2011 APR QMT500

7/30/2019 2011 APR QMT500

1/8

CONFIDENTIAL CS/APR2011/QMT500

UNIVERSITI TEKNOLOGI M ARAFINAL EXAMINATION

COURSECOURSE CODEEXAMINATIONTIME

STATISTICS FOR ENGINEERINGQMT500APRIL 20113 HOURS

INSTRUCTIONS TO CANDIDATES1 . This question paper consists of five (5) questions.

Answer ALL questions in the Answer Booklet. Start each answer on a new page..3.

4.

Do not bring any material into the examination room unless permission is given by theinvigilator.Please check to make sure that this examination pack consists o f:

i) the Question Paperii) a two - page Appendix 1 (Key Formulas)iii) an Ans we r Booklet - provided by the Facultyiv) a Statistica l Tab le - provided by the Faculty

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SOThis examination pap er con sists of 6 printed pages

) Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

7/30/2019 2011 APR QMT500

2/8

CONFIDENTIAL 2 CS/APR2011/QMT500

QUESTION 1a) The proba bility that a compo nent intended for use in a compu ter passes a purity test is

0.038.i) In a batch of 10 randomly selected components find, to 3 decimal places, theprobability that n one of the components p asses the test. (2 marks)ii) Using a suitable approxima tion, estimate the probability that fewer than fourcom ponents in a batch of 100 pass the test. (3 marks)

b) A process of making plate glass produces small bubbles (imperfections) scattered atrandom in the glass, at an average rate of four small bubbles per 10 m 2. Assum e thatthe number of sma ll bubbles follows a Poisson distribution.i) Determine, to 3 decimal places, the probability that a piece of glass 2.2m x 3.0m willcontain exactly two small bubbles. (3 marks)ii) Show that the probability that five pieces of glass, each 2.5m x 2. 0m , will all be free

of small bubbles is e~10. (2 marks)iii) Find, to 3 decimal places, the probability that five pieces of glass, each 2.5m x 2.0 m,

will contain a total of at least ten small bubbles. (2 marks)c) A study discussed the percent purity of oxygen from a certain supplier. Ass ume thedistribution of percent purity was approximately normal with mean 99.61 and standarddeviation 0.08.

i) What percentag e of the purity values would you expect to be between 99.5 and99.7? (4 marks)ii) Below what purity value would you expect exactly 5% of the distribution to lie?(4 marks)

Hak Cipta Universiti Teknologi M ARA CONFIDENTIAL

7/30/2019 2011 APR QMT500

3/8


QUESTION 2a) The following d ata show s the altitude x metres above sea level, for 1 0 places and the

weight of a type of fruit (in grams) cultivated in each of the places.Altitude, x (m)Weight, y (g)

29.7

7 1210.7 9.9

4010.4

769.5

999.2

1359.2

1639.4

2358.7

3077.5

i) W rite the leas t squares line to estima te the weigh t of fruit cultivated at a place of xmetres abov e sea level. Explain the meaning of the estimated regressioncoefficients. (3 marks)ii) Estimate the weight of fruit cultivated at a place 200 m etres above the sea level.(1 mark)iii) Calcu late the coefficient of correlation and explain its value . (2 marks)iv) At the 1 % significance level, can we conclude that the linear regression m odel issignificant? (8 marks)

b) A paint ma nufacturer uses a large amount of titanium dioxide in its coatings. Titaniumdioxide is the primary white pigment used in paints and coatings. The whiteness of thepaint is measured using a scale of 0-30, with 30 being essentially white. A new vendorclaims that its titanium dioxide averages 25 on the whiteness scale, with a variance of0.4. The paint manufacturer doubts that the new vendor has a small amount variability inits product. A sa mp le of 10 readings on the whiteness of the paint is as follows:

24 25 27 25 2626 24 25 26 25Test the claim a bout the variability using a 0.05 significance level. (6 marks)

Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

7/30/2019 2011 APR QMT500

4/8


QUESTION 3a) Officials of a small transit syste m with only five buses want to evalua te four types of tires

with respect to wear. Each of the buses runs a different route so that terrain and drivingconditions differ from bus to bus. One tire of each type is placed on each bus, with thewheel positions being assigned randomly. The tires are run for 15,000 miles, after whichthe tread wear, in millimeters, is measured. The data obtained is as follows:Type of buses

ABCDEColumn total

Types o19. 113.415.611 .012.761 .8

217.120.324.618.219.8100

320.828.323.721 .425.1119 .3

F tires411 .816.016.214.115.873.9

Row total58.878.080.164.773.4355.0

i) What is the type of experime ntal design used in this study? (2 marks)ii) Construct the analysis of variance table. (6 marks)iii) Do the data provide sufficient evidence to indicate that a difference ex ists in the treadwear in different type s of tires? Use a 5% level of significance. (2 marks)iv) Is there sufficient eviden ce to indicate that tread wear of tires vary from bus to bus?Test at 5% level of significance. (2 marks)

b) Within a certain tribe it was found that 1 5% had blood that was rhesus negative. A studydone at a later date took samples of 10 people from different areas within the tribe'sboundary and found how many had rhesus negative blood in each of the samples, withthe following resultsNumber of rhesusnegativeFrequency

023

130

232

310

>35

Do the data provide sufficient evidence to conclude that the number of rhesus negativehas a Binomial distribution with n = 10 and p = 0.15? Test at a = 0.05. (8 marks)


7/30/2019 2011 APR QMT500

5/8


QUESTION 4a) One important aspect of com puting is the cpu time required by a particular algorithm tosolve a problem. A new algorithm is developed to solve zero-one multiple objectivesproblems in linear programming. It is thought that the new algorithm will solve problemsfaster than the previous algorithm. The data sets below represents a random sample ofcpu times using the old algorithm and the new ones.

Old 8.05 24.74 28.33 8.45 9.19 25.20 14.05 20.33 4.82 8.54New 0.71 0.74 0.74 0.77 0.80 0.83 0.82 0.77 0.71 0.72Is there evidence that the new algorithm is better than the old algorithm at a = 0.05?(7 marks)

b) The following table show s the num ber of trains arriving on tim e and the num ber of trainsthat were late was observed at three different stations.On time LateA 26 14Station B 30 10C 44 26

Test at 5% level of significance whether there is any assoc iation between the station andlateness. (7 marks)c) A utility company h as a large stock of voltmeters that are used interchangeably by manyemployees. A study is conducted to detect the differences among average readingsgiven by these vo ltmeters. A random sam ple of six meters is selected from the stock andfour readings are taken for each meter. A partially completed ANOVA table is givenbelow:

Source of variationVoltmeterErrorTotal

df Sum ofSquares7.66918.926

MeanSquare F

Complete the AN OV A tab le. H ence test at 1 % significance level if there is anysignificant difference in the means of voltmeter readings. (6 marks)

Hak Cipta Universiti Teknologi M ARA CONFIDENTIAL

7/30/2019 2011 APR QMT500

6/8

CONFIDENTIAL 6 CS/APR 2011 /QMT500

QUESTION 5Environmental testing is one of the attempts to test a component under conditions thatclosely simulates the environment in which the component will be used. An electricalcomponent is to be used in two different locations in Country A. Before environmentaltesting can be conducted, it is necessary to determine the soil composition in theselocalities. These data are obtained on the percentage of Si 0 2 by weight of the soil:

Soil Type 1 Soil Type 2Number of SampleSample MeanSample Variance1 364.949

1157.067.29Assu me that data is normally distributed for both popu lations.a) A researcher claims that the percentage of S i0 2 in Soil Type 1 is 64. Do the datacontradict the researcher's claim? Test at the 1 % significance level. (5 marks)b) Calculate a 99 % confidence interval for the percentage of S i0 2 in Soil Type 1 . Doesthe interval support your conclusion in (a)? Explain. (4 marks)c) Is there evidence at 5% significance level that there is a difference in the variancesin the percentage of Si0 2 in these two types of soil?

(5 marks)d) Using the result on the variances in part (c), find a 90% confidence interval for thedifference between the means percentage of Si0 2 in Soil Type 1 and in Soil Type 2.Does there appear to be a difference between the means? Explain based on theanswer. (6 marks)

END OF QUESTION PAPER


7/30/2019 2011 APR QMT500

7/8

CONFIDENTIAL APPENDIX 1(1)K E Y F O R M U L A S

CS/APR2011/QMT500

Binomial probability formula

Poisson probability formula

p(x) = P(X = x) = p x (1-p)n - x ; x = 0 , 1 , 2 n

p(x) = P(X = x) = ^ - ; x = 0 , 1 ,2

C O N F I D E N C E I N T E R V A L SParameter & description Two-sided (1 - a)100% confidence intervalMean |i of a normal distribution,variance a 2 unknown x t a/2

f S ^Vn"

d f = n - 1

Difference in means of twonormal distributions m - \i2,variances c^2 = G 22 and unknown

r" i i( x 1 - x 2 ) + t a / 2 s p + ;V n1 n 2d f = n ! + n 2 - 2 , s p = (ni-Vs-f + ( n 2 - 1 ) s 2 2n 1 + n 2 - 2

Difference in means of twonormal distributions m - p,2,variances a-i2 * a2 2 and unknown ( x i - x 2 ) t a / 2 K + ^]| n.| n 2

df _ ( s 1 2 / n 1 + s 2 2 / n 2 f

n 1 - 1 n 9 - 1

Mean difference of two normaldistributions forpaired samples, |add t a/2 Vn~ df = n - 1 where n is no, of pairs

Variance a of a normaldistribution ( n - l ) s 2 ( n - l ) s2Xa /2 Y 2x 1 - a / 2df = n - 1

Ratio of the variances a2/a22 oftwo norm al distributions U2

s 1 2 P iS 2 ' a / 2 ; v i ,V2 S 2 '2 ' a / 2 ; v 2 , v i vi = ni - 1 , v2 = r n 2 - 1


7/30/2019 2011 APR QMT500

8/8

CONFIDENTIAL APPENDIX 1(2) CS/APR2011/QMT500

H Y P O T H E S I S T E S T I N GNull Hypothesis Test statisticH 0 : (i = \x 0a2 unknown

t c a , c = ^ - ; df = n - isVn"H 0 : m - \x2 = Da-i2 = a22 and unknown

Icalc ~ (* 1 - * 2 ) - D .1 1 +" 1 " 2n 1 1 )s 1 2 + ( n 2 - 1 ) s 2 2d f = n i + n 2 - 2 , s p = . p - ! ' ^ y-+ -w \ ni + n 2 - 2

H 0 : Hi - ^2 = Dvariances a!2 * a 2 2 and unknown_ ( x 1 - x 2 ) - D .kale

n-] n 2

df = ( s 1 2 / n 1 + s 2 2 / n 2 fM / n i f | ( s 2 2 / n 2 rn i - 1 no - 1

H 0 : Hd = D loalc ~ d - DSd_Vn

df = n - 1 where n is no. of pairs

2 _ 2H 0 : CT = o-0 (or a = a 0 ) 2 _ ( n - l ) s;

^ c a l c df = n - 1

2 _ 2H Q : CT I - o"2' s 1f r c a l c = - o - : v ^ n i - 1 , v2 = n 2 - 1

T E S T I N G S I G N I F I C A N C E O F R E G R E S S I O N(Ana lys is of Var iance approach )

Total sum of squaress s Y = i ( y i - y ) 2 = y 2i=1 i=1 n

Regression sum of squares SSR = E ( y i - y ) 2 = p 1 Si=1

where S x y = X x i Y i -i= i

xy

Z X iU=1 l y in

Hak C ip ta Univers it i tekno log i MARA CONFIDENTIAL

Documents

2011 APR QMT500