NAME: AISHWARYA.A REG.NO:12E01 41 cars equipped with standard carburettors were for gas usage and yielded an average of 8.1 km/litre with a standard deviation of 1.2 km/litre.21 of these cars were then chosen randomly, fitted with special carburettors and tested, yielding an average of 8.8km/litre with a standard deviation of 0.9 km/litre . At the 5% level of significance , does the new carburettor decrease gas usage?

DATA GIVEN: n = 41 n = 21 x = 8.1 km/litre x =8.8 km/litre =1.2 km/litre = 0.9 km/litre = 5 %SOLUTION:

H : x = x ( no change ) H : x > x ( new carburettor decreases gas usage )Thus , from H , it is a right tailed test. Since = 5% , Z = 1.645 = = = - 9.5 Now, Z > Z at 5 % significance level. Thus H is rejected and H is accepted . Thus, the new carburettors decrease gas usage.

Name : K. Bala murugan.Reg No : 12E15.Question:Difference of Mean:1.Two companies produce resistors with a nominal resistance of 4000 ohms. Resistors from company A give a sample of size 9 with sample mean 4025 ohms and estimated standard deviation 42.6 ohms. A shipment from company B gives a sample of size l3 with sample mean 3980 ohms and estimated standard deviation 30.6 ohms. Resistances are approximately normally distributed. At 5% level of significance, is there a difference in the mean values of the resistors produced by the two companies?(12E15)Given: = 4000 ; X1 = 4025 n1 = 9 ; X2 = 3980 n2 = 13 ; S1 = 42.6 ; S2 = 30.6Soln: n1 < 30 ; n2 < 30So this is test for small sample and normally distributed. So, t-test for difference mean.H0 : X1 > X2 ( There is no difference in mean )H1 : X1 > X2 ( There is difference in mean )( One tail , Right tail test)LOS: = 5% = 0.05 t =( X1 X2) / (( (n1s12+ n2s22) / (n1+n2))((1/n1)+(1/n2) ) )1/2 t= (4025 - 3980) / ((9x(42.6)2 + 13x(30.6)2 )((1/9) + (1/13) ) )1/2 t = 2.883Table value: = n1 + n2 -2 = 20 One tail, so t(2,) = t0.1,20 = 1.73 Since t > t2, so rejected. NAME:S.AISWARYAREG.NO:12E021.An investigation of 2 kinds of photocopying machine showed that 80 failures of 1st kind of machine took average of 75.2 mins to repair with S.D of 20 mins when 80 failures of 2nd machine took average of 82.8 mins to repair with S.D of 22 mins.Test the null hypothesis against at 5% L.O.S.GIVEN:75.2=2080=82.8=22=80SOLN:1)[There is no difference between two means] 2)[two tail test]3)Level of significance 5%4)Test statistics ()= == =7.45)Table value:=1.96CONCLUSION:since|rejectTherefore, the difference between and is significant at 5%L.O.S

12E03 T.Angulakshmi 1.To test the claim that resistance of electric wire can be reduced by alloying. 32 numbers of standard wire yielded a reduction of mean resistance of 0.136ohm with standard deviation =0.004ohm and another 32 numbers of alloyed wire yielded a reduction of mean resistance of 0.0083ohm with standard deviation =0.005ohm. At 5% LOS does this support claim

Given: =0.136ohm =0.0083ohm s1=0.004 s2=0.005 n1=32 n2=32 H0: Set up null hypothesis H1:(RTT)LOS =5%

Zc = x1- x2/v(s1'/n1)+(s2/n2) =0.136-0.0083/v(0.004/32)+(0.005/32) =0.0180Z0.05=1.96Z0.05>ZcAccept Null Hypothesis

NAME:ANUPAMA.K 1.A manufacture of light bulbs claims that on the average 2% of the bulbs manufactured by him are defective.A random sample of 400 bulbs contained 13 defectives.on the basis of this sample can you support the manufacturers claim at 5% LOS.Solution:H0:P=0.02 i.e .2% of the products are defectiveH1:p>P.One tailed test(right tailed) test is to be used.Let LOS be 5%.therefore ,Z=1.645Z=(p-P)/((PQ)/n),where p= x/n=13/400=0.0325P=0.02, Q=1-P=0.98Z=(0.0325-0.02)/(0.02*0.98)/400Z=0.0125/0.007Z=1.785Z=1.79(approx)Z>ZTherefore H0 is rejectedtherefore the claim cannot be supported. NAME:M.APARNAA foundry produces steel forgings used in automobile manufacturing. We wish to test the hypothesis that the fraction conforming or fallout from this process is 10%. In a random samples of 250 forgings, 41 were found to be nonconforming. What are your conclusions using =0.05? Solution:H0:P=0.1 i.e.10%of the products are conforming.H1:pP.Two tailed test is to be used.Let LOS be 5% .Therefore, Z=1.96 Z=(p-P)/(PQ)/nWhere p=x/n=41/250=0.164P=0.1, Q=1-P=1-0.1=0.9Z=(0.164-0.1)/(0.1*0.9)/250Z=3.37|Z|=3.37|Z|>|Z|H0 is rejected and H1 is accepted.Thus the foundry produces steel forgings were found to be nonconforming.

P.V. Aravind12E06Question: A new rocket launching is considered for deployment of small and short range rockets. The existing system has 80% successful launches. A sample of 40 experimental launches is made with new system and 34 are successful. Would you claim that the new system is better?

Given: n=40, x=34, P=80% Soln:n>30 .So, this is Test for Large Samples and Test for Single proportion. P===0.8 Q=1-P =1-= =0.2 p= == = 0.8

1) Null Hypothesis (Ho) : There is no difference between proportions. p=P

2) Alternate Hypothesis (H1): There is difference between proportions. p>P This is Single Right tail test.

3) Level of Significance(LOS): =5% =0.05 (Assuming)

4) Test Statistic (Zc): Zc= Zc= Zc= Zc=Zc=0.7915) Conclusion: Z=Z0.05=1.645If =1%Z=2.33

Zc < Z So, Accept Null Hypothesis (Ho).There is no difference in Both System.

NAME : Aravind.S.REG.NO : 12E071.A study shows that 16 out of 200 submersible pumps produced on one assembly line required extensive adjustments before they could be shipped,while the same was true for 14 of 400 pumps produced on another assembly line.At 0.01 LOS,does this support the claim that the second production line does superior work?Solution:16 fails out of 200 14 fails out of 400

P1=0.92 P2=0.965 H0:P1=P2H1:P1|Z|The difference between P1 and P2 is significant.i.e.H0 is rejected and H1 is accepted.i.e., this does not support the claim that the second production line.

NAME : Aravindhan.R.REG.NO : 12E08.1.A study shows that 16 out of 200 capacitors produced on one assembly line required extensive adjustments before they could be shipped ,while the same was true for 14 of 400 capacitors produced on another assembly line.At 0.01 LOS, does this support the claim that the second production line does superior work?Solution:16 fails out of 200 14 fails out of 400

P1=0.92 P2=0.965 H0:P1=P2H1:P1|Z|The difference between P1 and P2 is significant.i.e.H0 is rejected and H1 is accepted.i.e., this does not support the claim that the second production line.

NAME :R.ARCHANAREG NO :12E09

1.57 out of 150 patients suffering with certain disease are cured by Allopathy medicine and 33 out of 100 patients with same disease are cured by Homeopathy medicine, is there reason to believe that Allopathy is better than Homeopathy at 5% LOS.(12E09)SOLUTION: denotes probability of success in allopathy medicine. denotes probability of success in homeopathy medicine. X denotes number of successors in n trails.: = :< (one tail LTT) =150-57 =93, =100-33=67 =150 =100 = / = / =93/150=0.62 =67/100=0.67 P= + /+ =(150)(0.62)+(100)(0.67)/150+100 =93+67/250 =0.64Z = - /=0.62-0.67/=0.806=-1.645

< ZReject null hypothesis,

CONCLUSION: Homeopathy is better than allopathy.

NAME :R.ARCHANAREG NO :12E09

1.57 out of 150 patients suffering with certain disease are cured by Allopathy medicine and 33 out of 100 patients with same disease are cured by Homeopathy medicine, is there reason to believe that Allopathy is better than Homeopathy at 5% LOS.(12E09)SOLUTION: P_1 denotes probability of success in allopathy medicine. P_2 denotes probability of success in homeopathy medicine. X denotes number of successors in n trails.H_0: P_1= P_2H_1: P_1< P_2(one tail LTT) X_1 =150-57 =93, X_2=100-33=67 n_1=150 n_2=100 P_1 = X_1/ n_1 P_2 = X_2/ n_2 =93/150=0.62 =67/100=0.67 P= ( n_1)( P_1)+ (n_2)(P_2)/(n_1 + n_2) =(150)(0.62)+(100)(0.67)/150+100 =93+67/250 =0.64Z = P_1 - P_2/{(PQ)(1/n_1+1/n_2)}1/2=0.62-0.67/{(0.64)(0.36)(1/150+1/100)}1/2=0.806Z_0.05=-1.645

Z_0.05< ZReject null hypothesis,H_0

CONCLUSION: Homeopathy is better than allopathy.

(On Test of Hypothesis on Small samples for Single Mean)ArunKumar.S.JReg.No:12E10Qn. High sulphur content in steel is very undesirable, giving corrosion problems among other disadvantages. If the sulphur content becomes too high, steps have to be taken. Five successive independent specimens in a steel-making process give values of % sulphur of 0.0307, 0.0324, 0.0314, 0.0311, and 0.0307. Do the data give evidence at 5% los that the true mean % sulphur is above 0.0300?Soln. =0.0300Null Hypothesis H0: =0.0300Alternative Hypothesis H1: 0.0300Calculation of sample mean and standard deviation:xx-(x-)2

0.0307-0.000563.136*10-7

0.0324+0.001141.2996*10-6

0.0314-0.000141.96*10-8

0.0311-0.000162.56*10-8

0.0307-0.000563.136*10-7

=0.1563=-2.8*10-4=19.72*10-7

Formula: t= (-) (S/n)=0.03126; n=5; =0.0300S2=19.72*10-7/4 = 49.3* 10-8t= (0.03126-0.0300)(49.3*10-8/5)t= 4.03t>t0.05=2.31Conclusion: Hence, the Null Hypothesis is rejected. The mean of data differ from true mean % sulphur of 0.0300

Reg. No. 12E11T. ARUN RAJQuestion:Two chemical processes for manufacturing the same product are being compared under the same conditions. Yield from Process A gives an average value of 96.2 from six runs, and the estimated standard deviation of yield is 2.75. Yield from Process B gives an average value of 93.3 from seven runs, and the estimated standard deviation is 3.35. Yields follow a normal distribution. Is the difference between the mean yields statistically significant? Use the 5% level of significance, and show rejection regions for the difference of mean yields on a sketch. (12E11)Solution:

Two tail test to be used. LOS = 5%.

t = 1.55

Also

,

From the t-table,

So H0 is accepted and H1 is rejected.That is two sample means are do not differ significantly at 5% LOS.Graph:

(Reject)-2.2 (Accept) 2.2 (Reject)

ROLL NO: 12E12SMALL SAMPLE- SINGLE MEAN :The average daily amount of scrap from a particular manufacturing process is 25.5 kg with a standard deviation of 1.6 kg .A modification of the process is tried in an attempt to reduce this amount .During a 10 day trial period ,the kg of scrap produced each day were 25,21.9,23.5,25.2,22,23,24.5,25,26.1,22.8. From the nature of the modification no change in day to day variability of the amount of scrap will result .The normal distribution will apply .A first glance at the figures suggest that the modification is effective in reducing he scrap level .Does a significant test confirm this at the 1% level.SOLUTION:=25.5 kgs=1.6 kgn=10The mean of the sample is given by,X===23.9H0: X=H1: XTwo tail test is to be used .Let LOS be 1%.t==t==-3.00t0.01=3.25=n-1=9; 3 t2, so rejected. Maths AssignmentName: P.BharathiReg. no: 12E16Difference of Means2. A new composition for car tires has been developed and is being compared with an older composition. Ten tires are manufactured from the new composition, and ten are manufactured from the old composition. One tire of the new composition and one of the old compositions are placed on the front wheels of each of ten cars. Which composition goes on the left-hand or right-hand wheel is determined randomly. The wheels are properly aligned. Each car is driven 60,000 km under a variety of driving conditions. Then the wear on each tire is measured. The results are:car no.12345678910

wear new comp.2.41.34.23.82.84.73.24.83.82.9

Wear of old comp.2.71.94.34.23.04.83.85.33.73.1

Do the results show at the 1% level of significance that the new composition gives significantly less wear than the old composition?

Let X1 be the wear of the new composition and X2 be the wear of old compositionNull hypothesis H0: X1=X2Alternate hypothesis H1: X1 FratioSo Accept null HypothesisCONCLUSION:Yes,There is no difference in means

5) A manufacturer of television sets is interested in the effect on tube conductivity of four different types of coating for color picture tubes. The following conductivity data are obtained.Coating TypeConductivity

1143141150146

2152149137143

3134136132127

4129127132129

Test the null hypothesis that , against the alternative that at least two of the means differ. Use = 0.05Solution:

Coating TypeConductivity

1143141150146

2152149137143

3134136132127

4129127132129

Assume null hypothesis Assume alternative hypothesis H1=At least two of means differn=Total no data=16Correction Factor C.F=T2/n =68.06Origin=140

Coating TypeConductivity Ti

1 3 1 10 6 20

2 12 9 -3 3 21

3 -6 -4 -8 -13 -31

4 -11 -13 -8 -11 -43

Tj-2-7-9-15-33

SST = xij2- T2/n =1149-68.06 =1080.94SSB = Ti 2/ni -T2/n= 912.75-68.06 = 844.69SSW = SST - SSB=1080.94 844.69 =236.25

ANOVA TABLESquare of variation Sum of Squares DOF Mean square Fratio

Between SampleSSB = 844.69 3 MSB = 281.56 14.30

Within SampleSSW = 236.25 12 MSW = 19.69 _

TotalSST = 1080.94 15 MST = 72.06 _

MSB =SSB/3 =844.69/3 =281.56MSW = SSW/12 =236.25/12 =19.69F(3,12) =3.49 at 5% LOSFratio > F(3,12) So Reject null HypothesisCONCLUSION:Yes,There is difference in means

An experiment was designed to study the performance of four different detergents for cleaning injectors. The following cleanliness readings were obtained with specially designed equipment for 12 tanks of gas distributed over three different models of engines.Engine 1Engine 2Engine 3

Detergent A454351

Detergent B474652

Detergent C485055

Detergent D423749

Obtain appropriate ANOVA table and test at 1% LOS whether there are differences in the detergents on in the engines.

SOLUTION:Between detergents:H0 : There are no differences in the performance of detergentsH1 : There are differences in the performance of detergentsBetween engines:H0 : There are no differences in the performance of enginesH1 : There are differences in the performance of enginesEngine 1Engine 2Engines 3Ti

Detergent A45-40 =531119120.33155

Detergent B761225208.33229

Detergent C8101533363389

Detergent D2-39821.3394

Tj221647T=85=867

12164552.2525

142154571=867

No of rows , r=4No of columns c=3= 1%=867T=25 ; Correction Factor SST = = 867-602.08=264.92SSR= = 110.98SSC = =135.17SSE=SST SSR - SSc=264.92-110.92-135.17= 18.83R-1 = 4-1 =3C-1 = 3-1 =2(R-1) (C-1) =3*2=6RC-1= 12-1 =11MSR = = MSC = = = 67.59MSE = = ANOVA TABLESource of variationSum of SquaresDegrees of freedomMean squareF ratio

Between detergentsSSR=110.92R-1 = 3MSR=36.97

Between enginesSSC=135.17C-1 = 2MSC=67.59

ErrorSSE=18.33(R-1)(C-1)=6MSE = 3.14

TotalSST=264.92RC-1 = 11

(i) F,(R-1),(R-1)(C-1)=F0.01,3,6=4.76F> F,(R-1),(R-1)(C-1)Reject H0There are significant differences in the performance of detergents(ii) F,(C-1),(R-1)(C-1)=F0.01,2,6=5.14F> F,(C-1),(R-1)(C-1)Reject H0There are significant differences in the performance of the engines.CONCLUSION:There are significant differences in the performance of both detergents and the engines NAME:T.KALAIVANI(12E39)

Q: An industrial engineer tests four different shop-floor layouts by having each of six work crews construct a sub assembly and measuring the construction (minutes) as follows.Layout 1Layout 2Layout 3Layout 4

Crew A 48.2 53.1 51.2 58.6

Crew B 49.5 52.9 50.0 60.1

Crew C 50.7 56.8 49.9 62.4

Crew D 48.6 50.6 47.5 57.5

Crew E 47.1 51.8 49.1 55.3

Crew F 52.4 57.2 53.5 61.7

Test at 1% LOS whether the four layouts produce different assembly times and whether some of the work crews are consistently faster in constructing this assembly than the others.A:Take =1% Shift the origin to 50.Layout1Layout2Layout3Layout4

crewA-1.83.11.28.611.130.8088.25

CrewB-0.52.9010.112.539.06110.67

crewC0.76.8-0.112.419.898.0164.82

crewD-1.40.6-2.57.54.24.4140.55

crewE-2.91.8-0.95.33.32.72206.74

crewF2.47.23.511.724.8153.76

-3.522.41.255.675.7328.76

2.0483.630.24515.23601.14

20.11119.720.76550.96711.53

= - = 711.53-238.77 = 472.76. =328.76-238.77 =89.99. =601.14-238.77 =362.37. = - - =472.76 -89.99 -362.37 =20.4.Source of variationSum of squaresDegrees of freedomMean squaresF ratio

Between row89.99517.99F==13.23

Between column362.373120.79F==88.82

Error20.4151.36

Total472.7618

=4.56=5.42Both the value>Result: Reject null hypothesis.REGISTER NO: 12E40NAME: V.U.KalpanaQUESTION:An industrial engineer is conducting an experiment on eye focus time. He is interested in the effect of the distance of the object from the eye on the focus time. Four different distances are of interest. He has five subjects available for the experiment. Because there may be differences among individuals, he decides to conduct the experiment in a randomized block design. The data obtained follow.Subject

Distance (ft)12345

4106666

676616

853325

Can we say distance affects the eye focus time at 5% l.o.s? (12E40)ANSWER: SUBJECTSDistance12345TiTi2/CXij2

410666634231.2244

67661626135.2158

8533251864.872

Tj22151591778431.2

Tj2/R161.375752796.3434.6

Xij17481814197474

CORRECTION FACTOR = T2 / (R*C) = 782 / ( 3 * 5 ) = 405.6Total sum of Squares: SST = Xij2 - T2 / (R*C) = 474 405.6= 68.4BETWEEN ROWS :SSB = Tj2/R - T2 / (R*C) = 434.6 405.6 = 29BETWEEN COLUMNS :SSC = Ti2/C - T2 / (R*C) = 431.2 405.6 = 25.6ERROR:SSE = SST SSC SSB = 68.4 29 25.6 =13.8HYPOTHESIS :Between rows :H0 = Distance doesnt affect the Eye focus.H1 = Distance affects the Eye focus.Between column :H0 = There is no significant difference between the subjects.H1 = There is significant difference between the subjects.ANOVA TABLE :Source ofVariationSum of squaresDegrees of freedomMean square

Fratio

BetweenRows29214.68.44

Between columns25.646.43.70

Error13.881.73

Total68.412

From table :F0.05,2,8 = 19.37 >Fratio Accept H0F005,4,8= 6.04 >FratioAccept H0Conclusion:At 5 % LOS we can conclude that Distance doesnt affect eye Focus.

NAME: KAMALI.SREG.NO:12E419.Prior to submitting a quotation for a construction project, companies prepare a detailed analysis of the estimated labour and materials costs required to complete the project. A company which employs three project cost assessors, wished to compare the mean values of these assessors' cost estimates. This was done by requiring each assessor to estimate independently the costs of the same four construction projects. These costs, in 0000s, are shown in the next column. (12E41)

SOLUTION: ABCTiTi/CXij

PROJECT1-8-5-10-2363189

PROJECT28952256.67170

PROJECT3-400-4516

PROJECT41214935140.33421

Tj818430Ti/C=134.67=796

Tj/R7275.551.5Tj/R=563.5

Xij288302206=796

T/RC=75SST=ij Xij-(T/RC) =796-75 =721SSR=i(Ti/C)-T/RC =134.67-75 =59.67SSC=i(Tj/R)-T/RC =563.5-75 =488.8SSE=SST-(SSC+SSR) =172.83Source of variationSum Of SquaresDegree of FreedomMean squareFraction ratio

b/w rows59.67319.8912.29

b/w column488.82244.41.45

Error172.83628.8

Total72111-

From the F Table, F5%(v1=6,v2=3)=8.94F5%(v1=6,v2=2)= 19.37With respect to rows, F0>F5% (F0 is rejected)With respect to columns, F0F5% for the rows, There is a difference in the mean cost estimatesAnd since F0F(r-1),(r-1)(c-1)FR>F0.05(2,8)FR=10.12F0.05(2,8)=4.46 So reject H0.H0=No effect due to column factorH1=An effect due to column factorFC>F(c-1),(r-1)(c-1)Fc=2.04F0.05(4,8)=3.87So accept Ho.CONCLUSION: There is difference between arrangement A&C .so it is not a good option to replace arrangement C with A.A varietal trial was conducted on wheat with 4 varities A,B,C,D in a latin square design. The plan of experiment and the plot yield are given below:C25 B23 A20 D20A19 D19 C21 B18B19 A14 D17 C20D17 C20 B21 A15 analyse data and interpret the result.use 5%LOS.Solution: we subtract 20 from the given values and work out with the new values of x ij i/j12 34T iT^2 i /nx^2 ij

1C5B3A0D08 1634

2A-1D-1C1B-2-32.257

3B-1A-6D-3C0-102546

4D-3C0B1A-5-712.2535

T j0-4-1-7-12T i^2/n=55.5122

T^2 j/n04.2512.25 T j^2/n=16.5

i x^2 ij36461129122

Rearraning the data according to the letters,we haveLetterX kT kT k^2/n

A0-1-6-5-1236

B3-2-1110.25

C510069

D0-1-3-3-712.25

TOTAL-1257.5

Q=X ij^2-T^2/N=122-12^2/16=113 Q1=1/nT i^2-T^2/N=55.5-9=46.5 Q2=1/nT j^2-T^2/N=-16.5-9=7.5 Q3=1/nT k^2-T^2/N=57.5-9=48.5 Q4=Q-Q1-Q2-Q3=113-(46.5+7.5+48.5) =10.5

ANOVA TABLE:S.VS.S.d.f.M.S.F0

Between rowsQ1=46.5n-1=315.5

Between columnsQ2=7.5n-1=32.516.16/1.75=9.2

Between lettersQ3=48.5n-1=316.16_

Residual Q4=10.5(n-1)(n-2)=61.75_

Total Q=113n^2-1=15__

From the F-tables,F 5%(v1=3,v2=6)=4.76 Since F0(=9.2)>F 5%(=4.76) with respect to the letters,the difference between the methods of cultivation is significant. Submitted by, K.KARTHICK(12E43), 11.A varietal trial was conducted on wheat with 4varieties A, B,C D. in a Latin square design.The plan of experiment and the plot yield are given below.C25B23A20D20A19D19C21B18B19A14D17C20D17C20B21A15Analyze the data and interpret the result. Use 5 % LOS.(12E43)SOLUTION:TiTiniTi/n

A1914201568462441156

B1923211881656141640.25

C2520212086739641849

D1719172073532941332.25

Ti=308Ti=23910n=16Ti/n= 5977.5

Xij (361+196+400+225)+(361+529+441+324)+(625+400+441+400)+(289+361+289+400) =6042Correction Factor=T/n=(308)/16 =5929SST=Xij-CF =6042-5929 =113SSB =Ti/n-CF =5977.5-5929 =48.5SSW=SST-SSB =64.5ANOVA TABLE:Source of VariationSum of SquaresDegree of FreedomMean SquareFraction Ratio

B/w varieties48.5316.163.0

Within varieties64.5125.38-

Total11315--

From the F table F5% (v1=3,v2=12)=3.49 (F0 30 and n2 > 30. So, This is test for Large Samples and Difference in means.Ho:x1=x2 (There is no difference in two means). H1:x1< x2 (There is difference in two means). So, Left Tail test.LOS: =0.01 Test Statistic:Zc=Zc=Zc=

Zc== =-4.540Zc = -4.54

Conclusion:=1%=0.01Z= Z0.01=-2.33Zc30, n2>30.So, This is test for Large Samples and difference in two means.Ho:x1=x2. There is no difference in two means.H1:x1>x2. There is difference in two means.Right Tail Test.

LOS:=0.05Test Statistic:Zc=

Zc=

Zc=Zc====1.491Zc=1.491Conclusion:Right Tail Test.Z=Z0.05=1.645Zc 30 and n2 > 30. So, This is test for Large Samples and Difference in means.Ho:x1=x2 (There is no difference in two means). H1:x1< x2 (There is difference in two means). So, Left Tail test.LOS: =0.01 Test Statistic:Zc=Zc=Zc=

Zc== =-4.540Zc = -4.54

Conclusion:=1%=0.01Z= Z0.01=-2.33Zc30, n2>30.So, This is test for Large Samples and difference in two means.Ho:x1=x2. There is no difference in two means.H1:x1>x2. There is difference in two means.Right Tail Test.

LOS:=0.05Test Statistic:Zc=

Zc=

Zc=Zc====1.491Zc=1.491Conclusion:Right Tail Test.Z=Z0.05=1.645Zc 1 the curve is tangential to x-axis at origin. The probability the value of from a random sample will exceed is given by P= dx And it is tabulated for various value of P and r=1,2,..30. For r > 30, were approximately become normal curve. The mean of distribution is mean = r and variance = 2r = r = 2rTest statics = = (n-1) d.o.fOi = Observe FrequencyEi= Excepted FrequencySubmitted by,S.ARIVUSELVI13LE01 QUESTION NO:10NAME; N.ENAMUL HASANREG.NO; 13LE07Discuss randomized block design briefly:Let us consider an agricultural experiment using which we wish to test the effect of k fertilizing treatments on the yield of a crop. We assume that we know some information about the soil fertility of the plots. Then we divide the plots into h blocks, according to the soil fertility, each block containing k plots. Thus the plots in each block will be of homogeneous fertility as far as possible. Within each block, the k treatments are given to the k plots in a perfectly random manner, such that each treatment occurs only in any block. But the same k treatments are repeated from block to block. This design is called randomized block design.Analysis of variance for two factors of classification: Let the N variance values {xij} (representing the yield of paddy) be classified according to two factors. Let there be h rows (blocks) representing one factor of classification (soil fertility) and k columns representing the other factor (treatment),so that N=hk. We wish to test the null hypothesis that the rows and columns are homogeneous viz.., there is no difference in the yields of paddy between the various rows and between the various columns.Let xij be the variate value in the row and jth column.Let x be the general mean of all the N values ,xi* be the of the k values ith row and x*j be the mean of the h values in the jth column Now XIJ = (XIJ - I*- *J+ )+( I*- )+( *J ) ( XIJ- ) =( XIJ- I*- *J+ )2+ ( I*- )2+( *J-X)2+2( XIJ-I*-*J +)(I*-) +2( XIJ-I*-*J+)(*J-) +2(I*-)(*J-)Now ,the fourth member in the R.H.S of(1)=2 (X I*- ) ( IJ- I*- *J+ )=2 ( I*- ) (k IJ- k I*- k *J+k )=0Similarly, the last two members in the R.H.S of (1) also become zero eachAlso ( *J- )2 = k ( *J- )2 =Q1 Say ( *J- )2 = h ( *J- )2 = Q2 SayLet Q= ( X*J- )2 andQ3 =2( XIJ-I*-*J+)2 Using all there in (1), we getQ=Q1+Q2+Q3 where Q= Total variationQ1=Sum of the squares due to the variations in the rowsQ2= that in the columnsQ3= that due to the residual variations.proceeding as in one factor of classification, we can prove that Q1/h-1,Q2/k-1 Q3/(h-1)(k-1) and Q/hk-1 are unbiased normal, all these estimates of the population variance with degrees of freedom h-1,k-1,(h-1) (k-1) respectively. If the sampled population is assumed normal, all these estimates are independent. (Q1/h-1)/( Q3/(h-1)(k-1)) follows a F-distribution with [h-1,(h-1) (k-1)]degrees of freedom and (Q2/k-1)/ (Q3/(h-1)(k-1)) follows a F- distribution with [k-1,(h-1),(k-1) degrees of freedom. Then the F-tests are applied as usual and the significance of difference between rows and between columns is analysed.TABLE; The ANOVA table for the two factors of classificationsS.VS.Sd.fM.SF

Between rows Q1h-1Q1/(h-1)(Q1/h-1)/( Q3/(h-1)(k-1))

Between columnsQ2k-1Q2/(k-1)(Q2/k-1)/ (Q3/(h-1)(k-1))

ResidualQ3(h-1)(k-1)Q3/(h-1)(k-1) --

TotalQhk-1 -- --

QUESTION NO:10Given; n=150 X=49solu; p=49/150 =0.327 P=40/100 =0.4 H0; No difference in proportion P=0.4 H1; P0.4 =5% Z=p-P/ =(0.327-0.4)/) Z= -0.894

Z0.05= 1.96 Z0.05 Accept H0

QUESTION NO:11Given; N=20 Sample size (n)=4FORMULAS; = 1/N i =1/N i UCL= +A2 LCL= -A2

Soundara Pandi13LE24Question:45 standard reinforcing bars were tested in tension and found to have mean yield strength of 31,500 psi with a sample variance of 25 x 104 psi2. Another sample of 35 bars composed of a new alloy gave a mean and coefficient of variation of 32,000 psi and 23 x 104 psi2 respectively. Yield strengths follow a normal distribution. At the 1% level of significance, does the new alloy give increased yield strength?Given:n1=45, x1=31500, =250000, n2=35, x2=32000 = 230000, =1%Soln:n1 > 30 and n2 > 30. So, This is test for Large Samples and Difference in means.Ho:x1=x2 (There is no difference in two means). H1:x1< x2 (There is difference in two means). So, Left Tail test.LOS: =0.01 Test Statistic:Zc=Zc=Zc=

Zc== =-4.540Zc = -4.54

Conclusion:=1%=0.01Z= Z0.01=-2.33Zc30, n2>30.So, This is test for Large Samples and difference in two means.Ho:x1=x2. There is no difference in two means.H1:x1>x2. There is difference in two means.Right Tail Test.

LOS:=0.05Test Statistic:Zc=

Zc=

Zc=Zc====1.491Zc=1.491Conclusion:Right Tail Test.Z=Z0.05=1.645Zc] FOR ,(k-1,n-k) REJECT IF CALCULATED f< FOR f (k-1,n-k) ACCEPT

(18)SUPPOSE THAT THE FIVE MEMBERS 0.44,0.81,0.14,0.05,0.93 WERE GENERATED.PERFORM A TEST FOR UNIFORMITY USING KOLMOGORO-SMIRNOV TEST WITH 5% LOS.SOLUTION:: THE NUMBERS ARE UNIFORMLY DISTRIBUTED : NOT UNIFORMLY DISTRIBUTED %R(i)0.050.140.440.810.93

0.20.40.60.81.0

R(i)0.150.260.16-0.07

R(i)(i1)/N0.05-0.040.210.13

={0.26} ={0.21}D=max{,}=0.26 =0.565 ACCEPT

NAME:S.DINESHKUMAR NO: 13LE06 1.Distinguish between level of significance and degrees of freedom? LEVEL OF SIGNIFICANCE DEGREES OF FREEDOM

1. The propability level below which we reject the hypothesis is called level of significance.And it is denoted by .

1.In mathematics degrees of freedom is any of the number of independent quantities necessary to express the value of all the variable properties of a system

2.Mention the properties of F distribution

1.The probability curve of the F-distribution is roughly sketched in fig 2.The square of the t-variate with with n degrees of freedom follows a F- distribution with 1and n degree of freedom3.The mean of the F-distribution is

4.The variance of the F-distribution is

3.In a sample of 600 parts manufactured by a factory, the number of defective parts was found to be 45. The company claims that only 5% of their products is defective. Is the claim tenable at 5% LOS.?Sol: ;

;

1.No difference in propotion 2.Right tailedOne-tailed (right-tailed) test is to be used.Let LOS be 5%. Ie

The difference betweet p and P is significant. i.e.., isrejected and is accepted

3.In a sample of 600 parts manufactured by a factory, the number of defective parts was found to be 45. The company claims that only 5% of their products is defective. Is the claim tenable at 5% LOS.?Sol: ;

;

1.No difference in propotion 2.Right tailedOne-tailed (right-tailed) test is to be used.Let LOS be 5%. Ie

The difference betweet p and P is significant. i.e.., isrejected and is accepted

M.GOTHAIREG.NO.13LE09

1.Discuss latin square design. Solution: Three Way Classification: A B C D

B C D A

C D A B

D A B C

Rows=n Columns=n Treatment=n Sum of k observations of treatments Tk = xij k Between sum of squares, SSTk =((Tk*Tk)/n (T*T)/n) Error sum of residuals SSE=SSE=SSR SSC SSTk ANOVA TABLE:Sources ofVariationSum of SquaresDegrees of FreedomMean Square F-Ratio

BetweenRows SSR n-1 MSR F.R

Columns SSC n-1 MSC F.C

TreatmentError SSTk SSE n-1(n-1)(n-2) MSTk Ftk

Total SST ((n*n)-1) MSE

Discuss the ANOVA table for1)One way classification2)Two way classification3)Three tableSolution:One way classification:Completely randomized design in which treatments are randomly assigned to the experiment units or in which independent random samples ofExperimental units or selected for each treatment.This test procedure compares the variation in observations between samples to the variation within samples. Here all the observations are classified into one factor. This is exhibited column wise if you denote jth observations in the ith samples by xij. The general scheme for level scheme for one way classificationis as follows.Sample 1 sample 2 sample i sample kx11 x21 xi xk1` . . . . . . . . . . . . x1n x2n xin xknNotations: T= Sum of all observations No of samples(or levels) =k No of observations in the sample=ni,i=1,2,.k. Total no of observations = n =ni i=1 observations j in the ith sample = xij j=1ni sum of ni observations in the ith sample Ti=xijcomputational formulae: T=Ti=xij i jTotal sum of squares,SST=((xij*xij) (T*T) / n) i jBetween samples sum of squares,SSB=((Ti*Ti)/ni (T*T)/N) iWithin samples sum of squares,SSW=SST SSBTotal mean square,MST=SST/n-1Within sample square,MSW=SSW/n-kNo of degrees of freedom=(k-1)+(n-k)=n-1ANOVA TABLE:Source of variationSum of squaresDegrees of freedomMean squareF ratio

Between samplesSSBk-1MSBF=MSB/MSW

Within samplesSSWn-kMSW _

TotalSSTn-1

2.Two way classification: Two way ANOVA is an analysis method for study with quantity outcome and two variables and this design is called randomized block design.Assumptions: The population at each factor level combination is approximately normally distributed these population have a common variance 3. The effect of one factor is same at all the levels of the other factorNotations : No.of levels of row factor = r No.of levels of column factor = c Total no.of observation= r*cObservations in ijth cell xijith level of row factor= i =1,2.rjth level of column factor=j=1,2.csum of cobservation in ith row=TRi =xij j j = 1,2.cSum of r observations in jth column = Tcj = xij i i=1,2.rSum of all r*c observations = T = xij i j = TRi = TcjThere are the notations used in computational formulaeSST = xij*xij - T*T/rc i jBetween rows sum of squares,SSR = ((TRi*TRi)/c)-(T*T/rc) iBetween columns sum of squares,SSC= ((TRj*TRj)/c)-(T*T/rc) iError sum of squares,SSC = SST SSR - SSC

ANOVA TABLE:SOURCE OF VARIATIONSUM OF SQUARESDEGREES OF FREEDOMMEAN SQUAREF RATIO

Between rowsSSRR-1MSRMSR/MSE

Between columnsSSCC-1MSCMSC/MSE

ErrorSSCr-1*c-1MSC_

TotalSSTr*c-1

1.Explain the steps in random variate generation Inverse Transform Technique :The inverse transform technique can be used to sample from:- 1.exponential distribution, 2.uniform distribution, 3.Weibull distribution,4.triangular distribution and 5.empirical distri-Jbutions

EXPONENTIAL DISTRIBUTION :The exponential distribution, discussed as before has probability density function (pdf)given byf(X)= {e-x , x 0 0, x < 0}and cumulative distribution function (cdf) given by f(X)= - x f(t) dt = 1 e x, x 0 0, x < 0

Steps:-

Step 1:- Compute the cdf of the desired random variable X. For the exponentialdistribution, the cdf is F(x) = 1 e , x > 0.

Step 2:- Set F(X) = R on the range of X.

Step 3:-Solve the equation F(X) = R for range of X. 1 e-X = R on the range x >=0. e-x = 1 R -X = log(1 - R) x=-1/ log(1 R)

Step 4:- Generate (as needed) uniform random numbers R1, R2, R3,... andcompute the desired random variates by Xi = F-1 (Ri) F (R) = (-1/)log(1- R) Xi = -1/ log ( 1 Ri)for i = 1,2,3,.... One simplification that is usually employed is to replace 1 Ri by Ri2.Mention three areas in which simulation is applied. Areas of Applications:-

1.Manufacturing Applications2.Military Applications3.Logistics, Transportation and Distribution Applications

1.Manufacturing Applications1. Analysis of electronics assembly operations2. Design and evaluation of a selective assembly station for highprecisionscroll compressor shells.3. Comparison of dispatching rules for semiconductor manufacturingusing large facility models.4. Evaluation of cluster tool throughput for thin-film head production.5. Determining optimal lot size for a semiconductor backend factory.6. Optimization of cycle time and utilization in semiconductor testmanufacturing.7. Analysis of storage and retrieval strategies in a warehouse.8. Investigation of dynamics in a service oriented supply chain.9. Model for an Army chemical munitions disposal facility.

Semiconductor Manufacturing1. Comparison of dispatching rules using large-facility models.2. The corrupting influence of variability.3. A new lot-release rule for wafer fabs.4. Assessment of potential gains in productivity due to proactiveretied management.5. Comparison of a 200 mm and 300 mm X-ray lithography cell.6. Capacity planning with time constraints between operations.7. 300 mm logistic system risk reduction.

Construction Engineering1. Construction of a dam embankment.2. Trench less renewal of underground urban infrastructures.3. Activity scheduling in a dynamic, multiproject setting.4. Investigation of the structural steel erection process.5. Special purpose template for utility tunnel construction.

2.Military Applications1. Modeling leadership effects and recruit type in a Army recruitingstation.2. Design and test of an intelligent controller for autonomousunderwater vehicles.3. Modeling military requirements for nonwarfighting operations.4. Multitrajectory performance for varying scenario sizes.5. Using adaptive agents in U.S. Air Force retention.

3.Logistics, Transportation and Distribution Applications1. Evaluating the potential benefits of a rail-traffic planningalgorithm.2. Evaluating strategies to improve railroad performance.3. Parametric Modeling in rail-capacity planning.4. Analysis of passenger flows in an airport terminal.5. Proactive flight-schedule evaluation.6. Logistic issues in autonomous food production systems forextended duration space exploration.7. Sizing industrial rail-car fleets.8. Production distribution in newspaper industry.9. Design of a toll plaza10.Choosing between rental-car locations.11.Quick response replenishment.

Business Process Simulation1. Impact of connection bank redesign on airport gate assignment.2. Product development program planning.3. Reconciliation of business and system modeling.4. Personal forecasting and strategic workforce planning.

Human Systems1. Modeling human performance in complex systems.2. Studying the human element in out traffic control.

3.Mention various components of rapid rail system.System:- Rapid railComponents:- 1. Entities:- Riders 2. Attributes:- Origination; destination 3. Activities:- Travelling 4. Events:- Arrival at station; arrival at destination 5.state variables:- Numbers of riders waiting at each station; numbers of Riders in transit.

M.Muthuselvi13 LE 15Explain briefly the different types of models.Solution: Models can be classified as being mathematical of physical. A mathematical model uses symbolic notation and mathematical equations to Represent a system. A simulation model is a particular type of mathematical model of a system. Simulation models may be further classified as being static or dynamic, deterministic or stochastic, and discrete or continuous. A static simulation model, sometimes called a Monte Carlo simulation, represents a system at a particular point in time. Dynamic simulation models represent systems as they change over time. The simulation of a bank from9:00 A.M. to 4:00 P.M. is an example of a dynamic simulation.

Simulation models that contain no random variables are classified as deterministic. Deterministic models have a known set of inputs which will result in a unique set of outputs. Deterministic arrivals would occur at a dentists office. In the case of the factory system, for example, the factors controlling the arrival of orders may be considered to be outside the influence of the factory and therefore part of the environment. However, if the effect of supply on demand is to be considered, there will be a relationship between factory output and arrival of orders, and this relationship must be considered an activity of the system. Similarly, in the case of a bank system, there may be a limit on the maximum interest rate that can be paid. For the study of a single bank, this would be regarded as a constraint imposed by the environment. In a study of the effects of monetary laws on the banking industry, however, the setting of the limit would be an activity of the system.How will you perform Kolmogorov-Smirnov test for uniformity of random numbers.Solution:Steps:Step1: Rearrange the data from smallest to largest. Let R(i) denote the I th smallest observation. So that R(1) 2 and is greater than 1, but it tends to 1 as v .

3. Properties of normal distribution curve: Normal Distribution Curves are symmetrical bell-shaped curves possessed of distinct characteristics. For example, the area under any given normal curve has the same proportional distribution to its total area; that is to say, the area from negative infinity to one S away from the mean is always the same: 84.13 percent (0.8413). The total area under a normal curve is always equal to one. Additionally, the curves never quite reach Y=0; curve extends to infinity and negative infinity. The first Illustration shows a few normal curves with equal means, but unequal standard deviations. The tall and skinny graph has a lesser S value than the others, and a greater frequency of values closer to the Mean. The second Illustration depicts curves with equivalent means and standard deviations, but unequal population sizes. As you can see, the shape that a Normal Curve takes is dependent upon its mean and standard deviation, and though it always has the properties listed above, the possibilities as to the shape that the curve may take are infinite. One particularly important Normal Curve is the Standard Normal Distribution.

F-table 0.05

Table of F-statistics P=0.05

df2\df1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 24 26 28 30 35 40 45 50 60 70 80 100 200 500 1000 >1000

df1/df2

3 10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.85 8.81 8.79 8.76 8.74 8.73 8.71 8.70 8.69 8.68 8.67 8.67 8.66 8.65 8.64 8.63 8.62 8.62 8.60 8.59 8.59 8.58 8.57 8.57 8.56 8.55 8.54 8.53 8.53 8.54 3

4 7.71 6.94 6.59 6.39 6.26 6.16 6.09 6.04 6.00 5.96 5.94 5.91 5.89 5.87 5.86 5.84 5.83 5.82 5.81 5.80 5.79 5.77 5.76 5.75 5.75 5.73 5.72 5.71 5.70 5.69 5.68 5.67 5.66 5.65 5.64 5.63 5.63 4

5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.70 4.68 4.66 4.64 4.62 4.60 4.59 4.58 4.57 4.56 4.54 4.53 4.52 4.50 4.50 4.48 4.46 4.45 4.44 4.43 4.42 4.42 4.41 4.39 4.37 4.37 4.36 5

6 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 4.03 4.00 3.98 3.96 3.94 3.92 3.91 3.90 3.88 3.87 3.86 3.84 3.83 3.82 3.81 3.79 3.77 3.76 3.75 3.74 3.73 3.72 3.71 3.69 3.68 3.67 3.67 6

7 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.60 3.57 3.55 3.53 3.51 3.49 3.48 3.47 3.46 3.44 3.43 3.41 3.40 3.39 3.38 3.36 3.34 3.33 3.32 3.30 3.29 3.29 3.27 3.25 3.24 3.23 3.23 7

8 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.31 3.28 3.26 3.24 3.22 3.20 3.19 3.17 3.16 3.15 3.13 3.12 3.10 3.09 3.08 3.06 3.04 3.03 3.02 3.01 2.99 2.99 2.97 2.95 2.94 2.93 2.93 8

9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.10 3.07 3.05 3.03 3.01 2.99 2.97 2.96 2.95 2.94 2.92 2.90 2.89 2.87 2.86 2.84 2.83 2.81 2.80 2.79 2.78 2.77 2.76 2.73 2.72 2.71 2.71 9

10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.94 2.91 2.89 2.86 2.85 2.83 2.81 2.80 2.79 2.77 2.75 2.74 2.72 2.71 2.70 2.68 2.66 2.65 2.64 2.62 2.61 2.60 2.59 2.56 2.55 2.54 2.54 10

11 4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 2.85 2.82 2.79 2.76 2.74 2.72 2.70 2.69 2.67 2.66 2.65 2.63 2.61 2.59 2.58 2.57 2.55 2.53 2.52 2.51 2.49 2.48 2.47 2.46 2.43 2.42 2.41 2.41 11

12 4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 2.72 2.69 2.66 2.64 2.62 2.60 2.58 2.57 2.56 2.54 2.52 2.51 2.49 2.48 2.47 2.44 2.43 2.41 2.40 2.38 2.37 2.36 2.35 2.32 2.31 2.30 2.30 12

13 4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 2.67 2.63 2.60 2.58 2.55 2.53 2.51 2.50 2.48 2.47 2.46 2.44 2.42 2.41 2.39 2.38 2.36 2.34 2.33 2.31 2.30 2.28 2.27 2.26 2.23 2.22 2.21 2.21 13

14 4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.57 2.53 2.51 2.48 2.46 2.44 2.43 2.41 2.40 2.39 2.37 2.35 2.33 2.32 2.31 2.28 2.27 2.25 2.24 2.22 2.21 2.20 2.19 2.16 2.14 2.14 2.13 14

15 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.51 2.48 2.45 2.42 2.40 2.38 2.37 2.35 2.34 2.33 2.31 2.29 2.27 2.26 2.25 2.22 2.20 2.19 2.18 2.16 2.15 2.14 2.12 2.10 2.08 2.07 2.07 15

16 4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 2.49 2.46 2.42 2.40 2.37 2.35 2.33 2.32 2.30 2.29 2.28 2.25 2.24 2.22 2.21 2.19 2.17 2.15 2.14 2.12 2.11 2.09 2.08 2.07 2.04 2.02 2.02 2.01 16

17 4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 2.45 2.41 2.38 2.35 2.33 2.31 2.29 2.27 2.26 2.24 2.23 2.21 2.19 2.17 2.16 2.15 2.12 2.10 2.09 2.08 2.06 2.05 2.03 2.02 1.99 1.97 1.97 1.96 17

18 4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 2.41 2.37 2.34 2.31 2.29 2.27 2.25 2.23 2.22 2.20 2.19 2.17 2.15 2.13 2.12 2.11 2.08 2.06 2.05 2.04 2.02 2.00 1.99 1.98 1.95 1.93 1.92 1.92 18

19 4.38 3.52 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.34 2.31 2.28 2.26 2.23 2.21 2.20 2.18 2.17 2.16 2.13 2.11 2.10 2.08 2.07 2.05 2.03 2.01 2.00 1.98 1.97 1.96 1.94 1.91 1.89 1.88 1.88 19

20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.31 2.28 2.25 2.23 2.20 2.18 2.17 2.15 2.14 2.12 2.10 2.08 2.07 2.05 2.04 2.01 1.99 1.98 1.97 1.95 1.93 1.92 1.91 1.88 1.86 1.85 1.84 20

22 4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 2.30 2.26 2.23 2.20 2.17 2.15 2.13 2.11 2.10 2.08 2.07 2.05 2.03 2.01 2.00 1.98 1.96 1.94 1.92 1.91 1.89 1.88 1.86 1.85 1.82 1.80 1.79 1.78 22

24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 2.25 2.22 2.18 2.15 2.13 2.11 2.09 2.07 2.05 2.04 2.03 2.00 1.98 1.97 1.95 1.94 1.91 1.89 1.88 1.86 1.84 1.83 1.82 1.80 1.77 1.75 1.74 1.73 24

26 4.23 3.37 2.98 2.74 2.59 2.47 2.39 2.32 2.27 2.22 2.18 2.15 2.12 2.09 2.07 2.05 2.03 2.02 2.00 1.99 1.97 1.95 1.93 1.91 1.90 1.87 1.85 1.84 1.82 1.80 1.79 1.78 1.76 1.73 1.71 1.70 1.69 26

28 4.20 3.34 2.95 2.71 2.56 2.45 2.36 2.29 2.24 2.19 2.15 2.12 2.09 2.06 2.04 2.02 2.00 1.99 1.97 1.96 1.93 1.91 1.90 1.88 1.87 1.84 1.82 1.80 1.79 1.77 1.75 1.74 1.73 1.69 1.67 1.66 1.66 28

30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 2.16 2.13 2.09 2.06 2.04 2.01 1.99 1.98 1.96 1.95 1.93 1.91 1.89 1.87 1.85 1.84 1.81 1.79 1.77 1.76 1.74 1.72 1.71 1.70 1.66 1.64 1.63 1.62 30

35 4.12 3.27 2.87 2.64 2.49 2.37 2.29 2.22 2.16 2.11 2.08 2.04 2.01 1.99 1.96 1.94 1.92 1.91 1.89 1.88 1.85 1.83 1.82 1.80 1.79 1.76 1.74 1.72 1.70 1.68 1.66 1.65 1.63 1.60 1.57 1.57 1.56 35

40 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12 2.08 2.04 2.00 1.97 1.95 1.92 1.90 1.89 1.87 1.85 1.84 1.81 1.79 1.77 1.76 1.74 1.72 1.69 1.67 1.66 1.64 1.62 1.61 1.59 1.55 1.53 1.52 1.51 40

45 4.06 3.20 2.81 2.58 2.42 2.31 2.22 2.15 2.10 2.05 2.01 1.97 1.94 1.92 1.89 1.87 1.86 1.84 1.82 1.81 1.78 1.76 1.74 1.73 1.71 1.68 1.66 1.64 1.63 1.60 1.59 1.57 1.55 1.51 1.49 1.48 1.47 45

50 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13 2.07 2.03 1.99 1.95 1.92 1.89 1.87 1.85 1.83 1.81 1.80 1.78 1.76 1.74 1.72 1.70 1.69 1.66 1.63 1.61 1.60 1.58 1.56 1.54 1.52 1.48 1.46 1.45 1.44 50

60 4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 1.99 1.95 1.92 1.89 1.86 1.84 1.82 1.80 1.78 1.76 1.75 1.72 1.70 1.68 1.66 1.65 1.62 1.59 1.57 1.56 1.53 1.52 1.50 1.48 1.44 1.41 1.40 1.39 60

70 3.98 3.13 2.74 2.50 2.35 2.23 2.14 2.07 2.02 1.97 1.93 1.89 1.86 1.84 1.81 1.79 1.77 1.75 1.74 1.72 1.70 1.67 1.65 1.64 1.62 1.59 1.57 1.55 1.53 1.50 1.49 1.47 1.45 1.40 1.37 1.36 1.35 70

80 3.96 3.11 2.72 2.49 2.33 2.21 2.13 2.06 2.00 1.95 1.91 1.88 1.84 1.82 1.79 1.77 1.75 1.73 1.72 1.70 1.68 1.65 1.63 1.62 1.60 1.57 1.54 1.52 1.51 1.48 1.46 1.45 1.43 1.38 1.35 1.34 1.33 80

100 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03 1.97 1.93 1.89 1.85 1.82 1.79 1.77 1.75 1.73 1.71 1.69 1.68 1.65 1.63 1.61 1.59 1.57 1.54 1.52 1.49 1.48 1.45 1.43 1.41 1.39 1.34 1.31 1.30 1.28 100

200 3.89 3.04 2.65 2.42 2.26 2.14 2.06 1.98 1.93 1.88 1.84 1.80 1.77 1.74 1.72 1.69 1.67 1.66 1.64 1.62 1.60 1.57 1.55 1.53 1.52 1.48 1.46 1.43 1.41 1.39 1.36 1.35 1.32 1.26 1.22 1.21 1.19 200

500 3.86 3.01 2.62 2.39 2.23 2.12 2.03 1.96 1.90 1.85 1.81 1.77 1.74 1.71 1.69 1.66 1.64 1.62 1.61 1.59 1.56 1.54 1.52 1.50 1.48 1.45 1.42 1.40 1.38 1.35 1.32 1.30 1.28 1.21 1.16 1.14 1.12 500

1000 3.85 3.00 2.61 2.38 2.22 2.11 2.02 1.95 1.89 1.84 1.80 1.76 1.73 1.70 1.68 1.65 1.63 1.61 1.60 1.58 1.55 1.53 1.51 1.49 1.47 1.43 1.41 1.38 1.36 1.33 1.31 1.29 1.26 1.19 1.13 1.11 1.08 1000

>1000 1.04 3.00 2.61 2.37 2.21 2.10 2.01 1.94 1.88 1.83 1.79 1.75 1.72 1.69 1.67 1.64 1.62 1.61 1.59 1.57 1.54 1.52 1.50 1.48 1.46 1.42 1.40 1.37 1.35 1.32 1.30 1.28 1.25 1.17 1.11 1.08 1.03 >1000

df2/df1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 24 26 28 30 35 40 45 50 60 70 80 100 200 500 1000 >1000

df1\df2

vt.eduF-table 0.05

MATHS ASSIGNMENTQn:The following data resulted from an experiment to compare 3 burners B1,B2,B3.A Latin square was used as test were made on 3 engines and were spread over 3 days. Engine 1Engine 2 Engine 3

Day 1 B1 16 B2 17 B3 20

Day 2 B2 16 B3 21 B1 15

Day 3 B3 15 B1 12 B2 13

Test the hypothesis that there is no difference between the burners at 5% L.O.S yields.

1) Inverse Transform Technique:Weibull distribution :F(x) ={ /(^) (x^(-1)) (e^-(x/) ,x 0 0 otherwise }Step 1 : compute cdf xF(x) = f(x) dx - 0 x= + (/^) x^-1 e^(-x/)^ dx- 0 x= -d e^(-x/) 0 F(x) =1-e^(-x/)Step 2: F(x) =1-e^(-x/) = R Step 3: solving x in terms of R 1-R = e^(x/)Log (1-R) = (-x/)X =[-log(1-R) ]^(1/)