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MINISTRY F EDUCATION ALAYSIAFOUR-YEAR EACHEREDUCATION ROGRAMME

FOR PRIMARY DUGATION XAMINATIONOO8(PR|MARYMATHEMATTCS)

STATISTICS

Code No. : MTE3105

lndex No.:

Date May2008

Duration: Hours 30 Minutes

Centre :

INSTRUCTIONSO CANDIDATES: :"t

1. This paper onsists f two sections;Section : 5 structured uestionsSection : 4 essay uestions

2. Answer llquestions n Section and answer questions nfy n Section .

3. Answer uestions:Section in he space rovidedn he est booklet; ndSection in he answer heets rovided.

4. All answers heets must be ied ogether ith he est booklet nd be handed n atthe end of he examination. nswer heets ot handed niat he end of heexamination illnot be marked y he examiner.

5. An additional 0minutes s given or you o read hrough he wholepaper eforeyou begin o answer.

r

FOR EXAMINER'SUSE ONLY

SECTION QUESTION UMBER MARKS TOTALA

(50marks)

1a 1b 1c 2a 2b 2c 3a 3b 4a 4b 5a 5b

B(50

marks

1a 1b 2a 2b 3 4a 4b

GRANDTOTAL

THISPAPER ONSISTS F 25 PRINTED AGES NCLUDING

THEAPPENDIX

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Code No.: MTE 3105 lndex No.:

SECTION(50 marks)

Answer ll questions n hissection n he space provided.

1. The Road Transport Department equires hat all 10-year ld carshave to undergo a performance est. The probabilities hatthree 10-year ld cars A, B and C willpass he performance est areI I .1j ,4*d rrespect ively.

(a) Explainwhether he events of the cars passing he test are

independent r dependent vents.{ '} (2 marks}

(b) From he information iven, draw a tree diagram or the testperformance esult f the cars A, B and C.

(3 marks)

q",,9

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Code No.: MTE 3105 lndex No. :

(c) (i) Find he probabilityhat onlyone of the hree ars passesthe performance est.

(3 marks)

(ii) Calculate he probabilityhat at most wo out of the hreecars willpass he performance est.{2marks)

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Code No.: MTE 3105 Index No.:

2. A machine manufactures 000 cylindrical omponents n a five-hoursday, the mean value of the diameter eing 0.500 cm with standarddeviation .30mm.

(a) A random ampte f 25 components s taken every 15 minutes.Determine he mean of the sampling ist ribution f the meansand the standard rror of the means, orrect o four significantfigures, or one day's output rom he machine, f the sarnpleswere aken

(i) without eplacement(3 marks)

(ii) with eplacement.(2 marks)

{

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Code No.:MTE3105 lndex No.:

(b) Another ample f 50 components s aken and he meandiameter s ound o be 0.550 m. Find he 95% confidenceinterval f the population ean.

(2 marks)

(c) (i) Findalso he 99% confidence nterval f the populationmean or he sample n (b).

(2 marks)

(ii) Compare ouranswers n (b)and cXi),what conclusioncan you make about hem?

3.

(1 mark

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Code No.: MTE 3105 lndex No.:

3. A focal magazine as run a survey and claims hat the trainees ofpublic ntitutions atch ess television han the general public. Thelocalaverage s 25.4 hours per week,with he"standard eviation f 2hours.

(a) List the steps or solving hypothesis-testing roblems.

(4 marks)

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Code No.:MTE3105 Index No.:

(b) A sample f 30 institute rainees akes part n the survey nd hasa mean of 19 hours. Using the steps isted n (a) , do you haveenough vidence o support he claimat c = A.U?

(6 marks)

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Code No. : MTE3105 lndex No.:

4. The trafficcontroldepartment f a town wishes o determine whetherthe number of summons ssued has an equal distribution mong hetrafficoffenders f variousage groups. The'followingable displaysthe age group distributionor a sample f people ummoned or traffic

offences.5.Age group 50

4ummoned 20 10

You will use the chi-square est to test whether he number ofsummons re equally istributed mong he age groups.

(a) (i) What s'Low Expected alues' n a chi-square est ?

({ mark)

(ii) What should ou do if (a) happen?

(1 mark)rb

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Code No.: MTE 3105 lndex No.:

(b) Test the claim that the distribution or each age groupsummoned s the same at q = 0.01 by using he following

steps:

(i) State he hypotheses o test he claim.

(2 marks)

(ii) State he degree of freedom n he est and he critical

value.

Critical alue:(2 marks)

Hr:

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l : ' | l

i t

iF

Gode No.:MTE3105 lndex No.;

(iii) Calculate he est-statistics yusingappropriate ormula.

(3 marks)

(iv) Writea conclusion f your est above.

(1 mark)

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Code No.: MTE 3105 lndex No.:

5. The manager f Welcorne upermarket ants o know f there s adifference n the average ime a customer as o wait n a check'outline n three different ounters. Analysis fdata collected n he check-out imes inminutes) or hree ounters s shown elow:

Counter A Counter B Counter C

4 5 1

3 .8 J

5 I 4

6 6 2

4 3 7

2 5 6

xl)si

-4 1=asl = 4'8

Xt =3'61

s2t 4'67)

GrandMean = 4.5556Grand ariance = 4.4967

(a) Whathypotheses ndclaimwilltheManager se n an ANOVAtest?

(2 marks)

(b) At q =0.01, est he hypptheses n (a):

(i) State he degree f freedom or numerator nddenominator, ence ind he critical alue

d.f. numerator):

d.f. deryominator):

Ho:

Hr:

CriticalValue:

(2 marks)

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Code No.: MTE 3105 Index No, :

(ii) Evaluate he variance etween roups nd he variancewithingroups, nd hence he F-test alue.

(5 marks)

(iii) Makea decision bout he est above.(1 mark)

,, .i

.L

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Code No.: MTE 3105

SECTION(50 marks)

Answer any two (2) questions in this section.

1. A sample f 50 bulbs has he mean ime o failure f 895 hourswith hestandard eviation f 49 hour. Construct 99% confidence nterval orthe mean ime o failure f the bulbs.

(a) Explain how he confidence ntervalwill e affected n hefollowing ases:

(i) A bigger ample f 500 ather han 50 was used.

(ii) lt is given hat he population f bulbs.has tandarddeviation f 45 hours

(12marks)

(b) lf the population f bulbshas a mean ime o failure f 900 hoursand standard eviation f 45 hours, etermine he probabilityhatthe mean ime o failureof a random ample of 36 bulbs willbebetween 90 hours and 915 hours.

(5 marks)

A production anager laimed hat he is 95% confident hat hemean ime o failure s between 52 o 948 hours, or any sampletaken rom he population, hen o = 45 hours Showwhether heclaim s true witha sample ize of 30 bulbs and he mean of 892

. hours. How arge must a sample be selected f he wants o be90% confident f the rue mean differs rom he sample mean by

5 hours?(8 marks)

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Code No.: MTE 3105

2. A construction ornpany urchases teel ables romsupplier . Asample f 14 cables s selected, nd he breaking trength inkg)ofeach s ound. The data s shown below.

907, 906, 908, 906, 909, 1000, 908,

1001,1002, 09, 907, 905, 908, 909

Given that the breaking strength of the lteel cables s normallydistributed ith he population tandard eviation eing 30 kg.

(a) Determine he mean and he standard eviation f the sample.The lot of cables s rejected f the standard deviation of thesample s greater han 30 kg. At q = 0.01, should he lot be hrejected?

(15marks)

(b) The construction ompany purchases more steel cables romanother supplier Y. The standard deviation f the breakingstrength in kg) for this population s 28 kg. A sample f 30cables s taken and t is found hat he mean s 930 kg. At o =0.05, an t be concluded hat here s no significant ifferencen

the breaking trength f the cables rom he wo suppliers?

(10marks)

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Code No. : MTE3105

3. Explain wo uses of the chi-square est and he conditions equired orthe use'

(6 marks

A study is being conducted o determine whether there is arelationship etween hysical xercise nd blood pressure. A randomsample of 206 subjects s selected, and' they are classified ntocategories s shown n the able below. At q = 0.01, est he claim ha tphysical xercise nd blood pressure re not related.

Exercise Status Blood Pressure

Exercise egularly

No exercise

Low

30

15

Moderate

56

60

High

20

25

(19 marks)

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APPENDIX 19

LlsToF FORMULAENSTATISTICS MTE 10s)

PROBABILITY

(1) General Addition Rule :

(21 Genral Multiplication Rule :

P(AU B) = P(A) + P(B) P(A nB)

P(AOB) = P(A)xP(B)

SAMPLING ND ESTIMATION HEORY

({) Standard error of the means (Samples drawn from a finite population)

(21

.o-W:"6i=(17){l, '_

o.rln

where

or - Standard eviation f he sampling istributionhe meanso - Standard eviation f he population,l/ - Population ize

n - Sample ize

Standard error of the means (Samples drawn from an infinitepopulation)

oi=

Transformation o z-score

Z_ X-po

wherea

Z - Standardized coreX - Value of variableIt - Mean f he distributiono - Standard eviation f he distribution

(3)

I

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Code No.: MTE 3105

4. (a) Define the ineof best it' n the regression nalysis.(3 marks)

(b) Data from a sample of 10 randomly elected students wascollected egarding he number of hours hey spent n a weekstudying for final examination and their scores on thatexamination. he data was as shown elow.

Drawa scatter plot or the data.Determine he regression quation nd use t to predict he scoreof a student ho studies 2 hours per week or he examination.

(20 marks)

Explain, n yourownwords, he meaning f residual f points,and he slope n your egression quation btained.

(3 marks)

@Government f Malaysia 008 07)

, i'i

Hours 6 8 5 I 11 5 10 6 15 4

Score 60 76 45 80 85 65 82 50 90 45

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APPENDIX /9

(4) The confidence imitsof the mean of the sample rom a finitepopulation

xt , .L . ly :-a t2Ji \ w - t

(5) The confidence imits of the mean of the santple rom an infinitepopulation

,6x * zo ' ' '-tn

(6) The confidence imitsof the mean alue of the population pl

/ t , + zarz . ozwhere

Ft - Mean of sampling distribution f the means

zat2 - Confidence oefficient

6i - Standard error of the means

(71 The confidence limits of the standard deviation of the population

,s + zatz 'o iwhere

s - Standard eviation f a sample

(8) The confidence imitsof the mean value of the population based on

small sample ize)s

x-t f .- 4n-l

where

t" - Confidenoe oefficient fsmall amples - Standard eviation f a samplen - Sample ize

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\

APPENDIX 19

CHI-SQUARE YPOTHESIS EST

(1) When expected requencies are equal

nH -_u-

kwhere

n - Sum of allobserved requenciesk - Number f categories

i (21 When expected requencies re not all equal

E=npwhere

n - Sum f allobservedrequenciesp - Frobabilityor he category

(3) Ghi-Square est Statistic

, ,2 T(ro-n> ' ;1"[E)

where

O - ObservedrequenciesE - Expectedrequencies

ANALYSTS FVARIANCEANOVA)

(1) Test Statistic or One-Way NOVA Catculations ith equal samplesizes)

O _Variancebetweensamples _yVariancewithinsamples s',

wheret

sj - Variance f sample means

) -s; - Pooled ariance r he mean f he sample ariances

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APPENDIX /9

(2) Test Statistc or One-Way NOVA Calculations ith unequal amplesizes)

r - Varianc e b etween samp esVarianc e w thin s amp e s

F_

>(", -l)

where

i - rn?fiof allsample cores ombined

k - number f population eans eing ompared

t1i - number f values n he rth sample

sf - variance f values n the r'th ample

(3) Key Components of ANOVA Method

SS(Total) = lx ,_(zn'//SS (Betweeo) =

SS(Within)

Where

= SS(Total) Ss(Between)

N - Total umber f elements n all he groups

Ti - Sum of elem,entSn each group

n - Number f elements n each group

z(n, r) s

z,:n

(Zxt'.Ar

1r",6,-;ILft{l

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i

APPENDIX 19

LINEAR EGRESSION

(1) Prediction quation n linear egression

Y=b, ,+brxwhere

b,= 'Za NZ"

and b,,= Y-bri

(21 Goefficient f determination R2

"Zxr - (IxXZr If t=

HYPOTHESIS ESTING

(1) Testing claims about a population tandard deviation or variance

, (n- 1) lt - --;-

o-

where

n - Sample ize

.r' - Sample variance

o' - Population ariance

Testing laims bouta population ean based n smallsample singStudent - distribution

X-Fx

(21

sr

ln

t

l"Zx' - (Zx)' l"ZY'- (I rl'

t -

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3)

APPENDIX /9

Testing claims about a population mean based on big sample usingstandard Normal distribution

i

Testing a claimabout a population roportion

where

n - number f rials

t, - x ln (sample roportion)

p - population roportionused n he nullhypothesis)

q - 1-p

) Testing he difference between wo population means

__(in- iu)-(pn-ltn)t ) , ,

lol , o;. l--1--

VM, NB

where

,l/ - Sample ize

i - Samplemean

P l i

X-ltxor

"rln

4)

p-p

lpq1tYn

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." \j

APPENDIX 19

TABLES

1. TABLE OF THE STANDARDNORMAL z) DISTIBUTION

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.02390.0279 0.03 190.03590 .1 0.0398 0 .0438 0 .0478 0.0517 0.0557 0.05960.0636 0 .0 67 5 0 . 07 1 4 0.0753o. 2 0.07930.0832 0.0871 0.0910 0.0948 0.0987 0.1 26 0.10640,1103 0.1141

0. 3 0.1179 0.1217 0 .1 55 0 .1293 0.'1331 0.13680.1406 0.'t443 0.1480 0.15170.4 0.1554 0.1 91 0.1628 0.1664 0.1700 0.1736 0.1772 0 .1 80 8 0 .1 84 4 0 .1 7 9

0.5 0.1 1 0.1 50 0.1 85 0.2019 0.20540.20880.2123 o.2 157o.2190 0.2224

0.6 0.22570.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 o.2642 0.2673 0.2704 0.27340.2764 0.2794 0.2823 0.28520.8 0.2881 0.29100.2939 0.2967 0.2995 0.30230.30510.30780.31060.31330.9 0.31s9 0.3186 0.3212 0.3238 0.3264 0.32890.331 50.3340 0.3365 0.33891.0 0.34130.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 n ?6()0 0.36211.1 0.3643 0.36650.3686 0.3708 0.3729 0.3749 0.37700.3790 0.38 100.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.39620.3980 0 .3997 0 .40151.3 0.4032 0.4049 0.4066 0.4082 0 .4 09 9 0 .4 11 5 0 . 41 3 1 0 .4 14 7 0 .4 16 2 0 .4 17 7

1.4 0.4192 o_4207o.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 4.4357 0.4370 0.4382 0 .4 39 4 0 .4 40 6 0 .4 41 8 0 .4 42 9 0 .4 44 11.6 0.4452 0.4463 0 .4474 0 .4484 0 .4 49 5 0 .4 50 5 0 . 45 1 5 0 .4 52 5 0 .4 5 35 0.4545l- t 0.4554 0.4564 0.4573 0.4582 0.45910.4599 0 .4608 0 .4616 0.4625 0.46331.8 0.4641 0.4649 0.46s6 0.4664 0.4671 0.4678 0.4686 0 .4693 0 .4699 0 .47061. 9 o .4713 0 .4719 o.4726 0.4732 0 .4 73 8 0 .4 74 4 0 .4 75 0 0 .4 75 6 0 . 47 61 0 .4 76 72.0 0.4772 0.4778 0 .4783 0 .4788 0 .4793 0.4798 0.4803 0.4808 0.48120.48172.1 0.4821 0.4826 0.4830 0.4834 0 .4 83 8 0 .4 84 2 0 .4 84 6 0 .4 85 0 0 .4 85 4 0 .4 85 7

2.2 0.4861 0.4864 0 .4868 0 .4871 0.4875 0 .4 87 8 0 .4 88 1 0 .4 88 4 0 .4 88 7 0 .4 89 02.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.49110.4913 0.4916

2.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.49310.4932 0.4934 0.49362.5 0.4938 0.4940 o.4941 0.4943 0.4945 0 .4946 0 .4948 0.4949 0_49510.4952z- o 0.4953 0 .4955 0 .4956 0 .4957 0 .4 95 9 0 .4 96 0 0 . 49 61 0 .4 96 2 0 .4 96 3 0 .4 96 42.7 0 .4 96 5 0 .4 96 6 0 .4 96 7 0 .4 96 8 0 .4 96 9 0 .4 97 0 0 .4 97 'l 0 .4 97 2 0 .4 97 3 0 .4 97 42.8 0.4974 0.4975 0.4976 0.4977 o.4977 0 .4 97 8 0 . 49 79 0 .4 97 9 0 .4 98 0 0 . 49 812.9 0.4981 0.4982 0.4982 0.4983 0 .4984 0.4984 0 .4985 0.4985 0.4986 0.49863.0 0.4987 0.4987 0.4987 0.4988 0 .4988 0 .4989 0 .4989 0 .4989 0.4990 0.49903.1 0.4990 0-4991 0 .4991 0 .4991 0.4992 0.4992 0.4992 0.4992 0.4993 0.49933.2 0 .4 99 3 0 .4 99 3 0 .4 99 4 0 .4 99 4 0.4994 0.4994 0.4994 0.499s 0.49950.49953.3 0 .4 99 5 0 .4 99 5 0 .4 99 5 0 .4 99 6 0 .4996 0 .4996 0 .4996 0 .4996 0.4996 0.49973. 4 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0.4997 0 .4997 0 .4998

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APPENDIX /9

2. CH|-SQUAREx') TABLE

T,t * E 4nfffiittdl* 3*Ofc$ffinnG*Ylf*-G*ffi"s*rffittrffiFrflEE*d

{t f*rVo-i:

triil

# ,*tT734f6u*t

,*TItllit*i5t6 ?tIt9st l :[g?}{,*sswtst9}B

3,S*t,.91t?"stt*,,1{g

fi.ffireffix*sdtIS#rtfi"$lf**.3gtr**"sfsxt#sw&7" LSF'*rgfs$6itf* ?srtffi,8fit0^t{*$t,{FI?SrrIt93*IitnS6tiIs}Tff*&t*t

{e}t}*t.t57et5?{$.??}

6S1t9Jr*

tts*tILIfrTr3#Ht,6.fts.18.+?*affiat"*66*3ffix.f.25a&zt73?-6f$2I"l'tlmJ?*JAffi$"s$.mfr&l*rnffi3&ryX{e m4t.f3*ll}*ffi+{"}t+rfeerE

*69StJ&A'Ery"st5U.r$t

#ilr.# n ir ftbrr-[]. fr&+ Axee* ltrFSh* ftffit* "*|***il#[ rrd rtu&rg ]ffi*,*t ui" **,p ffi tt,r*"anCx @{arfr pr|-ffir*trtrEt* * *tr* Rfidkd&mi**:r"* tf F.r"ta&s*o* tiE ffikr* rn***SiMt*,

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APPENDIX /9

3. F. DISTRIBUTION ABLE

F- Oictributi*nff: S.$ in the Right uil|

dflfl

q

+7tsI

I*1l

tl

t,{

41Ft"l98503;4.t 163t"r*8.1f.t5fr .?4tn.146I tl5*t$J6lr0,t 4496*60f.l;+e9S73$s"$6id$5*31s,5sI0s" Fgt$,r85{E.lssg8"0+soSllr*67.9454?-s*l?Jlt*?-?*e$'r-72r,?*676?t.oJ)c?J9?7?J6i}?"3t4t.7.fr77'6.S50S6,tl4s

.{989.5 54{3r9S,0SS T9,t6+I0.st7 I9,{5?IS,mO ffi,6,9dlt,t?{ 2,060ts.*ts $,??959,546{t ff.*5t38.firl 7.53108,{Ht5 d,S$tf7.5ss* *,5* 3?,ess? 4,il6?6;V2*& t,9$t$*,islo sJ3p46,s14* 5,56J*s*15fr9 5dJ?06,1 62 33s126.r t l s , rE6.$la3 SrlBt?J.E 5C 5.0lSJt.f,4$S ,1.*38I

5,?#t4 {.8?'{S5.?l 9 ,t.Slf$3,&Sl? ,1.76S*s.*136 { ,7181r.56#8 4.*?5F5,S?63 .t,6rd$sd*Br 4[6fl95,451* ,*.5*6t5.,r?*4 .t537&5.1003 4.td?5.r7ff5 {.31t6{.*??rl {.1e594.?$65 *,gd+t*"*$tt 3.1*16

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