UNIVERSITY OF KWAZULU-NATAL

SCHOOL OF STATISTICS & ACTUARIAL SCIENCE

EXAMINATIONS

17 NOVEMBER 2007

COURSE AND CODE STATISTICAL METHODS (STAT140)

_________________________________________________________________________ DURATION: 3 hours TOTAL MARKS: 100 INTERNAL EXAMINER: Dr H Mwambi EXTERNAL EXAMINER: Mr S Ramroop, UKZN-P THIS EXAM PAPER CONSISTS OF A TOTAL OF 20 PAGES INCLUDING THIS

ONE. PLEASE SEE THAT YOU HAVE THEM ALL. INSTRUCTIONS:

1. This is a closed book exam. 2. ALL questions are compulsory. 3. Formula sheet can be found on pages 8-9 4. Statistical tables can be found on pages 10-20.

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 2 QUESTION 1 [12 MARKS] a) Suppose X is a continuous random variable with a continuous probability density

function, )(xf . State how you would calculate the probability that X lies in the interval ( c , d ). [2]

b) Suppose that X is a random variable best described by a uniform probability distribution with 20=a and 45=b . i) Find the mean and standard deviation of X

ii) Find )22( σμσμ +≤≤− XP [3] c) Suppose that X is normally distributed with mean 11=μ and 2=σ . Find )6.128.7( ≤≤ XP [2] d) A microwave oven manufacturer is trying to determine the length of warranty period it

should attach to its magnetron tube, the most critical component in the oven. Preliminary testing has shown that the length of life (in years), X of a magnetron tube has an exponential probability distribution with 16.0=λ . (i) Find the mean and standard deviation of X

(ii) Suppose a warranty of 5 years is attached to the magnetron tube. What fraction of tubes must the manufacturer plan to replace, assuming the exponential model with 16.0=λ is correct. [Hint: Find )5( ≤XP ] [4]

QUESTION TWO A cigarette manufacture advertises that its new low-tar cigarette “contains on average no more than 4 milligrams of tar.” You have been asked to test the claim using the following sample information 10=n , 16.4=x milligrams, 30.0=s . a) State the appropriate null and alternative hypotheses in order to test the claim by the manufacture. [1] b) Test the claim at .05.0=α What do you conclude? [2]

c) Calculate a 95% confidence interval for the unknown mean μ . [2]

d) Suppose the manufacturer was also interested to test the hypothesis 0625.0: 2

0 ≤σH against 0625.0: 2 >σaH . i) Calculate the value of the test statistic for this problem

ii) Test at 05.0=α . Comment. [4]

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 3 QUESTION THREE Independent random samples were selected from two binomial populations. The sizes and number of observed successes for each sample are shown in the table below:

Sample 1 Sample 2 2001 =n 1101 =x

2002 =n 1302 =x

a) Test the null hypothesis 210 : ppH = against the alternative hypothesis 21: ppH a ≠ . Use .05.0=α [4] b) Calculate a 95% confidence interval for 21 pp − . Comment. [3] c) What sample size would be required if we wish to use a 95% confidence interval of width 0.14 to estimate 21 pp − . Comment. [3] d) Suppose the data was re-arranged according to the following format:

Population 1 Population 2 Successes Failures

110 90

130 70

Use the Chi-square test to test the hypothesis that the proportion of successes is constant across the two populations [Hint: Calculate expected frequencies]. Use

.05.0=α [3] QUESTION FOUR Independent random samples were selected from two normally distributed population namely ),( 2

11 σμN and ),( 222 σμN respectively. The sample sizes, means, and variances

are shown in the tables below:

Population 1 Population 2 121 =n

8.171 =x 2.742

1 =s

142 =n 3.152 =x 5.602

2 =s a) Test 2

2210 : σσ =H against 2

221: σσ ≠aH . Use 05.0=α . [2]

b) Would you be willing to use a t -test with a pooled variance to test the equality of the two means? Why? [2] c) If your answer in (b) is yes carry out the appropriate t -test. Use 05.0=α . [3]

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 4 QUESTION FIVE A random sample of 10 pairs of observations were selected, one of each pair from a population with mean ,1μ the other from a population with mean .2μ The data are shown below:

PAIR

VALUE FROM POPULATION 1

VALUE FROM POPULATION 2

1 2 3 4 5 6 7 8 9 10

60 40 78 53 67 88 77 60 64 75

63 38 77 50 74 96 80 70 65 75

a) Test the null hypothesis 0:0 =DH μ against 0: ≠DaH μ using a paired t-test, where 21 μμμ −=D . Use .05.0=α [4] b) Find a 95% confidence interval for .Dμ Comment. [2]

c) Suppose you do not wish to apply the paired t-test but rather use a non-parametric test.

Use the Wilcoxon matched-pairs signed rank test to test 0H against aH . Use .05.0=α Comment. [4]

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 5 QUESTION SIX A public policy polling group is investigating whether people living in the same household tend to make independent political choices. They select 200 homes where exactly three voters live. The residents are asked separately for their opinion (“yes” or “no”) on a city charter amendment. If their opinions are formed independently, the number saying “yes” should be binomially distributed. The data is displayed in the table below.

No. saying “yes” ( x )

Frequency ( f )

0 1 2 3

30 56 73 41

Total 200 a) Calculate the expected frequencies assuming the distribution of those saying “yes” in a household of three is Binomially distributed with 3=n and 55.0=p . [4] b) Test the goodness of fit in (b) using an appropriate test statistic. Use .05.0=α Comment. [3] QUESTION SEVEN The analysis of variance for a randomized block design produced the ANOVA table with partial entries shown below:

SOURCE DF SS MS F Treatments 3 28.2 Blocks 5 13.80 Error 34.1_____________ Total

a) Complete the ANOVA table. [3]

b) Does the data provide sufficient evidence to indicate a difference among the treatment means? Test using 05.0=α . [2] c) Does the data provide sufficient evidence to indicate that blocking was a useful design strategy? Use 05.0=α . [2] d) If the sample mean for treatment A and B are 7.9=Ax and 1.12=Bx , respectively, find a 90% confidence interval for )( BA μμ − . Comment. [3] e) Calculate the LSD for this data and show how to use it assuming the sample means of treatment A and B are as given in (d). Use .10.0=α [2]

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 6 QUESTION 8 a) Consider the following sample of ten measurements 12.1 8.5 15.8 17.7 9.6 4.0 25.2 10.3 6.2 13.9

Use the Wilcoxon’s signed rank test to test the claim that these values came from a population with a median value greater than 10. Use .05.0=α [4]

b) The tax Commission for a certain state conducted a study to determine whether there is

a difference in median deductions taken for charitable contributions depending on whether a tax return is filed as single or joint return. A random sample was selected, with the following results

Single Joint

61 =n

∑ = 431R ∑ =

=

628

2

2

Rn

(i) State the null and alternative hypotheses for this problem

(ii) Based on the data, what should the tax Commission conclude? Use 05.0=α .

[4] QUESTION 9 You are given the following summary statistics from a simple regression analysis:

840000

134679)(Re40.145

878.01200ˆ

==

==

+=

nx

gSSSSE

xy

a) Calculate a 95% confidence interval estimate for the mean of Y , given 40000=px . [3] b) Calculate a 95% prediction interval estimate for a particular Y , given 40000=px . [2] c) (i) What would happen to the precision of the estimate if the value of ?43000=px (ii) What value of px would produce the greatest precision? [2] d) Suppose now you wish to include a categorical variable at two levels. Explain how to achieve this and interpret the meaning of the additional regression coefficients. Note: No calculations needed here. [3]

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 7 QUESTION 10 A multiple regression model was fitted with Y as the dependent variable and three independent variables, 1X , 2X and 3X using a statistical computing software. The output is shown below. Use the results of the analysis to answer the question that follow. Assume

05.0=α in all statistical tests. Summary of analysis Source d.f. s.s. m.s. v.r. F pr. Regression 3 1419.04 473.013 88.77 <.001 Residual 8 42.63 5.328 Total 11 1461.67 132.879 Percentage variance accounted for 96.0 Standard error of observations is estimated to be 2.31. Estimates of parameters Parameter estimate s.e. t(8) t pr. Constant -3.84 4.08 -0.94 0.374 X1 0.053 0.544 0.10 0.925 X2 0.0032 0.0153 0.21 0.840 X3 0.9562 0.0798 11.99 <.001

(a) Can 0: 3210 === βββH be rejected or not? [2]

(b) Test 0:0 =iH β against 0: ≠iaH β for 3,2,1=i . [3]

(c) State the adjusted multiple coefficient of determination. Interpret. [2]

(d) Calculate the un-adjusted multiple coefficient of determination. Interpret. [2]

(e) Calculate the multiple correlation coefficient. Interpret. [2]

(f) Test 0:0 =ρH against 0≠ρ , where ρ is the population multiple correlation coefficient. [2]

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 8

LIST OF FORMULAE

1. ,1)(ab

xf−

= bxa ≤≤ ; 2

ba +=μ ,

12)( 2

2 ab −=σ .

2. λ=)(xf e xλ− , .0>x λ

μ 1= ,

λσ 12 =

3. ns

xtcalc /μ−

= , 1.. −= nfd

4. Confidence interval: ).(.2/ xestx ×± α

5. 2

22 )1(

σsnX ×−

= , 1.. −= nfd

6.

⎟⎠⎞

⎜⎝⎛

−=

nqppp

Z00

0ˆ

7.

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

21

21

11

ˆˆ

nnqp

ppZ

8. Confidence interval: )ˆˆ.(.)ˆˆ( 212/21 ppesZpp −×±− α

9. 22221

22/ ][)(4

WqpqpZ

n+

= α , =W width of the %100)1( ×−α interval.

10. 22

21

ss

Fcalc = , where 22

21 ss >

11. 2

)1()1(

21

222

2112

−+×−+×−

=nn

snsns p

12.

⎟⎟⎠

⎞⎜⎜⎝

⎛+×

−=

21

2

21

11nn

s

xxt

p

calc

13. Confidence interval: ).(.)( 212/21 xxestxx −×±− α

14. ns

dtd

calc /=

15. Confidence interval: ).(.2/ destd ×± α

16. ∑=

−=

k

i i

iicalc E

EOX

1

22 )(

on 1−k degrees of freedom

17. ∑∑= =

−r

i

c

j ij

ijij

EEO

1 1

2)( on )1()1( −×− cr degrees of freedom

18. SST = ∑ −ij

ij NGTx

22 )( , =N total number of observations.

19. SSB = ∑=

−k

i i

i

NGT

nT

1

22 )( , =iT total group i with in observations

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS 17 November 2007 Statistical Methods (Stat140) 9

20. SSBL = ∑=

−b

i

i

NGT

kB

1

22 )( , =iB block i total and =k number of groups.

21. kNkcalc FMSWMSBF −−= ,1~

22. =calcF kNbFMSWMSBL

−− ,1~

23. nMSWtLSD ×

×=2

2/αα

24. ∑−+×+×= 1

11211 2

)1( RnnnnU

25. ∑−+×+×= 2

22212 2

)1( RnnnnU

26. xx

xy

SS

=1β̂ , xy 10ˆˆ ββ −=

27. SSR = xyS×1β̂

28. Case 1: ⎟⎟⎠

⎞⎜⎜⎝

⎛ −+×=

xx

p

Sxx

nsyes

22 )(1)ˆ.(.

29. Case 2: s.e.( ⎟⎟⎠

⎞⎜⎜⎝

⎛ −++×=

xx

p

Sxx

nsy

22 )(11)ˆ

30. Confidence or prediction intervals: )ˆ.(.ˆ 2/ yesty ×± α

31. xxS

ses2

1 )ˆ.(. =β

32. Confidence interval: ).(.ˆ12/1 ββ α est ×±

33. 1,~ −−= knkcalc FMSEMSRF

34. yyS

SSRR =2 %100×

35. kkn

RRFcalc

11 2

2 −−×

−=

36. ∑=

+×−+×

=k

i i

i nnR

nnH

1

2

)1(3)1(

12 , =iR rank sum for group i .

37. ( )

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−×−

= ∑ ∑nx

xn

S ii

222

11 same formula for differences id .