Statistics exercise solution

19
MSc in Finance and Investment Presessional Maths and Stats. Stats Exercises Solutions (Tutorials) 1. The discrete Poisson distribution has one parameter λ. Its density function is given by f(y) = ! y e y y = 0, 1, 2,.............where y! = 1x2x3....xy. (a) Given that this density function is valid, what properties must it satisfy? 0 i i y ! y e i = 1 and ! y e i y i 0 i = 0,1,2,........ You might care to initiate a discussion about whether these hold in this case. There is a well known proof of the first which students do not have to know, the second is fairly obvious. Also E(y) = λ and Var(y) = λ for the Poisson. (b) Assuming λ = 2, calculate the following; (i) Prob{y = 2} = ! 2 e 2 2 2 = 0.27 (ii) Prob{y < 2} = ! 0 e 2 2 0 + ! 1 e 2 2 1 = 0.135 + 0.27 = 0.405 (iii) Prob{y > 2) = 1 - (0.27 + 0.405) = 0.325 (c) Sketch out the shape of the density function if λ = 2. Is it skewed or symmetric? Explain. f(0) = 0.135, f(1) = 0.27, f(2) = 0.27, f(3) = 0.18 Its skewed to the right. 3. (a) Given the density function:- f(x) = x 0 x ½ = (1 x) ½ x 1 (i) Find As the density function has a kink at x = ½, we have to integrate in two steps. The first step is from 0 to ½. The second from ½ to 1. Thus must solve the following equation;

description

For practice

Transcript of Statistics exercise solution

Page 1: Statistics exercise solution

MSc in Finance and Investment

Presessional Maths and Stats.

Stats Exercises Solutions (Tutorials)

1. The discrete Poisson distribution has one parameter λ. Its density function is given

by

f(y) = !y

ey y = 0, 1, 2,.............where y! = 1x2x3....xy.

(a) Given that this density function is valid, what properties must it satisfy?

0i i

y

!y

ei

= 1 and !y

e

i

yi ≥ 0 i = 0,1,2,........

You might care to initiate a discussion about whether these hold in this case.

There is a well known proof of the first which students do not have to know, the

second is fairly obvious. Also E(y) = λ and Var(y) = λ for the Poisson.

(b) Assuming λ = 2, calculate the following;

(i) Prob{y = 2} = !2

e2 22

= 0.27

(ii) Prob{y < 2} = !0

e2 20

+ !1

e2 21

= 0.135 + 0.27 = 0.405

(iii) Prob{y > 2) = 1 - (0.27 + 0.405) = 0.325

(c) Sketch out the shape of the density function if λ = 2. Is it skewed or symmetric? Explain.

f(0) = 0.135, f(1) = 0.27, f(2) = 0.27, f(3) = 0.18

Its skewed to the right.

3. (a) Given the density function:-

f(x) = x 0 x ½

= (1 – x) ½ x 1

(i) Find

As the density function has a kink at x = ½, we have to integrate in two steps. The

first step is from 0 to ½. The second from ½ to 1. Thus must solve the following

equation;

Page 2: Statistics exercise solution

2/1

0

xdx +

1

2/1

dx)x1( = 1

We find each of these terms in turn;

2/1

0

xdx = 2/1

02 ]2/x[ =

8

1

2/1

dx)x1( = 12/1

2 ]2/xx[ = )8/2/(2/ = 8

So 8

+ 8

= 1 and 14/ or 4

Thus f(x) = 4x 0 x ½

= 4(1 – x) ½ x 1

(ii) Derive the cumulative distribution function F(s)

F(s) = s

0

sds4 = 2s2 0 s ½

F(s) = F(1/2) +

s

2/1

ds)s1(4 = ½ + s 2/1

2s2s4 =

½ + 4s – 2s2 – 2 + ½ = 2s(2 – s) – 1 1/2 s 1

(iii) Sketch f(x) and by inspection what is the mean.

The distribution is symmetric about the mode x = ½. Thus the mean is 1/2

(iv) Check that the value of the mean in part (iii) is the same as E(x).

dxx4 2 = 2/1

03 ]3/x4[ =

6

1

dx)x1(x4 = 12/1

32 ]3/x4x2[ = )24/42/1(3/42 = 3

1

Thus E(X) = ½

(v) Find Var(x)

Var(x) = E(2x ) – 2)x(E

Page 3: Statistics exercise solution

E(2x ) =

2/1

0

3dxx4 +

1

2/1

2 dx)x1(x4

= 2/10

4 ]x[ + 12/1

3 ]3/x4[ – 12/1

4 ]x[

= 1/16 + 4/3 – 4/24 – 1 + 1/16 = 1/8 + 1/3 – 1/6 = 7/24

Var (x) = 7/24 – 2)]x(E[ = 1/24

4. If 12XE and 16XVar , find E(Y) and Var(Y) where Y is the

following :

(i) 34X5Y (ii) 9

4

XY (iii)

4

17X3Y

(iv)

4

12XY

.

(i) E(Y) = 3460 Var(Y) = 400 (ii) E(Y) = 12 Var(Y) = 1 (iii) E(Y) =

4

116 Var(Y) = 9

(iv) E(Y) = 0 Var(Y) = 1

5. A continuous random variable x has a distribution with probability density function

(pdf) given by

)x4(16

3)x(f 2 0 2 x

= 0 otherwise

(a) Check that f(x) is a proper density function.

dx)x4(16

3 2

2

0

=

2

0

3

16

x

16

x12

= 1.5 − 0.5 = 1 it is “proper”

(b) Find the following probabilities:

(i) P ( X > 1)

dx)x4(16

3 2

2

1

=

2

1

3

16

x

16

x12

= 1 − 11/16 = 5/16

(ii) P ( 0.5 X 1.7)

Page 4: Statistics exercise solution

dx)x4(16

3 2

7.1

5.0

=

7.1

5.0

3

16

x

16

x12

= 1.275 − 0.3071 − 0.375 + 0.0078 = 0.5851

(b) Find E(x)

dx)x4(16

x3 2

2

0

=

2

0

42

64

x3

64

x24

= 96/64 − 48/64 = 48/64 = 0.75

(c) Find the variance of x

)x(E 2 = dx)x4(

16

x3 2

2

0

2

=

2

0

53

80

x3

64

x16

= 2 − 96/80 = 0.8

Variance = 0.8 − 0.5625 = 0.2375

6. Suppose there is a sample of data, ix i = 1,2,3,....N

The sample mean is defined as N

xx

i

The sample variance is defined as N

)xx( 2

i

Derive a simpler expression for the sample variance which can be used in

calculations.

Sample variance = N

)xx( 2

i = ])x(xx2x[N

1 2i

2i

= N

)x( 2

i −

N

)x(x2 i +

N

)x( 2

N

)x( 2

i −

2)x(2 + 2)x( =

N

)x( 2

i −

2)x(

Page 5: Statistics exercise solution

7. If z = ax + by where E(x) = 1 ,E(y) = 2 , var(x) = 2

1 and var(y) = 2

2

and cov(x,y) = 2

12 , a and b are constants, show that

the variance of z = E[z – E(z)2] =

2

1

2a + 2

2

2b + 12ab2 .

The long proof is;

Var(z) = E[(ax + by – (a 1 + b 2 )2) ] = E(

2

2

22

1

22222 baybxa

+ 212

2

121

2 abybyabxab2xa2abxy2 ) =

2

2

22

1

22222 ba)y(Eb)x(Ea + 2abE(xy)

21

2

2

2

2112

2

1

2 ab2b2ab2ab2a2

= ))x(E(a 2

1

22 + ))y(E(b 2

2

22 + 2ab( 21)xy(E )

= )xvar(a 2 + )bvar(b2

+ 2abcov(x,y)

A shorter proof starts from

Var(z) = E[(ax + by – (a 1 + b 2 )2) ] = E[(a(x – 1 )+b(y – 2 )

2) ] =

2

12 )x(Ea +

22

2 )y(Eb + 2 )y)(x(abE 21 =

)xvar(a 2 + )bvar(b2

+ 2abcov(xy)

10.

X

f(y)

0 1 2

0 0.05 0.10 0.03 0.18

Y 1 0.21 0.11 0.19 0.51

2 0.08 0.15 0.08 0.31

f(x)

0.34

0.36

0.30

1.00

Page 6: Statistics exercise solution

(a) Compute the following three probabilities:

P(Y < 2) = P(Y = 0) + P(Y = 1) = 0.18 + 0.51 = 0.69.

P( Y < 2 , X > 0) = P( Y = 0 , X = 1) + P( Y = 0 , X = 2) + P( Y = 1 , X =1) + P( Y = 1 , X = 2) = 0.10 + 0.03 + 0.11 + 0.19 = 0.43

P(Y = 1 and X 1) = P(Y = 1 , X = 1) + P(Y = 1 and X = 2) = 0.11 + 0.19 = 0.30.

(b) Find the marginal distributions of X and Y Given above

(c) Calculate E(X), E(Y), Var(X), Var(Y), Cov Y,X and )YX(E 2.

x

ii )x(fxXE 0(0.34) + 1(0.36) + 2(0.30) = 0.96

y

ii )y(fyYE 0(0.18) + 1(0.51) + 2(0.31) = 1.13.

22 XEXEXVar where

x

i

2

i

2 xfxXE 20 (0.34) + 21 (0.36) +

22 (0.30) = 1.56

So 296.056.1 XVar = 0.6384

22 YEYEYVar where

y

i

2

i

2 yfyYE 20 (0.18) + 21 (0.51) +

22 (0.31) = 1.75

So 213.175.1 YVar = 0.4731

YEXEXYEYXCov , 13.196.0XYE

where yxpxyXYEy x

, .

See the table below – values in the brackets are the values of xy.

Page 7: Statistics exercise solution

X

0 1 2

0 0.05

(0)

0.10

(0)

0.03

(0)

Y 1 0.21

(0)

0.11

(1)

0.19

(2)

2 0.08

(0)

0.15

(2)

0.08

(4)

)08.0)(4()15.0)(2()19.0)(2()11.0)(1( XYE = 1.11

So 0252.00848.111.1, YXCov .

(d) Find the conditional distribution of Y given xX .

Need to find xXyYP for x = 0, 1 or 2 and y = 0, 1 or 2 where

)(

,

xXP

yYxXPxXyYP

Consider the conditional distribution of Y given X = 0.

00 XYP

0

0,0

XP

YXP

34

5

34.0

05.0 using the joint table and (b).

01 XYP

34.0

1,0 YXP

34

21

34.0

21.0 “ “ “

02 XYP

34.0

2,0 YXP

34

8

34.0

08.0 “ “ “

So the conditional distribution of Y given X = 0 is given by

Y 0 1 2 Total

0Xyf

or 0yf

34

5

34

21

34

8

1.0

Page 8: Statistics exercise solution

Consider the conditional distribution of Y given X = 1.

10 XYP

10,1

XP

YXP

18

5

36.0

10.0 using the joint table and (b).

11 XYP

36.0

1,1 YXP

36

11

36.0

11.0 “ “ “

12 XYP

36.0

2,1 YXP

12

5

36.0

15.0 “ “ “

So the conditional distribution of Y given X = 1 is given by

Y 0 1 2 Total

1Xyf

or 1yf

18

5

36

11

12

5

1.0

Consider the conditional distribution of Y given X = 2.

20 XYP

2

0,2

XP

YXP

10

1

30.0

03.0 using the joint table and (b).

21 XYP

30.0

1,2 YXP

30

19

30.0

19.0 “ “ “

22 XYP

30.0

2,2 YXP

15

4

30.0

08.0 “ “ “

So the conditional distribution of Y given X = 2 is given by

Y 0 1 2 Total

2Xyf

or 2yf

10

1

30

19

15

4

1.0

Page 9: Statistics exercise solution

You could put all 3 conditional distributions together in one table to give the conditional distribution of Y given X = x as follows :

X

0 1 2

0 34

5 18

5 10

1

Y 1 34

21 36

11 30

19

2 34

8 36

15 15

4

What is the conditional mean 1XYE and the conditional variance 1XYVar ?

1XYE = the mean of the conditional distribution of Y given X = 1

= y

1Xyfy

= 0

18

5 + 1

36

11 + 2

36

15=

36

41 or 1.1389 to 4 dec. places.

1XYVar = the variance of the conditional distribution of Y given X = 1

= 2

2 11 XYEXYE

where y

XypyXYE 11 22

18

502

+

36

1112

+

36

1522

=

36

71.So

6752.036

41

36

711

2

XYVar to 4 dec. places.

Page 10: Statistics exercise solution

11. Grosvenor Motors Ltd. is developing a marketing plan to better target advertising and

sales promotion to subgroups. As part of the market research they have prepared the

table given below which indicates the probabilities for subgroups defined by age of

car and owner age group. For example, X = 1 indicates that the car owners are aged

16 – 25 years and Y = 2 indicates that the car is between 2 and 4 years old.

Age of car (Y)

1

(16 – 25 yrs)

Age Group (X)

2

(26 – 45 yrs )

3

(46 – 65 yrs)

1 ( 2 yrs) 0.05 0.17 0.06

2 ( 2 – 4 yrs) 0.15 0.20 0.07

3 ( 5 yrs ) 0.10 0.08 0.12

(a) What age group would you concentrate your advertising and sales promotion

on if you were attempting to sell cars that are over 5 years old?

As Y is the age of the car, when Y = 3 the car is at least 5 years old. When Y = 3, the X value with the highest associated probability is X = 3 since 0.12 > 0.10 > 0.08, so the age-group you would concentrate on is 46 – 65 years.

(b) Calculate the following probabilities : P( X = 2, Y = 3), P( X = 3, Y =

1), 32 YXP , 32 XYP , 2,2 YXP and define

each probability in words.

(i) P( X = 2, Y = 3) = 0.08 = the probability that the car owner is aged 26 – 45 and the car is at least 5 years old.

(ii) P( X = 3, Y = 1) = 0.06 = the probability that the car owner is aged 46 – 65 and the car is at most 2 years old.

(iii) 15

4

30.0

08.0

)3(

3,232

YP

YXPYXP = the probability that the car owner

is aged 26 – 45 years old if his/her car is at least 5 years old.

(iv) 25

7

25.0

07.0

)3(

2,332

XP

YXPXYP = the probability that the car is 2

– 4 years old if the car owner is aged 46 – 65.

Page 11: Statistics exercise solution

(v) 2,2 YXP 21,21 orYorXP = P( X = 1, Y = 1) + P( X = 1, Y = 2) +

P( X = 2, Y = 1) + P( X = 2, Y = 2) = 0.05 + 0.15 + 0.17 + 0.20 = 0.57 = the probability that the car owner is no more that 45 years old and his/her car is at most 4 years old.

(c) Find the marginal distributions of X and Y and hence calculate 2XP and

2YP . See the joint table above with row and column totals.

The marginal distribution of X is given by

x 1 2 3 Total

xpX 0.30 0.45 0.25 1.00

The marginal distribution of Y is given by

y 1 2 3 Total

ypY 0.28 0.42 0.30 1.00

2XP = 0.30 + 0.45 = 0.75 or 2XP = 1 – P(X = 3) = 1 – 0.25 = 0.75.

2YP = 0.28 + 0.42 = 0.70 or 2YP = 1 – P(Y = 3) = 1 – 0.30 = 0.70.

(d) Calculate E(X), E(Y) and cov( X , Y).

x

X xxpXE 1(0.30) + 2(0.45) + 3(0.25) = 1.95

y

Y yypYE 1(0.28) + 2(0.42) + 3(0.30) = 2.02

YEXEXYEYXCov , 02.295.1XYE

where yxpxyXYEy x

, .

See the table below – values in the brackets are the values of xy.

Page 12: Statistics exercise solution

X

1 2 3

1 0.05

(1)

0.17

(2)

0.06

(3)

Y 2 0.15

(2)

0.20

(4)

0.07

(6)

3 0.10

(3)

0.08

(6)

0.12

(9)

XYE 1(0.05) + 2(0.17) + 3(0.06) + 2(0.15) + 4(0.20) + 6(0.07) + 3(0.10) +6(0.08) +

9(0.12) = 3.95

So 011.0)02.2(95.195.3, YXCov

(e) Explain why the random variables X and Y are not independent.

If X and Y were independent, then YXCov , would be 0 so as it is not 0, it means

that X and Y cannot be independent.

12. Let f(x,y) = )yx(k 22 , 0 x 1, 0 y 1 be the joint probability density

function of x and y where k is a constant. Find the following;

(a) value of k.

1

0

1

0

22 dxdy)yx(k =

1

0

1

0

32 dx)

3

kyykx(

= dx)

3

kkx(

1

0

2

=

1

0

3

3

kx

3

kx

= 2k/3 = 1 Thus k = 3/2

(b) the marginal density function of x.

f(x) =

1

0

22

2

dy)yx(3 =

1

0

32

2

y

2

yx3

=

2

)13( 2 x

(c) the marginal expectation of x.

Page 13: Statistics exercise solution

E(x) = dx2

x

2

x31

0

3

=

1

0

24

4

x

8

x3

= 5/8

(c) the marginal variance of x.

)( 2xE = dx2

x

2

x31

0

24

=

1

0

35

6

x

10

x3

= 3/10 + 1/6

= 14/30 Var(x) = 14/30 - 25/64 = 0.467 – 0.391 = 0.076

(d) the conditional distribution of y given x. [6 marks]

f(y | X) = )x(f

)y,x(f =

)1X3(2

)yX(62

22

13. Suppose 1̂ and 2̂ are independent unbiased estimators of the same unknown

population parameter . It is known that var( 1̂ ) = 2

1 and var( 2̂ ) = 2

2 . A

statistician proposes a new estimator

3̂ = a 1̂ + (1− a) 2̂

where a is a constant whose value is chosen by the statistician.

(a) Show that 3̂ is unbiased.

E( 3̂ ) = aE( 1̂ ) + (1− a)E ( 2̂ ) = a + (1− a) =

(b) Derive the variance of 3̂ .

Var( 3̂ ) = var( 1̂ ) + var( 2̂ ) + 2a(1-a)cov( 1̂ 2̂ )

Since the covariance term is zero,

Var( 3̂ ) = 2

1 + 2

2

(c) Find the value of a that minimises the variance of 3̂ (5 marks)

First order condition gives 2a2

1 - 2(1-a) 2

2 = 0 or

Thus optimal a = 2

2

2

1

2

2

Page 14: Statistics exercise solution

16 A random sample ( )x,....,x,x n21 of size n is drawn from a population of known mean

and unknown variance 2 . The following estimator of

2 is proposed

2̂ = 2)ix(

n

1

(a) Is 2̂ biased or unbiased? Give details.

Short proof; since E( 2

i )x(n

1) = ))x((E

n

1 2

i =

2

n

1 =

2 thus unbiased

Slightly longer proof, E[ 2

i )x(n

1]

= n

)x(E 2i

– n

)x(E2 i +

2 = )x(E 2i

– 2 =

)x(Var i = 2

(b) The sample variance is defined as

2s = 2

i )xx(n

1

where x is the sample mean. Explain why 2s is a biased estimator of

2 .

2s = 2

i )xx(n

1 = 2

ixn

1 − ix

n

x2 +

2x

= 2ix

n

1 −

2x

E(2s ) = )x(E

n

1 2i − E(

2x ) = E(2ix ) − E(

2x )

= var( ix ) + 2

i )]x(E[ − (var( x ) + 2)]x(E[ )

= var( ix ) − var( x )

Thus because 2s contains x and not , the E(

2s ) depends on the variance of x

as well as the var( ix ) . The variance of x unlike the variance of is non zero.

Page 15: Statistics exercise solution

17. Construct confidence intervals for the population mean when the sample mean is 11.5

and unbiased estimate of the population variance is 100.

(a) Use a level of confidence of 95 % and a sample size of 25

critical value of t(24) is 2.064. Standard deviation of the sample mean is

25/100 = 2 Thus 95% confidence interval is given by 11.5 (2.064x2)

7.372 15.628 where is the population mean

(iii) a sample size of 1,000,000

critical value of t(999,999) is 1.96 approx. Standard deviation of the sample

mean is 000,000,1/100 = 0.01 Thus 95% confidence interval is given by

11.5 (1.96x0.01)

11.48 11.52 where is the population mean

(b) Use a level of confidence of 90 %

(i) a sample size of 25 critical value of t(24) is 1.711. Standard deviation of the sample mean is

25/100 = 2 Thus 90% confidence interval is given by 11.5 (1.711x2)

8.078 14.922 where is the population mean

(iii) a sample size of 1,000,000

critical value of t(999,999) is 1.645 approx. Standard deviation of the

sample mean is 000,000,1/100 = 0.01 Thus 90% confidence

interval is given by 11.5 (1.645x0.01)

11.48 11.52 where is the population mean

18. Suppose X is a normally distributed random variable with mean = 200 and standard

deviation = 20.

(a) Calculate P(180X220)

P(180X220) = P(

220X

180 ) = P(

20

200220X

20

200180

)

= P(-1 X 1) = 1 – 0.1587 – 0.1587

= 0.6826

Page 16: Statistics exercise solution

(b) Calculate P(160X240)

P(160X240) = P(

240X

160 ) = P(

20

200240X

20

200160

) =

P(–2 X 2) = 1 – 0.0228 – 0.0228

= 0.9544

20 Explain what is meant by a Type 1 and a Type 2 error in hypothesis testing. (a) Consider the following decisions and indicate whether you think a type 1 or a type

2 error is more important

i. A doctor tests a patient for a serious infectious disease. The null hypothesis is that the patient has the disease.

ii. A bank is deciding whether or not to make a large loan. The null hypothesis is that the customer will default on the loan.

iii. The Bank of England is deciding on interest rate policy. The null

hypothesis is that the rate of inflation will be greater that the target level set by the government.

(b) Discuss whether you think a type 1 or a type 2 error is more important in general.

Type 1 error is rejecting a true null, type 2 is accepting a false one. The answer to (a) depends on the investigator’s loss function which I have not discussed except by implication. This is an opportunity for a discussion, There is no right or wrong answer. (b) In general there is no reason to favour either type of error. It all depends on the context. 21Test the null hypothesis that the population mean is 10 against a two sided alternative if the sample mean is 11.5 and an unbiased estimator of the population variance is 100 (as in 1 above).

(a) Set the probability of making a type 1 error is 5 per cent.

(i) if N = 25

test statistic is 5/10

)105.11( = 0.75

critical value of t(24) is 2.064, thus do not reject null.

Page 17: Statistics exercise solution

(ii) if N = 100

test statistic is 10/10

)105.11( = 1.5

critical value of t(99) is 1.99, thus do not reject null.

(iii) if N = 1,000,000

test statistic is 1000/10

)105.11( = 150.0

critical value of t(999,999) is 1.96, thus reject null.

(b) Calculate approximate p values for each of these tests.

(i) t table value corresponding to 0.75 (24 degrees of freedom) is 0.23 approx. Thus p value is 0.23x2 = 0.46

(ii) t table value corresponding to 1.5 (99 degrees of freedom) is 0.16 approx. Thus p value is 0.16x2 = 0.32

(iii) t table value corresponding to 150 (999,999 degrees of freedom) is 0.000 approx. Thus p value is 0.000.

(c) Would your results in (a) be any different if the population variance was known to be 100. Explain in detail.

It might make a difference for the two cases N = 4 and 25. Note in all cases

where the sample size is small (below 30 approx) the population must be

assumed to be normally distributed.

Critical values would then be 1.96, but the decision not to reject null would be

the same.

Page 18: Statistics exercise solution

22. Suppose you wish to test a hypothesis about the mean growth rate of a certain kind

of companies. A random sample of 51 companies give you an estimator of the mean

growth rate of 8.7 per cent. The estimated standard error of this estimator is 0.9. You

know that the estimator has a normal distribution.

(a)Test the hypothesis that the mean growth rate is 10 per cent using a 5 per cent

level of significance.

0H population mean = 8.7

1H population mean 8.7

t = (8.7−10)/0.9 = − 1.444 this has a t distribution with 50 degrees of freedom.

Critical value is 2.009. Thus do not reject null.

(b)Test the null that the mean rate of growth is 6 per cent using a 5 per cent level of

significance

0H population mean = 6

1H population mean 6

t = (8.7−6)/0.9 = 3 this has a t distribution with 50 degrees of freedom.

Critical value is 2.009. Thus reject null.

(c) A new larger sample becomes of 200 companies. The estimator of the mean

growth is 8.5 and standard error is 0.3. Test the null that the mean growth rate is

10 per cent

0H population mean = 10

1H population mean 10

z = (8.5 −10)/0.3 = − 5 thus reject the null at 5 % and at 1%

23 (a) The mean of the annual average rate of return (in per cent) for the stock of

the public companies in the smallest decile (by size) in the US for 33 years

(1970-2002) is 13.30. The standard deviation of the sample is 24.91. If you were

to test the hypothesis that the mean annual rate of return is at least 10 per cent,

write down the null and alternative hypotheses and the test statistic.

0H population mean = 10

1H population mean > 10

Test statistic is 33/91.24

103.13 = 0.7611. This has a t(32) if the null is true. 5%

critical value for a one sided test is 1.629 approx. Thus do not reject.

Page 19: Statistics exercise solution

(b) The mean annual average rate of return for the largest decile for the same

period is 11.87 per cent. The sample standard deviation is 18.23. Test the

null hypothesis that the annual average rate of return is 11 per cent. Use a

significance level of 5 %.

0H population mean = 11

1H population mean 11

Test statistic is 33/23.18

1187.11 = 0.274 do not reject but note relevant critical

value for a two sided test is 2.035 (approx)

Note: the standard deviations given are the square root of an unbiased

estimator of the population variance.