = =

È ˘+ = + = +Î ˚Ú Ú Ú Ú 1=

So:

220

4 12

112

xc x x c

c

=È ˘+ = =Î ˚

fi =

1

[1]



Now ,|

( , )( , )

( )= = X YX Y y

Y

f x yf x y

f y:

2 2

2

00

1 1 1 2( ) ( 2 ) 2 (1 2 )12 12 2 12==

È ˘= + = + = +Í ˙Î ˚ÚYxx

f y x y dx x xy y [1]

|

1 ( 2 ) 212( , ) 2 2(1 2 )(1 2 )12

=

+ += =++

X Y y

x y x yf x yyy

[1]

[Total 3] Note: Again students do not actually need to calculate the value of c to get the answer and full marks.



Solution X2.11

(i) Probability using normal approximation Let X be the number of squares moved up the drain pipe each turn. Hence:

16

512

13

112

1

0

1

2

X

Ï-ÔÔÔÔ= ÌÔ+ÔÔ+ÔÓ

[½]

To use a normal approximation, we first need to calculate and ( )E X var( )X : 1 1 1

6 3 6( ) ( ) 0E X xP X x= = = - + + + =Â 13 [½]

2 2 51 1 16 3 3 6

2 2 5 16 9 18

( ) ( ) 0

var( ) ( ) ( )

E X x P X x

X E X E X

= = = + + + =

fi = - = - =

Â13 [1]

Using the Central Limit Theorem, we know that the approximate distribution of the total movement after 15 turns is: ( ) ( )13 51

3 18~ 15 ,15 5 ,10X N N¥ ¥ =Â 6 [1]

So to obtain ( )8P X >Â , the probability that a spider has moved up more than 8

squares, we first use a continuity correction ( ) ( )8P X P X> Æ >Â Â 8.5 . [1]

Standardising, by setting

56

8.5 510

1.063z -= = , gives:

( )8.5 ( 1.063) 1 ( 1.063)

1 0.85611 0.14389

P X P Z P Z> = > = - <

= - =

Â [1]

[Total 5]



(ii) Suitability of normal approximation The Central Limit Theorem requires n to be large. Fifteen turns is not large, therefore this will be a poor approximation. [1] Solution X2.12

(i) Moment generating function

The aggregate claim amount, Y is given by . The moment generating function of Y is given by:

1 nY X X= + +

( ) ( )( ) | using ( ) [ ( | )]tY tYYM t E e E E e N E Y E E Y NÈ ˘= = =Î ˚ [1]

( )( ) ( )

( )

1

1

1

using independence

( ) ( )

( )

N

N

N

tXtX

tXtX

X X

NX

E E e e

E E e E e

E M t M t

E M t

…

…

…

È ˘= Î ˚

È ˘= Î ˚

È ˘= Î ˚

È ˘= Í ˙Î ˚ [1]

[ ]ln ( ) ln ( )XN M tN XE e M M tÈ ˘= =Î ˚ [1]

[Total 3] (ii) Coefficient of skewness

For the gamma distribution the mean and variance are al and 2al

respectively. Hence,

2150 and 50 450 and 3a al l

a= = fi = l = . [1]

So the moment generating functions are:

( )4501

3

12( 1)

45013

12 (1 ) 1

( )

( ) 1

( )

teN

X

tY

M t e

M t t

M t e-

-

-

È ˘- -Î

=

= -

= ˚ [1]



The cumulant generating function is given by:

45013( ) 12 (1 ) 1YC t t -È ˘= - -Î ˚

4511

3( ) 1,800(1 )YC t t -fi = -¢

45213

45313

( ) 270,600(1 ) var( ) (0) 270,600

( ) 40,770,400(1 ) skew( ) (0) 40,770,400

Y Y

Y Y

C t t Y C

C t t Y C

-

-

= - fi = =¢¢ ¢¢

= - fi = =¢¢¢ ¢¢¢ [1]

So the coefficient of skewness is given by:

1.540,770,400270,600

0.290= [1]


(i) Approximate probability when n = 400 We have . Using a Poisson approximation: ~ (400, 0.01)X Bin ~ (400, 0.01) ~ (4)X Bin Poi [1] We require . Using the cumulative Poisson tables on pages 176 and 180: (9 16)P X≤ ≤ [1] (9 16) ( 16) ( 9) ( 16) ( 8)P X P X P X P X P X≤ ≤ = ≤ − < = ≤ − ≤ [1] 1.00000 0.97864 0.02136= − = [Total 3] A normal approximation is not valid here as 4 5np = < . Students should only receive 1 mark for using this method and obtaining the answer of 0.0119. (ii) Approximate probability when n = 3,000 We have . Using a normal approximation: ~ (3000, 0.01)X Bin [1] ~ (3000 0.01, 3000 0.01 0.99) (30, 29.7)X N N¥ ¥ ¥ = This approximation is valid since and . 30 5np = > (1 ) 2,970 5n p- = >



Using a continuity correction: [1] (9 16) (8.5 16.5)£ £ ª < <P X P X Evaluating this:

( )8.5 30 16.5 3029.7 29.7

(9 16)

( 3.945 2.477)

P X P Z

P Z

- -£ £ = < <

= - < < - [1]

( 2.477) ( 3.945)

[1 ( 2.477)] [1 ( 3.945)]

( 3.945) ( 2.477)

P Z P Z

P Z P Z

P Z P Z

= < - - < -

= - < - - <

= < - < [1] 0.99996 0.99337 0.00659= - = [Total 4] Using the complete decimal when interpolating would result in 0.00658. A Poisson approximation of is not tabulated – it is not the best method! (30)Poi Solution X2.14

(i) Sum of two independent gamma random variables Let , and 1 1~ (X Gamma a l, ) 2 2~ (X Gamma a l, ) 21= +Z X X . Then:

( ) ( ) ( )( ) ( )

1 2 1 2

1 2

1 2

( )

by independence

( ) ( )

+= = =

=

=

tX tX tX tXtZZ

tX tX

X X

M t E e E e E e e

E e E e

M t M t [1]

Therefore:

1 2 1( )

( ) 1 1 1- - -Ê ˆ Ê ˆ Ê ˆ= - - = -Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯Z

t t tM ta a a

l l l

2+a

))

This is the MGF of a distribution. Hence, by the uniqueness property of MGFs, . [1]

1 2( ,+Gamma a a l

1 2~ ( ,+amma a a lZ G [Total 2]



(ii) MGF Generalising the result of part (i), where , gives: 1 = = =na a a ( )~ , ~ (Â Âi iX Gamma Gamma na l a l, ) [1]

So the MGF of Â iX is 1nt a

l

-Ê ˆ-Á ˜Ë ¯. Hence:

( )22 ( ) (2 )

21 (1 2-

-

Â= =Â Â

Ê ˆ= - = -Á ˜Ë ¯

ii i

t XX X

nn)

M t E e M t

t t

ll

aa

l

ll

[1]

This is the MGF of a distribution. Hence, by the uniqueness property of MGFs,

. [1]

22nac

222 ~Â i nX al c

[Total 3] (iii) Probability of sample mean exceeding 40 Here 2

10 205 10 ~ ~= fi fiÂ iX X X Xal c c 2 . [1] Using the percentage points for the 2c distribution on page 169 of the Tables: 2

20( 40) ( 40) 0.005> = > =P X P c [1] [Total 2]



Solution X2.15

(i)(a) Show X and Y not independent If X and Y are independent, then: ( , ) ( ) ( ) for all ,P X x Y y P X x P Y y x y= = = = = This is not this case. For example: 3 91

2 4 16( 0, 0)P X Y= = = ≠ × [1]

Any suitable counterexample will be sufficient to obtain the mark, provided that the student explains why it shows dependence. (i)(b) Show X and Y not uncorrelated If X and Y are uncorrelated, then: corr( , ) 0 cov( , ) 0 ( ) ( ) ( )X Y X Y E XY E= ⇒ = ⇒ = X E Y We have: ( ) ( ) ( )3 3 1

4 16 16( ) ( ) 0 1 2x

E X xP X x= = = × + × + × 516=∑ [1]

( ) ( ) ( )9 1

16 8 16 4( ) ( ) 0 1 2y

E Y yP Y y= = = × + × + × 5 3=∑ [½]

( ) ( ) ( )

( ) ( )12

1 18 16

14

( ) ( , )

0 0 0 1 0 2 2 0

1 1 1 2

x yE XY xyP X x Y y= = =

= × × + × × + + × ×

= × × + × ×

=

∑∑

[1]

Since 5 31

4 16 4( ) ( ) ( )E XY E X E Y= ≠ × = , they are not uncorrelated. [½]

[Total 4]



(ii) When is relationship true? This relationship is always true (provided all the quantities involved make sense). [1] We’ve included this question to emphasise that you can always use this result. There is no requirement for X and Y to be independent, uncorrelated, normally distributed or whatever. (iii)(a) ( + | = 1)E X Y X Since 3

16( 1)P X = = :

[ | 1] ( ) ( , |

( 1, )( )( 1)

x y

x y

E X Y X x y P X x Y y X

P X Y yx yP X

+ = = + = = =

= == +=

ÂÂ

ÂÂ

1)

1 8 1 162 33 16 3 16

= ¥ + ¥ [1]

2 13 3

13

2 3

2

= ¥ + ¥

= [1]

(iii)(b) ( | = 2)E X Y Since 5

16( 2)P Y = = :

[ | 2] ( | 2)

( ,( 2)

x

x

E X Y xP X x Y

P X x YxP Y

= = = =

= ===

Â

Â 2)

1 4 1 16 00 1 25 16 5 16 5 16

= ¥ + ¥ + ¥ [1]

( ) ( )4 15 5

15

0 1 (2= ¥ + ¥ + ¥

=

0)

[1]



(iii)(c) var( | = 2)X Y

2var( | 2) [ | 2] ( [ | 2])X Y E X Y E X Y= = = - = 2 [1]

But: ( ) ( )2 2 2 24 15 5[ | 2] ( | 2) 0 1

xE X Y x P X x Y= = = = = ¥ + ¥ =Â 1

5 [1]

So: ( )21 1 45 5 2var( | 2)X Y = = - = 5 [1]

[Total 7]

(iv) Calculate ( )È ˘Î ˚

2 |E E Y X

2[ | ]E Y X depends on the value of X . So we need to work out 2[ | ]E Y X x= for each

possible value of x :

( ) ( )

( ) ( )

2 2 2 22 13 3

2 2 2 22 13 3

2 2

[ | 0] ( | 0) 0 2 1

[ | 1] ( | 1) 1 2 2

[ | 2] ( | 2) 0

y

y

y

E Y X y P Y y X

E Y X y P Y y X

E Y X y P Y y X

= = = = = ¥ + ¥ =

= = = = = ¥ + ¥ =

= = = = =

Â

Â

Â

13

Combining these, we have:

13

2

1 if

[ | ] 2 if 1

0 if

X

E Y X X

X

Ï =ÔÔ= ÌÔ

=ÔÓ

0

2

= [1]

To calculate 2[ ( | )]E E Y X , we need to apply the probabilities of the possible values of X : [1] 2 2[ ( | )] [ | ] ( )

xE E Y X E Y X x P X x= =Â =

( ) ( ) ( )3 313 4 16 16 81 2 0= ¥ + ¥ + ¥ = 31 1 [1]

[Total 3]




(v) Calculate ( )2E Y

The marginal probabilities for Y are: 91 1

2 16 16( 0)P Y = = + =

18( 1)P Y = =

and 51 14 16 16( 2)P Y = = + =

So: ( ) ( ) ( )2 2 2 2 29 116 8 16 8[ ] ( ) 0 1 2 1

yE Y y P Y y= = = ¥ + ¥ + ¥Â 5 3=

2

[1]

As expected, . [1] 2[ ( | )] [ ]E E Y X E Y= [Total 2]



Using 2

212

( 1) ~ nn S cs -- , we can obtain a 90% confidence interval for from: 2s

2

2 21;0.95 1;0.052

( 1)0.9 n nn SP c cs- -

Ê ˆ-= < <Á ˜Ë ¯

which gives:

2 2

2 21;0.05 1;0.95

( 1) ( 1),n n

n s n sc c- -

Ê ˆ- -Á ˜Ë ¯

[½]

From the data, we have:

2

2 1 1281,168 16 9.615 16

sÈ ˘Ê ˆ= - ¥ =Í Á ˜Ë ¯Í ˙

˙Î ˚

[½]

This gives:

15 9.6 15 9.6, (5.76,19.8)25.00 7.261¥ ¥Ê ˆ =Á ˜Ë ¯

[1]

However, we require a 90% confidence interval for the standard deviation, so taking the square root: [1] (2.4, 4.45) [Total 3]



Solution X3.2

(i) State the value of k Since we know that the distribution is defined on the values 1, 2, 3, ... we need to use a negative binomial distribution which is defined on these values, ie we should use a Type 1 negative binomial with . [1] 1k = (ii) Estimate the value of p

If we calculate the sample mean, we find that 104 1 350677

x .= = . The mean of this

distribution is k p . [1] So, if we use the method of moments and equate the sample and population means, and put , we get: 1k =

1 1 3506 0 7404. p .p= fi = [1]

[Total 3] Students may try to use the method of moments to estimate both k and p. Using

( ) 1.3506E X x= = and 2 2 182177( ) 2.3636inE X x= = =Â gives and

. Alternatively, using

0.7146p =

0.9652k = ( ) 1.3506E X x= = and gives and . However since k must be a positive integer this method

should receive only 1 mark in total.

2 0.54648= =var( )X s0.7119p = 0.9616k =



Solution X3.3

(i) Confidence interval for mean (known variance)

We know that Xnm

s- has a distribution. So a symmetrical 95% confidence

interval will be given by:

(0,1)N

0.95 ( 1.96 1.96) or 1.96 or 1.96xP Z xn nm s

s-= - < < = ± ± [1]

For the data values given, 47.6x = . So the confidence interval is:

2047.6 1.96 47.6 3.92 (43.68,51.52)5

± ¥ = ± = [1]

[Total 2] (ii) Confidence interval for mean (unknown variance)

Since XS n

m- has a distribution, we will need to use the t tables. We want the

value of the distribution for which . From the Tables this value is 2.776.

1nt -

0.025a =

So the confidence interval is 2.776 sxn

± , where: [1]

2 21 11,746 5 47.6 104.34

s È ˘= - ¥ =Î ˚

So the confidence interval is 47.6 . [1] 12.68 (34.92,60.28)± = [Total 2]



Solution X3.4

Now: 2 4( an odd number)P X p pq pq= = + + +

This is an infinite geometric series with and , using a p= 2r q=1

aSr• =

- we get:

21 ( an odd number)

(1 )(1 ) 11p pP X

q qq= = =

- + +- q=

+

[1]

Similarly: 3 5( an even number)P X pq pq pq= = + +

This is an infinite geometric series with and , using a pq= 2r q=1

aSr• =

- we get:

2( an even number)(1 )(1 ) 11

pq pqP Xq qq

= = =- + +-

qq

= [1]

Alternatively, and more simply, students may use:

1( an even number) 1 ( an odd number) 11 1

qP X P Xq q

= = - = = -+ +

=

So the likelihood of equal numbers of even and odd values of X is:

22 21

1 1 (1 )

n n n

nn nq q

n n+q +q +qÊ ˆ Ê ˆ Ê ˆ Ê ˆ

=Á ˜ Á ˜ Á ˜ Á ˜Ë ¯ Ë ¯ Ë ¯ Ë ¯ [1]

[Total 3]

Students may also give the alternative answer of 2

2 22

(1 )

n n

nn p q

n qÊ ˆÁ ˜Ë ¯ -

.



Solution X3.5

(i) Likelihood The likelihood function is:

[ ] [ ] [ ]30 15 5( ) constant (0 1) (1 2) (2 )L P X P X P Xl = ¥ < < < < < < • [½] where the constant arises from the different possible permutations of the 50 observations amongst the 3 groups (as we don’t know exactly which 30 values are in the first group, etc). [½] The value of the constant is 50!

30!15!5! . Using from page 11 of the Tables: ( )F x (0 1) (1) 1P X F e l-< < = = -

2(1 2) (2) (1) l- -< < = - = -P X F F e e l

el

e

[½] 2( 2) 1 (2)P X F e l-> = - = Using these probabilities: [½] 30 2 15 2 5( ) constant (1 ) ( ) ( )L e e el l l ll - - - -= ¥ - -

[½] 30 15 15 10constant (1 ) ( ) (1 ) ( )e e e el l l- - - -= ¥ - -

25 45constant (1 )e el l- -= ¥ -

ln ( ) constant 25 45ln(1 )L ll l -fi = - + - [½] [Total 3] (ii) Maximum likelihood estimate Differentiating the log-likelihood with respect to l gives:

45ln 251

eLe

l

l∂∂l

-

-= - +-

[1]



Setting this derivative equal to 0:

25 5 14ln 1.03045 14 51

e ee

ll

l l-

-- = fi = fi = =

- [1]

Checking the second derivative:

( )( ) ( )

2

2 2 2

45 1 45 45ln 0 max1 1

- - - - -

- -

- - - -= =- -

e e e e eLe e

l l l l l

l l

∂∂l

< fi [1]

[Total 3]



Solution X3.6

(i) Confidence interval for the probability a missile hits target Let be the number of hits and p be the probability of a single missile hitting. Then: S ( )~ (100, ) ~ 100 ,100 (1 )S B p N p p p- or:

( )(1 )100ˆ ~ , p pp N p -

Estimating p in the variance as 72

100ˆ 0 72p .= = , we have:

72 100 0.72or20.16 0.2016 100

p pZ Z-= = - [1]

So: 0.90 ( 1.6449 1.6449)P Z= - < < Rearranging the inequality:

72 1.6449 20.16 72 1.6449 20.16100 100

p- +< <

0.2016 0.2016100 100or 0.72 1.6449 0.72 1.6449p- < < + [1]

Hence, our confidence interval for p is 0.72 . [1] 0.074 (0.646,0.794)± = [Total 3] (ii) Confidence interval for the probability that the target plane is destroyed The probability that a target plane is destroyed is given by a 31 (1 )pa = - - . [1] So a 90% confidence interval for can be obtained by inserting the limits on the 90% confidence interval for

ap obtained in (i) into the formula for given above. a

Hence, the required confidence interval is:

( ) ( )( )2 21 1 0.646 , 1 1 0.794 (0.956, 0.991)- - - - = [1]

[Total 2]



Solution X3.7

(i) MLE of mean The likelihood is:

1

1 1

1

1

( ) ( ) ( )

! !

! !

n

i

n nxx

nx

n

n

L P X x P X x

e ex x

ex x

m m

m

m

m m

m

- -

-

= = ¥ ¥ =

= ¥ ¥

Â=

…

…

… [1]

Taking logs: 1ln ( ) ln ln( ! !)i nL x x x nm m= - -Â … m [½] Differentiating and setting equal to zero:

ln ( ) ixd Ld

mm m

= -Â n [½]

ˆˆ

i ix xn x

nm

mfi = fi =Â Â =

Checking that this gives a maximum:

2

2 2ln ( ) 0 maxixd Ld

mm m

= - £ fiÂ [½]

So the estimator is ˆ Xm = . [½] [Total 3] (ii) Bias and MSE The bias of m is given by ˆ( )E m m- . Now: ( )1 1 1 1ˆ( ) ( ) ( )i in n n nE E X E X E X nm m m= = = = = =Â Â Â m

Hence the bias is zero. [1]



Students must show the working to receive the mark. The MSE of m is given by 2ˆ ˆvar( ) ( )biasm m+ . Now:

( ) 2 2 21 1 1 1ˆvar( ) var( ) var var( )i in nn n n

X X X n mm m= = = = =Â Â Â m =

So the MSE of m is nm . [1]

[Total 2] Students must show the working to receive the mark. Half marks to be given if students give the answer 2 ns (and fail to utilise the fact that the variance of ( )Poi m is m ). (iii) Variance attains CRLB

The CRLB of m is given by 22

1

ln ( )dd

E Lm

m-

È ˘Í ˙Î ˚

. Using the second derivative from

part (i), we get:

2

2 2 2 2

2 2

1 1ln ( ) [ ]

1 1

ii i

XdE L E E X Ed

nn

mm m m m

m mmm m

È ˘È ˘È ˘= - = - = -Í ˙Í ˙ Î ˚Í ˙ Í ˙Î ˚ Î ˚

= - = - = -

Â Â Â

Â

X

[1]

So the CRLB is given by nm . This is the same as the variance of m . Hence the

variance attains the CRLB. [1] [Total 2]



Solution X3.8

(i) Normal approximation confidence interval Since ~ ( )X Poi l , the distribution of X can be approximated by . So ( )N , l l

ˆX

nl

l-

has approximately a distribution. [1] (0, 1)N

So 0.99 2 5758 2 5758ˆ

XP . < < .n

l

l

Ê ˆ-= Á- ˜Á ˜Ë ¯

. [1]

Using ˆ 0.9 Xl = = as an approximation for l in the denominator, and rearranging:

0.9 0.910 100 9 2 5758 0 9 2 5758. . < . + .l- <

which gives a confidence interval for l of: 0.9 [1] 0.773 (0.127,1.673)± = [Total 3] (ii) Comment The approximation is not brilliant as:

• the Poisson parameter is small (the approximation is better for large values) [1]

• the sample size of 10 is small (the approximation is better for large samples) [1]

• a continuity correction is not used (we have used a continuous distribution to approximate a discrete one) [1]

• an estimate for l is used in the variance. [1] [Maximum 2]



(iii) Accurate confidence interval Since , the equations required to obtain the accurate confidence interval are:

~ (10iX Poi lÂ )

9 10

0

(10 ) 0.005!

x

x

ex

l l-

==Â to obtain the upper limit [1]

8 10

0

(10 ) 0.995!

x

x

ex

l l-

==Â to obtain the lower limit [1]

[Total 2]



Solution X3.9

(i)(a) MLE The likelihood function based on a sample of observations is given by: n

, [1] 2( ) 2( )

12 2i i

nx x nn

iL e eq q- - - Â -

== =’ 1provided , ,

( min )n

i

x xie x

qq

≥≥

i

So , the MLE of , is the value of that maximises q q q 2( )2 ix nne q- Â - subject to the condition that (otherwise the likelihood is zero). m i q £ in x

x When , we have: min i q £

2( )2ln ln 2 2( )

ln 2 0

ix nn

i

L eL n x n

L n >

q

q∂∂q

- -Â=fi = - Â -

fi = [½]

ie increases as increases. L q Markers: please give credit for appropriate alternative solutions (as usual). So the MLE of is the highest value of subject to the condition that q q min ixq £ , ie ˆ min iXq = is the maximum likelihood estimator of . [1] q

From this sample, the maximum likelihood estimate of is 1.56. [½] q [Total 3]



(ii) Method of moments estimate Now has an X q- (2)Exp distribution, so:

1[ ] [ ]2

E X E X +q q q= - + = [1]

Alternatively, from first principles:

2( )

2( ) 2( )

2( )

( ) 2

by parts

1 12 2

x

x x

x

E X xe dx

xe e dx

e

q

q

q qq

q

q

qq q

•- -

••- - - -

•- -

=

È ˘= - +Î ˚

È ˘= + - = +Í ˙Î ˚

Ú

Ú

Equating to the sample mean, ( )E X x , gives:

1 1 2.252 ix x

nq + = = Â = [½]

So the method of moments estimate of is . [½] q 2.25 0.5 1.75- = [Total 2] (iii) Comment One of the observed values was less than the method of moments estimate of ! So the method of moments gives an estimate of that is not “possible” in this case. [½]

qq

This contrasts with the situation for maximum likelihood estimators, which, provided they exist, must, by definition, give feasible estimates. [½] [Total 1]



Solution X3.10

(i) Formula for MLE The likelihood function is:

( ) ( )21 88 8 22 0.5

1 1 1

1 exp 22 0.5

iX

i ii i i

L f x e C Xm

mp

-Ê ˆ- Á ˜Ë ¯

= = =

Ê ˆ= = = - -Á¥ Ë ¯

Â’ ’ ˜ [1]

where C is a constant. So the log-likelihood is:

( )8

2

1ln ln 2 i

iL C X m

== - -Â [½]

Differentiating with respect to m :

( )8 8

1 1ln 4 4 32i i

i iL X Xm m

m = =

∂ = - = -∂ Â Â [½]

Setting this equal to 0 gives:

8

18i

iX m

==Â

So:

8

1

18 i

iX Xm

== Â = [½]

Differentiating again:

2

2 ln 32 0 maxLm∂ = - < fi∂

[½]

So m , the maximum likelihood estimator of m , is equal to X . [Total 3]



(ii) Maximum likelihood estimate of q By the invariance property of MLEs, the maximum likelihood estimate of is: q ( )ˆ ˆ3% | ~ ( ,0.25)P X X Nq = > m [½] From the given data: ˆ 2.1875%xm = = [½] So:

( )

( )( )

ˆ 3% | ~ (2.1875,0.25)

3 2.18750.5

1.625

1 1.625

1 0.9479

0.0521

= >

-Ê ˆ= >Á ˜Ë ¯

= >

= - F

= -

=

P X X N

P Z

P Z

q

[1] [Total 2]



Solution X3.11

(i) Confidence interval for mean Let the true mean age of wood found at this site be m . Then:

1~ nX tS n

m-

-

46 960 4 69610

= =,x , [½]

2 1{220 772 800 10 4,696 } 27,6279

= - ¥ =s , , 2 [½]

Now, using the distribution: 9t

2.262 2.262XS n

m-- < < [1]

2.262 2.262S SX Xn n

m- < < +

This gives a confidence interval for m of . [1] 4,696 119 (4,577, 4,815)± = [Total 3] (ii) Dotplot The dotplot for the ages of the sample:

4300 4400 4500 4600 4700 4800 4900

age (years)

[1] For our confidence interval in part (i), we require an assumption of underlying normality (or a large sample size). The line plot indicates a normal distribution, except for the extreme value of 4,300, which may be an outlier. However, more data would be needed to ascertain whether this is an outlier or not. If the value of 4,300 is an outlier and the underlying distribution is not normal, then the calculation of the confidence interval is not valid for a sample this small. [1] [Total 2]



(iii) Minimum sample size needed We require:

1;0.0251;0.0252 200 0.60164n

ntst

n n-

-¥ £ fi £ [1]

Trial and improvement leads to 13;0.025 0.577314

t= . Therefore a sample size of at least

14 is required. [2] [Total 3] (iv) Confidence interval for difference between means Using Xm and Ym to denote the true mean age of wood at the first and second sites respectively. Then:

22

( ) ( ) ~1 1 X Y

X Yn n

PX Y

X Y t

Sn n

m m+ -

- - -

Ê ˆ+Á ˜Ë ¯

Now:

2 21 {162,280,000 8 4,500 } 40,0007Ys = - ¥ =

2 9 27,627 7 40,000 528,640 33,04010 8 2 16Ps ¥ + ¥= =

+ -= [1]

Now, using the distribution: 16t

2

( ) ( )2.120 2.1201 1

10 8

X Y

P

X Y

s

m m- - -- < <Ê ˆ+Á ˜Ë ¯

[1]

( )2 21 1 1 1196 2.120 196 2.12010 8 10 8P X Y Ps sm mÊ ˆ Ê ˆ- + < - < +Á ˜ Á ˜Ë ¯ Ë ¯

+

This gives a confidence interval for X Ym m- of . [1] 196 182.8 (13.2, 379)± = [Total 3]



(v) Confidence interval for ratio of variances

Using 2 2

1, 12 2 ~X Y

X Yn n

X Y

S S Fs s - - , we have a distribution with upper and lower bounds

of 3.677 and

9,7F

13.293

. [1]

2 21 27,627 40,000 3.677

3.293 X Ys sfi < <

2

227,627 40,000 27,627 40,000 3.293

3.677X

Y

ss

fi < < ¥ [1]

This gives a confidence interval for 2

2X

Y

ss

of . [1] (0.188,2.27)

Since this confidence interval contains 1, the assumption of equal variances used in the confidence interval in part (iv) looks reasonable. [1] [Total 4]



Solution X3.12

(i) Proof that sample mean has a chi-square distribution

If the iX ’s are exponentially distributed with parameter l , then ( ) 1( ) 1

it

XM t l-

= - .

( ) ( )21 1( ) 222

2 ( )n t

nn nX X tXtXn tXn XM t E e E e E e e

l llll

+ +Ê ˆ= = =Ë ¯ [½]

( ) ( )1 22 by independencentXtXE e E e ll= [½]

1(2 ) (2 )

nX XM t M tl= l [½]

[ ](2 ) nXM tl=

[½] (1 2 ) nt -= - This is the MGF of a 2

2nc , hence by the uniqueness property of MGF’s 222 ~ nn Xl c .

[1] [Total 3] (ii) Confidence interval for mean Here, , so using the result from part (i), we have 10n = 2

2020 ~Xl c .

We want a confidence interval for the mean, so we need a confidence interval for 1l

.

We have: [½] 2

200.90 (10.85 31.41)P c= < <

(10.85 20 31.41)10.85 31.4120 20

P X

PX X

l

l

= < <

Ê ˆ= < <ÁË ˜ [½]

So a 90% confidence interval for l is given by:

10.85 31.41, (0.000798, 0.00231)20 680 20 680

Ê ˆ =Á ˜Ë ¯¥ ¥ [1]



Therefore a 90% confidence interval for the mean, 1m l= , is given to 3 SF by: (433, 1250) [1] [Total 3] (iii) Likelihood function and MLE of the mean Writing the pdf of the exponential distribution in terms of its mean, m :

1

1( )x

f x e mm

-=

So the likelihood function based on n observations is:

1 1

1

1 1( )in iX X

ni

L e emmm m

- -

=

Â= =’ m [1]

Taking logs:

1ln ( ) ln iL nm mm

= - - Â X

Differentiating with respect to m :

2ln ( ) iXnL mm m m∂ -= +∂

Â [1]

Setting the derivative equal to 0:

2iXn

m m= Â [½]

This rearranges to give:

iXX

nm = =Â [½]



The second derivative of the log-likelihood is:

2

2 2 3

2 2ln ( ) iXn nL mmm m m m∂ = - =∂

Â3

nX- [1]

Evaluating this at the point Xm = gives:

2

2 3 22ln ( ) 0 max

X

nX nX nLX Xm

mm =

∂ -= = - < fi∂

So the maximum likelihood estimator of m is X . [1] [Total 5] (iv)(a) CRLB The formula for the CRLB of the mean, m , is:

2

2

1

ln ( )CRLB

E L mm

= -È ˘∂Í ˙∂Í ˙Î ˚

From part (iii), we have:

2

2 22

ln ( ) i3XnL m

m m∂ = -∂

Âm

So:

( )2

2 3 2 3

2 3

2 3

2 2

2

2 2ln ( )

2 ( )

2

2

ii

i

Xn nE L E E

n E X

n n

n n

n

mm m m m m

m m

mm m

m m

m

È ˘È ˘∂ = - = -Í ˙Í ˙∂Í ˙ Í ˙Î ˚ Î ˚

= -

= -

= -

= -

Â Â

Â

X

[1]



Hence, 2

( )CRLBnmm = . [1]

(iv)(b) Estimated standard error The estimated standard error of the mean is:

2 2ˆ 680 46, 240 215.0

10CRLB

nm= = = = [1]

[Total 3] (v)(a) Confidence interval for mean The asymptotic distribution of ˆ Xm = is ( , )N CRLBm . Therefore:

~ (0,1X NCRLB

m- ) [1]

and using this approximate distribution:

1.6449 1.6449 0.9XPCRLB

mÊ -- < < =ÁË ¯ˆ˜ [1]

This can be rearranged to: ( )1.6449 1.6449 0.9P X CRLB X CRLBm- < < + =

So an approximate 90% confidence interval for m is: 1.6449 680 354 (326, 1034)x CRLB± = ± = [1]



Alternatively, using the result ~ (0,1)X NCRLB

m- , we could say:

1.6449 1.6449

1.6449 1.6449

0.9 1.6449 1.6449/

1.6449 1.64491 1

1 11

1 1

n n

n n

XPn

P Xn n

PX X

X XP

mm

m m

m

m

Ê ˆ-= - < <Á ˜Ë ¯

Ê ˆÊ ˆ Ê ˆ= - < < +Á ˜ Á ˜Á ˜Ë ¯ Ë ¯Ë ¯

Ê ˆ- +Á ˜= < <Á ˜Ë ¯

Ê ˆÁ ˜= < <

+ -Á ˜Ë ¯

So an approximate 90% confidence interval for m is:

( )1.6449 1.6449 1.6449 1.644910 10

680 680, , 447.3, 14171 1 1 1

n n

x xÊ ˆ Ê ˆÁ ˜ Á ˜= =

+ - + -Á ˜ Á ˜Ë ¯ Ë ¯

(v)(b) Compare confidence intervals The exact confidence interval of (433, 1250) from part (ii) is very different to the approximate confidence intervals of (326, 1034) or (447, 1417) from part (v)(a). The normal approximation used in part (v)(a) requires a large sample. We have a sample of only 10 values. Hence the approximation is not very good. [1] [Total 4]



Solution X3.13

(i) Mean and variance

The mean of X is 2( ) 21

E X/

qq

= = . [1]

The variance of X is 22

2var( ) 21

X qq

= = . [1]

[Total 2] (ii) MLE and MSE The likelihood function is given by:

1 2 1

1 22 2 2 2 21

1 1 1 1 1( )n

ii

xx xn xxi n n

iL x e x e x e ... x e const eq q q q qq

q q q q q

- - - --

=

Â= = = ¥’

[1] Taking logs, differentiating and setting to zero gives:

1ln ( ) constant 2 lniL xq qq

= - -Â n

22ln ( ) ix nL∂ q

∂q qqfi = Â - [1]

22ˆˆ

1ˆ as required2 2

i

i

x n

xX

n

qq

q

fi =

fi = =

Â

Â [1]

Check that this is indeed a maximum by differentiating again:

2

2 32 2log ix n L +

q q q∂ = -∂

Â2

Substituting in 1

2 Xq = :

3 2 3 2 22 2 16 8 8 0 mix n n X n n + + ax

X X Xq q- = - = - < fiÂ [1]



The mean square error of is given by . But since: q 2ˆ ˆ ˆ( ) var( ) + [ ( )]MSE Biasq q q= ( ) 1 1 1

2 2 2 2ˆ[ ] [ ] [ ] 2XE E E X E Xq = = = = ¥ =q q [1]

this estimator is unbiased. So we require the variance of . q Using the fact that the individual values of iX are independent:

( ) 2

2 2

1 1 12 4 4

214 2

ˆvar( ) var var( ) n

n n

X X s

q q

q = = =

= ¥ =

¥ [1]

and so the MSE of is q2

2nq . [1]

[Total 7] (iii)(a) MSE of aX Using the definition of MSE for aX :

2

2 2

( ) var( ) [ ( )]

var( ) ( [ ] ) var( ) + ( [ ] )

MSE aX aX Bias aX

aX E aX a X aE Xq

= +

= + - = - 2q [1]

222

222

2 + [ (2 1)]

[2 + (2 1 ])

a an

a n an

q q

q

= ¥ -

= - [1]

(iii)(b) Minimum MSE Differentiating the MSE respect to : a

2

[4 4 (2 1)]MSE a + n aa n

q∂ =∂

- [1]

Setting this equal to 0 gives 2 1

nan+

= . [1]




Differentiating the MSE a second time with respect to a , we obtain:

[ ]2 2

2 4 8MSE nnaq∂ = +

∂

which is positive, and so this does give a minimum value for the MSE. [1] [Total 5] (iv) Compare MSEs

We first need to find the MSE of the estimator 2 1

n Xn+

. With the optimal value of a

we find that the MSE is 2

1 2nq+

. Substituting in the suggested values of , we obtain: n

n 2 1

nan

=+

MSE of q MSE of optimal aX

1 0.33333 20.5q 20.33333q 5 0.45455 20.1q 20.09091q

100 0.49751 20.005q 20.00498q [2] For small sample sizes, the MSE of the optimal aX is smaller than that of the MLE. However, aX is a biased estimator whereas the MLE is unbiased. So for small samples, there is a trade-off between unbiasedness and low variance. [1] For larger samples the difference between the two MSEs becomes less significant. As

tends to infinity, n2 1

n an+

= tends to 12

, so that for large n the two estimators are

nearly the same, and asymptotically both are unbiased and have minimum variance. [1] [Total 4]



First calculate the values of xxs , xys and : yys 2 2 2126 4 5 26xxs x nx= - = - ¥ =Â [½]

210.1 4 5 8.275 44.6xys xy nx y= - = - ¥ ¥ =Â [½]

44.6ˆ 1.71526

xy

xx

sb

sfi = = = [1]


2 2 2350.49 4 8.275 76.5875yys y ny= - = - ¥ =Â [½] The estimate for the variance parameter is given by:

2 2

2 1 1 44.6ˆ 76.58752 2

xyyy

xx

ss

n ss

Ê ˆ

26È ˘

= - = -Í ˙Á ˜- Í ˙Ë ¯ Î ˚ [1]

[½] 0.0407= [Total 2]



Solution X4.3

We know that 22

ˆ~

ˆ

-

xx

b b tSs

. So, using the tables of the distribution: 2t

( )

2 2

2 2

ˆ0.95 ( 4.303 4.303) 4.403 4.303

ˆ

ˆ ˆ4.303 4.303ˆ ˆ

xx

xx xx

b bP t PS

P b S b b S

s

s s

Ê ˆ-= - < < = - < <Á ˜Á ˜Ë ¯

= - < < + [1]

Substituting in the numerical values, , and , we obtain 1 as the appropriate confidence interval. [2]

ˆ 1.715=b 2ˆ 0.04067=s 26=xxS.715 0.170 (1.54, 1.89)± =


(i) Sample correlation coefficient The sample correlation coefficient is given by:

44.6 0.999526 76.5875

xy

xx yy

sr

s s= = =

¥ [Total 1]

(ii)(a) Confidence interval for correlation coefficient Using Fisher’s approximation with the full value of : r 1tanh 4.116619rZ r-= =



We know that , where . So ~ ( ,1)r pZ N Z 1tanhZr r-= ~ (0,1)1

-rZ ZNr . Using the

appropriate percentage points from the Tables: [1]

[1]

0.95 ( 1.96 1.96)

(2.1566 6.0766)

rP Z Z

P Z

r

r

= - < - <

= < <

( )12.1566 tanh 6.0766

(0.974 1.000)

P

P

r

r

-= < <

= < <

Therefore a 95% confidence interval for r is ( . [1] 0.974,1.000) Using gives a confidence interval of ( . 0.995r = 0.975,1.000) Alternatively, students may use the transformation ( )11

2 1log rr rZ +

-= for full marks.

(ii)(b) Coefficient of determination The coefficient of determination is given by: [1] 2 2 20.9995 99.9%R r= = = The coefficient of determination gives the proportion of variability explained by the model. Since nearly all of the variability is explained, this tells us that the linear regression model is a very good fit. [1] [Total 5]



Solution X4.5

The size of a test is the probability of rejecting , given that is true. Here, assuming that , we have:

0H 0H0.4=p

0(reject ) ( 25)= ≥P H P X Using a normal approximation to the binomial distribution, where

, with a continuity correction, we have: (50,0.4) ~ (20,12)Bin N [1] [ 25] [ 24 5≥ ªP X P X > . ]

24 5 2012

[ 1 299]

Ê ˆ-= Á ˜Ë ¯

=

.P Z >

P Z > . [1] From the tables of the normal distribution: 1 (1 299) 1 0 90303 0 09697- F = - =. . . So the approximate size of the test is 9.7%. [1] [Total 3]



Solution X4.6

(i) Type I error A Type I error occurs if you reject when is true. [1] 0H 0H Award no marks for definitions that include probabilities. (ii) Type II error A Type II error occurs if you don’t reject when is false. [1] 0H 0H Award no marks for definitions that include probabilities. (iii) Size The size of a test is the probability of rejecting when is true (ie the probability of a Type I error). [1]

0H 0H

(iv) Power The power of a test is the probability of rejecting when is false (and may be a function of an unknown parameter). [1]

0H 0H

[Total 4] Alternatively, it could be defined as 1 ( type II error- P .)



Solution X4.7

(i) Testing the mean We are testing: 2

0 1: 9 : 9 ( unknown)= πH vs H m m s

Under , the statistic 0H XS n

m- has a distribution. 9t

Using the data given: 10.5x =

2 1{1,232.46 10 10.5 } 14 449

s = - ¥ =2 . [1]

So the value of our test statistic is:

10 5 9 1 2482714.44 10

. .- = [1]

Comparing this with the tables of the distribution, we find that we have a probability value in excess of , as the test here is two-sided.

9t0 1 2 20%¥ =.

Alternatively, we could say that, testing at the 5% significance level, the critical values are and the value of the test statistic lies between the two critical values. 2.262± So we have insufficient evidence to reject at the 5% level, therefore it is reasonable to assume that the population mean is 9. [1]

0H

[Total 3] (ii) Testing the variance We are now testing: 2 2

0 1: 8 : 8= πH vs Hs s

Under , the statistic 0H2

2( 1)-n S

s has a 2

9c distribution. So here our test statistic is:

9 14.44 16 2458

¥ = . [1]



Comparing this value with the tables of the 29c distribution, we find that we have a

probability value slightly in excess of 10% (since the test is two-sided). [1] Alternatively, we could say that, testing at the 5% significance level, the critical values are 2.700 and 19.02. So the value of the test statistic lies between the two critical values. So we have insufficient evidence to reject at the 5% level, therefore it is reasonable to assume that the population variance is 8. [1]

0H


(i) Testing the proportion We are testing: 0 1: 0.42 : 0.42H p vs H p= > If X is the number of people who supported the government then: ~ (5000, ) ~ (5000 , 5000 )X Bin p N p pq Hence our statistic is:

ˆ5,000 ~ (0,1) or ~ (0,1)

5,000 5,000X p p pN

pq pq− − N [1]

where ˆ5,000

Xp = .

Using a continuity correction, the value of our test statistic is:

2,184.55,000 0.422,184½ 2,100 2.421 or 2.421

1,218 0.2436 5,000

−−= = [1]

Comparing this with the normal distribution tables, we find that we have a probability value approximately equal to 0.8%. So we have sufficient evidence to reject at the 0.8% level, therefore it is reasonable to assume that the proportion of those who would vote for the current government is greater than 42%. [1]

0H

[Total 3]



Alternatively, we could say that the upper 5% point of is 1.6449 and the upper 1% point is 2.3263. So we would reject the null hypothesis at both the 5% and 1% significance levels and hence reach the conclusion stated above.

(0,1)N

(ii) Testing the change in the proportions Using 1X and 2X to represent the number of people supporting the government before and after the scandal. Then: 1 1 2~ (5000, ) and ~ (3000, )2X Bin p X Bin p We are testing: 0 1 2 1 1: :H p p vs H p p= π 2

The statistic given in Chapter 12, Section 4.3 is:

1 2 1 2ˆ ˆ( ) ( ) ~ (0,1)ˆ ˆ ˆ ˆ(1 ) (1 )5,000 3,000

p p p p Np p p p− − −− −

+ [1]

where 2,185 1,191 3,376ˆ 0.4225,000 3,000 8,000

p += = =

+.

We have 1 22,185 1,191ˆ ˆ0.437 and 0.3975,000 3,000

p p= = = = which gives:

0.437 0.397 3.5070.422 0.578 0.422 0.578

5,000 3,000

−=

× ×+

[1]

Comparing this with the normal distribution tables, we find that we have a probability value of less than (since it is a two-sided test). So we have sufficient evidence to reject at the 0.1% level, therefore it is reasonable to assume that the proportion of those who would vote for the current government is not the same as before the scandal. [1]

2 0.05% 0.1%× =

0H

[Total 3] Alternatively, testing at the 5% significance level gives critical values of , and testing at the 1% significance level gives critical values of . Since the test statistic is more extreme than the critical values, we reject the null hypothesis at both the 5% and 1% significance levels and reach the conclusion stated above. Note that technically, we would have to carry out a one-sided test to say it was less than before.

1.96±2.5758±



Solution X4.9

(i) Contingency table test We are testing: there is no association between sex and colour-blindness (ie they are

independent) 0 :H

there is an association between sex and colour-blindness (ie they are not independent)

1 :H

The expected values on the contingency table are: 2 2¥

male female normal 950.5 950.5 1,901 colour-blind 49.5 49.5 99 1,000 1,000 2,000

[1]

Since (row total) (column total) 1,901 1,000 950.5grand total 2,000

¥ ¥= = , etc.

The degrees of freedom are ( . [1] 2 1) (2 1) 1- ¥ - =

Using 2

2 ( )-= Â O EE

c we get:

2 2 2

21

(908 950.5) (993 950.5) (92 49.5) (7 49.5)950.5 950.5 49.5 49.5

c - - - -= + + +2

[1]

[1]

1.9003 1.9003 36.490 36.490

76.78

= + + +

= Since this exceeds even the 0.05% critical value of 12.12, we have overwhelming evidence at the 0.05% level to reject . We therefore conclude that there is an association between sex and colour blindness. [1]

0H

[Total 5]



(ii) Goodness of fit test Using the maximum likelihood estimate of in the formulae, we obtain the following numbers:

ˆ 0.0895=q

male female normal 910.5 992.0 colour-blind 89.5 8.010

[1] We are testing: the model is a good fit. 0 :H the model is not a good fit. 1 :H Notice here that we are no longer dealing with a contingency table despite the presentation of the data. We are doing a 2c goodness of fit test. The degrees of freedom are which is

since q was estimated using the data. [1] (no. of groups) 1 (no. of parameters estimated)- -

4 1 1 2- - =

2 2 2

22

(908 910.5) (993 992) (92 89.5) (7 8.01)910.5 992 89.5 8.01

c - - - -= + + +2

[1]

[1]

0.00686 0.00101 0.06983 0.12735

0.205

= + + +

= Since this is less than the 5% critical value of 5.991, we have insufficient evidence at the 5% level to reject . We therefore conclude that the model is a good fit. [1] 0H [Total 5]



Solution X4.10

(i)(a)(1) Dotplot The dotplot is as follows: Existing Desiccant (A)

4.4 4.6 4.8 5 5.2 5.4 5.6 5.8

4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 New Desiccant (B) [1] The plots suggest that the new desiccant may extract more water. The spread of values is similar for each desiccant. [1] (i)(a)(2) Test equality of variances We use the test to test for equality of variance. The hypotheses are: F 2 2 2 2

0 1: := πA B AH vs H s s s s B

Under , the statistic 0H2 2

2 2A A

B B

SS

ss

has an distribution. 7,7F

Calculating the sample variances:

2 21{201.1574 8 5 } 0.165347As = - ¥ = [½]

2 1{210.3659 8 5.11375 } 0.166067Bs = - ¥ =2 [½]

So our test statistic is 0 16534 1 0.99570 16606

=. .

. [1]

Since this lies between the critical values of 4.995 and 0.2002 we have insufficient evidence to reject at the 5% level. Therefore it is reasonable to assume that the two population variances are equal. This is not surprising, since the spreads on our two data plots looked very similar. [1]

0H



Alternatively, we could have used 2 2

2 2B B

A A

SS

ss

to get a ratio of 1.0043. This also has an

distribution and we would make the same conclusion as before. 7,7F (i)(b) Test if new desiccant extracts more moisture We now use a two-sample t test. The hypotheses are: 0 1: := >A B BH vs H Am m m m This test is one-sided. The sample means are 5 5 11375= =A , B . .

2 2

2 7 7 2.3198 0.165714 14+

= = =A BP

s ss [1]

Our test statistic is:

2

( ) ( ) 0 11375 0.5590.0414251 1

8 8

- - -= =

Ê ˆÁ ˜Ë ¯

B A

P

B A .Ts +

m m= [1]

which has a distribution under . 14t 0H We compare this with the 5% point of the distribution, which is 1.761. Our value lies well within the acceptance region. We have insufficient evidence to reject on the basis of this data. It does not appear that the new desiccant extracts any more moisture than the existing one. [1]

14t

0H

[Total 8] (ii) Paired t test We now perform a paired t test, looking at the differences. [1] Let represent the difference in moisture level extracted, so that , and the values of are:

D = -D B AD

[1] 0.16, 0.02, 0.17, 0.23, 0.24, 0.10, 0.13, 0.10- -



We must test whether these can be considered to come from a normal population with mean zero. We have the hypotheses: 0 10 0=D DH : vs H : >m m Again, the test is one-sided. Calculating the sample mean and variance: = 0 11375= -D B A .

{ }2 2 22 1 1{0.2023 8 0.11375 } 0 014117 7D D nD .s = - = - ¥ =Â [1]

Under , 0H 08D

DS

- has a distribution. So the observed value of our test statistic is: 7t

0 11375 2 70830.01411 8

. .= [1]

Comparing this with (critical value at 5% = 1.895), we have a result significant at the 5% level (and indeed at the 2½% level). This suggests that the new desiccant is more efficient at extracting moisture than the old one. [1]

7t

[Total 5] (iii) Comment The paired test shows that there was a significant difference between the two dessicators, whereas the two-sample test does not indicate any significant difference. [1] The small but significant difference between the two desiccants is masked in the two-sample test because the test statistic for the two-sample test is calculated using the pooled variance (which is 0.1657) rather than the sample variance of the differenced data (which is 0.01411). A smaller variance leads to a larger test statistic, which means we are more likely to reject the null hypothesis. In other words, the increased power of the paired test enables a significant difference to be identified. [1] [Total 2]



Solution X4.11

(i) Diagram Line plots for the four sets of data are:

A I -100 -50 0 50 100 150

A II -100 -50 0 50 100 150

B I -100 -50 0 50 100 150

B II -100 -50 0 50 100 150 [2] The centres of the distributions do appear to differ. In general, houses of type A appear to exceed the initial asking price by more than houses of type B. [1] Similarly, within the house types, dwellings in area I tend to exceed their initial asking price by more than the corresponding figures for comparable dwellings in Area II. [1] [Total 4] (ii) ANOVA For these data: 1 2 3 44 7= = = = = =k n n n n n 28 1 2 3 4394 221 44 58 601= = = = - =y y y y yi i i i ii

2 49,799ijy =ÂÂ

260149,799 36,899

28TSS = - = [1]

2 2 2 2 2394 221 44 ( 58) 601 17,011

7 7 7 7 28BSSÊ ˆ-= + + + - =Á ˜Ë ¯

[1]

36,899 17,011 19,888RSSfi = - =



We are testing: the mean is the same for each category 0 :H against: the means are not the same for all of the categories 1 :H The ANOVA table is:

Source of variation DF Sum of squares Mean sum of squares

Between categories Residual

3 24

17,011 19,888

5670 828.7

Total 27 36,899 [1 for correct MSS values]

Under , 0H 5,670 6.843828.7

F = = , using the distribution. [1] 3,24F

The 1% critical point is 4.718, so we have sufficient evidence to reject at the 1% level. Therefore it is reasonable to conclude that there are underlying differences between the categories. [1]

0H

[Total 5] (iii) Confidence interval for standard deviation The unbiased estimate of is given by: 2s

2 19,888ˆ 828.66724

RSSn k

s = = =-

[1]

Now:

2

22

ˆ( ) ~ n kn k s cs -- [1]



So, using the critical values of 12.40 and 39.36, we have:

219,88812.40 39.36s

< <

This gives a confidence interval for of (505.3 . Hence, the confidence interval for the standard deviation is . [1]

2s22.5,

, 1604)( 40.0)

[Total 3] (iv) Analysis of the mean differences

Since, 1 2 3 4394 221 44 5856.3, 31.6, 6.3, 8.37 7 7 7i i i iy y y y= = = = = = = - = - we can

write: 1 2 3 4y y y yi i i> > > i [½] From part (iii), we have . The least significant difference between any pair of means is:

2ˆ 828.667σ =

1 1 1 124,0.025 7 7 7 7

ˆ 2.064 828.667 31.76t s + = ¥ ¥ + = [1]

Now we can examine the difference between each of the pairs of means. If the difference is less than the least significant difference then there is no significant difference between the means. We have: 1 2 2 3 3 424.7 25.3 14.6y y y y y yi i i i i i- = - = - = [½] Showing this on a diagram, by underlining those pairs that have no significant difference, we get:

1 2 3 4i i i iy y y y> > > [1]

Examining to see if the first two groups can be combined: 1 3 50y y− =i i There is a significant between means 1 and 3, so we cannot combine the first two groups. [½]



Examining to see if the last two groups can be combined: 2 4 39.9y y− =i i There is a significant between means 2 and 4, so we cannot combine the last two groups. Therefore the diagram remains as before. [½] [Total 4]



Solution X4.12

(i)(a) Derive least squares estimators Since = fi = -i i i i i - iy a +bx + e e y a bx , the sum of squares is: 22 ( )= = - -Â Âi iQ e Y a bxi

Differentiating with respect to and , we have: a b

2( ) ( 1)

2( ) ( )

∂ = - - ¥ -∂

∂ = - - ¥ -∂

Â

Â

i i

i i

Q Y a bxa

Q Y a bx xb i

[1]

Setting these two expressions to zero, we have:

2

(1)

(2)

= +

=

Â ÂÂ Â Â

Y na b x

xY a x +b x

…

… Multiplying equation (1) by Â x and equation (2) by n , we obtain:

2

2

( ) (3)

(4)

= +

=

Â Â Â ÂÂ Â Â

x Y na x b x

n xY na x + nb x

…

… Subtracting equation (4) from equation (3), we obtain: 2 2[( ])- = -Â Â Â Â Âx Y n xY b x n x [1] Rearranging this gives:

2(2

ˆ)

x Ynx

n

xYb

x

Â Â-=

Â-

ÂÂ

or ˆ = xy

xx

Sb

S [1]

and this is our least squares estimator for . The estimator for a is given by b

ˆa y bx= − (this can be obtained by rearranging equation (1)). [1]

Award the marks for the alternative method using substitution.



(i)(b) Maximum likelihood estimators The answer would not have differed at all. For a normal distribution, maximum likelihood and least squares obtain the same estimates. [1] Since = + +i i iy a bx e , where , gives the likelihood function using maximum likelihood estimation is:

2~ (0, )ie N s 2~ ( , )+i iy N a bx s

( )21 1

22

22

( , ) exp

1ln( ) ( )2

i iy a bx

ii

L a b

L constant a bxy

ss p

s

- -Ï ¸= -Ì ˝

Ó ˛

fi = - - -

’

Â

So maximising is equivalent to minimising , which is identical to the criterion used above to find the least squares estimators. So the MLEs for and b are equal to the least squares estimators in this case. [1]

L 2( - -Â ii a bxy )a

[Total 6] (ii) Regression coefficients

826ˆ 19.66742

xy

xx

sb

s= = = [1]

ˆˆ 105 19.667 4.5 16.5a y bx= - = - ¥ = [1] [Total 2] (iii) Testing slope parameter We wish to test: 0 1: 22 : 2= πH b vs H b 2

Under , the statistic 0H2

ˆ2ˆ

~xx

b bn

St

s-

- , where is given by: 2s

( )2 22 8261 12 6 42

ˆ 16 492 41 2222xy

xx

syyn Ss , s -

Ê ˆ= - = - =Á ˜Ë ¯. [1]

So the value of our test statistic here is:

22 19.667 2.3550.9907- = [2]



This is a two-sided test. So the probability value for this test statistic is about . So we have insufficient evidence to reject at the 5% level,

therefore it is reasonable to assume that the slope of the model is £22 per hour. [1] 0 03 2 6%¥ =. 0H

[Total 4] (iv)(a) Confidence interval for the average cost of a job lasting 4 hours Our estimate of the mean predicted cost is: [1] ˆˆ ˆ 4 16.5 19.667 4 95.17= + ¥ = + ¥ =a bm The standard error is given by:

21 (4 4.5) 41.22 5.3981 2.323

8 42Ê ˆ-+ =Á ˜Ë ¯

= [1]

This gives a confidence interval, using the tables, of: 6t [1] 95.17 1.943 2.323 95.17 4.51 (90.7, 99.7)± ¥ = ± (iv)(b) Confidence interval for an individual job lasting 6 hours Our estimate of the individual predicted cost is: [1] ˆˆ ˆ 6 16.5 19.667 6 134.5a bm = + ¥ = + ¥ = The standard error is given by:

21 (6 4.5)1 41.22 48.583 6.970

8 42Ê ˆ-+ + = =Á ˜Ë ¯

[1]

This gives a confidence interval of: [1] 134.5 1.943 6.970 134.5 13.54 (121,148)± ¥ = ± [Total 6]



(v) Comment The confidence interval for the individual job is wider (£27) than the confidence interval for the average cost (£9). So there is greater uncertainty over an individual result than an average one. [1]



Solution X4.13

(i) Show μ is an unbiased estimator We need to show ˆ[ ] =E m m :

1 1

1 1

1 1

1

1ˆ[ ] [ ]

1 [ ]

1 ( ) since [ ]

1 ( )

1 ( 0)

= =

= =

= =

=

È ˘Í ˙= =Í ˙Î ˚

=

= +

= +

= + =

ÂÂ

ÂÂ

ÂÂ

Â

i

i

i

nk

iji j

nk

iji j

nk

i ii j

k

i i ii

E E Y E Yn

E Yn

E en

n nn

nn

m

m t

m t

m m

ii

0=j

[1]

(ii) Show is an unbiased estimator ˆiτ We need to show : ˆ[ ] =i iE t t

1

1

1

1ˆ[ ] [ ] since [ ]

1 ( )

1 ( )

1 ( )

=

=

=

È ˘Í ˙= - = - =Í ˙Î ˚

= -

= +

= + -

=

Â

Â

Â

i

i

i

n

i i iji j

n

iji j

n

ii j

i i ii

i

E E Y Y E Y E Yn

E Yn

n

n nn

t m

m

m t

m t m

t

i ii ii m

[1]



(iii) Rewrite ˆ 2σ Starting with the definition of 2σ we get:

2 2 2

1 1 1 1

2 2

1 1 1 1 1 1

2 2

1 1 1 1

1 1ˆ ( ) ( 2

1 2 1

1 2

= = = =

= = = = = =

= = = =

È ˘ ÈÍ ˙ Í= - = -

- -Í ˙ ÍÎ ˚ Î

È ˘Í ˙= - +

- Í ˙Î ˚

È ˘Í ˙= - +

- Í ˙Î ˚

ÂÂ ÂÂ

ÂÂ Â Â Â Â

ÂÂ Â Â

i i

i i i

i

n nk k

ij i ij ij i ii j i j

n n nk k k

ij i ij ii j i j i j

nk k k

ij i i i ii j i i

Y Y Y Y Y Yn k n k

Y Y Y Yn k

Y Y Y n Yn k

s i i

i i

i i i

2 )˘˙+˙

i

[1]

Using the fact that 1=ii

Yni iY i gives:

2 2 2

1 1 1 1

2 2

1 1 1

1 1ˆ 2

1 1

= = = =

= = =

È ˘Í ˙= - +

- Í ˙Î ˚

È ˘Í ˙= -

- Í ˙Î ˚

ÂÂ Â Â

ÂÂ Â

i

i

nk k k

ij i ii ii j i i

nk k

ij iii j i

Y Yn k n n

Y Yn k n

s i i

i

21 Y

[1]



(iv) Show is unbiased ˆ 2σ We need to show . Using the expression for from part (iii): 2ˆ[ ] =E s s 2 2s

2 2

1 1 1

2

1 1 1

1 1ˆ[ ]

1 [ ] [ ]

= = =

= = =

È ˘Ê ˆÍ ˙= -Á ˜-Í ˙Ë ¯Î ˚

È ˘Í= -

- Í ˙Î ˚

ÂÂ Â

ÂÂ Â

i

i

nk k

ij iii j i

nk k

ij iii j i

E E Y Yn k n

E Y E Yn k n

s i

i

2

21 ˙

)

2

[1]

Using the relationship gives: 2 2( ) var( ) (= +E X X E X

[1]

( )

( )

( )

2 2

1 1 1 1

2 2

1 1

2 2

1

2

1

[ ] var[ ] [ ]

( )

( )

( )

= = = =

= =

=

=

= +

= + +

= + +

= + +

ÂÂ ÂÂ

ÂÂ

Â

Â

i i

i

n nk k

ij ij iji j i j

nk

ii j

k

i i ii

k

i ii

E Y Y E Y

n n

n n

s m t

s m t

s m t




Now consider the remaining term in the expression for : 2ˆ[ ]E s

( )2 2

1 1

2

1 1 1

2

1 1 1

22

1 1 1

1 1[ ] var[ ] [ ]

1 var

1 var[ ] [ ]

1 [ ]

1

= =

= = =

= = =

= = =

= +

Ê ˆÈ ˘ È ˘Í ˙ Í ˙= +Á ˜

Á ˜Í ˙ Í ˙Ë ¯Î ˚ Î ˚

Ê ˆÊ ˆÁ ˜= + Á ˜Á ˜Ë ¯Ë ¯

Ê ˆÊ ˆÁ ˜= + +Á ˜Á ˜Ë ¯Ë ¯

=

Â Â

Â Â Â

Â Â Â

Â Â Â

i i

i i

i i

k k

i i ii ii i

n nk

ij ijii j j

n nk

ij ijii j j

n nk

iii j j

ii

E Y Y E Yn n

Y E Yn

Y E Yn

n

nn

s m t

s

i i i

( )( )

( )

22

1

2 2

1

2 2

1

[ ]

[ ]

( )

=

=

=

+ +

= + +

= + +

Â

Â

Â

k

i ii

k

i ii

k

i ii

n

n

k n

m t

s m t

s m t [1]

Hence:

2 2 2

1 1 1

2 2 2 2

1 1

2 2

2

1 1ˆ[ ] [ ] [ ]

1 ( ) ( )

1

= = =

= =

È ˘Í ˙= -

- Í ˙Î ˚

È ˘Ê ˆ Ê= + + - + +Í ˙Á ˜ Á- Ë ¯ ËÍ ˙Î ˚

È ˘= -Î ˚-

=

ÂÂ Â

Â Â

ink k

ij iii j i

k k

i i i ii i

E E Y E Yn k n

n n k nn k

n kn k

s

s m t s m t

s s

s

i

ˆ˜

[1]


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You may not hire out, lend, give out, sell, store or transmitelectronically or photocopy any part of the study material.

You must take care of your study material to ensure that itis not used or copied by anybody else.

Legal action will be taken if these terms are infringed. Inaddition, we may seek to take disciplinary action through

the profession or through your employer.

These conditions remain in force after you have finishedusing the course.

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