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- 1. Most popular continuous probabilitydistribution is normal distribution.It has mean μ & standard deviation σDeviation from mean x- μZ = --------------------------- = --------------Standard deviation σGraphical representation of is called normalcurve
- 2. Standard deviation( σ )Standard deviation is a measure of spread( variability ) of aroundBecause sum of deviations from mean isalways zero , we measure the spread bymeans of standard deviation which is definedas square root of∑ (x- μ) 2∑ (x- xbar) 2Variance (σ2) = -------------= ----------------N n-1σ2 = variance
- 3. Interpretation of sigma (σ )1.Sigma (σ ) – standard deviation is a measure ofvariation of population2.Sigma (σ ) – is a statistical measure of theprocess’s capability to meet customer’srequirements3.Six sigma ( 6σ ) – as a managementphilosophy4.View process measures from a customer’spoint of view5.Continual improvement6.Integration of quality and daily work7.Completely satisfying customer’s needsprofitably
- 4. Use of standard deviation( σ )Standard deviation enables us to determine ,with a great deal of accuracy , where thevalues of frequency distribution are located inrelation to mean.1.About 68 % of the values in the populationwill fall within +- 1 standard deviation from themean2.About 95 % of the values in the populationwill fall within +- 2 standard deviation from themean3.About 99 % of the values in the populationwill fall within +- 3 standard deviation from themean
- 5. 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 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
- 6. 1.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
- 7. 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48902.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49162.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49642.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49812.9 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 0.4993 0.4993 0.4994 0.4994 0.4994 0.4994 0.4994 0.4995 0.4995 0.49953.3 0.4995 0.4995 0.4995 0.4996 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
- 8. SIGMA Mean CenteredProcessMean shifted( 1.5)Defects/million% Defects/million%1 σ 317400 31.74 697000 69.02 σ 45600 4.56 308537 30.83 σ 2700 .26 66807 6.684 σ 63 0.0063 6210 0.6215 σ .57 0.00006 233 0.02336 σ .002 3.4 0.00034
- 9. Thus , if t is any statistic , then by central limittheoremvariable value – AverageZ = --------------Standard deviation or standard errorx- μZ = --------------σ
- 10. Properties1. Perfectly symmetrical to y axis2. Bell shaped curve3. Two halves on left & right are same. Skewnessis zero4. Total area1. area on left & right is 0.55. Mean = mode = median , unimodal6. Has asymptotic base i.e. two tails of the curveextend indefinitely & never touch x – axis( horizontal )
- 11. Importance of Normal Distribution1. When number of trials increase , probabilitydistribution tends to normal distribution .hence, majority of problems and studies can beanalysed through normal distribution2. Used in statistical quality control for settingquality standards and to define control limits
- 12. Hypothesis : a statement about the populationparameterStatistical hypothesis is some assumption orstatement which may or may not be true ,about a population or a probabilitydistribution characteristics about the givenpopulation , which we want to test on thebasis of the evidence from a random sample
- 13. Testing of Hypothesis : is a procedure thathelps us to ascertain the likelihood ofhypothecated population parameter beingcorrect by making use of sample statisticA statistic is computed from a sample drawnfrom the parent population and on the basis ofthis statistic , it is observed whether thesample so drawn has come from thepopulation with certain specifiedcharacteristic
- 14. Procedure / steps for Testing a hypothesis1. Setting up hypothesis2. Computation of test statistic3. Level of significance4. Critical region or rejection region5. Two tailed test or one tailed test6. Critical value7. Decision
- 15. Hypothesis : two types1. Null Hypothesis H02. Alternative Hypothesis H1Null Hypothesis asserts that there is no differencebetween sample statistic and populationparameter& whatever difference is there it is attributable tosampling errorsAlternative Hypothesis : set in such a way thatrejection of null hypothesis implies theacceptance of alternative hypothesis
- 16. Null HypothesisSay , if we want to find the population mean hasa specified value μ0H0 : μ = μ0Alternative Hypothesis could bei. H1 : μ ≠ μ0 ( i.e. μ > μ0 or μ < μ0 )ii. H1 : μ > μ0iii. H1 : μ < μ0iv. R. A. Fisher “Null Hypothesis is thehypothesis which is to be tested for possiblerejection under the assumption that it is true
- 17. 4. level of significance :is the maximum probability( α ) of making a Type I error i.e. : P [ Rejecting H0when H0 is true ]Probability of making correct decision is ( 1 - α )Common level of significance 5 % ( .05 ) or 1 % ( .01 )For 5 % level of significance ( α = .05 ) , probabilityof making a Type I error is 5 % or .05 i.e. : P[ Rejecting H0 when H0 is true ] = .05Or we are ( 1 - α or 1-0.05 = 95 % ) confidence thata correct decision is madeWhen no level of significance is given we take α =0.05
- 18. 5.Critical region or rejection region :the valueof test statistic computed to test the nullhypothesis H0is known as critical value . Itseparates rejection region from theacceptance region
- 19. 6.Two tailed test or one tailed test :Rejection region may be represented by aportion of the area on each of the two sidesor by only one side of the normal curve ,accordingly the test is known as two tailedtest ( or two sided test )or one tailed ( or one sided test )
- 20. Two tailed test :where alternativehypothesis is two sided or two tailede.g.Null HypothesisH0 : μ = μ0Alternative HypothesisH1 : μ ≠ μ0 ( i.e. μ > μ0 or μ < μ0 )
- 21. One tailed test :where alternativehypothesis is one sided or onetailedtwo typesa. Right tailed test :- rejectionregion or critical region liesentirely on right tail of normalcurveb. Left tailed test :- rejection regionor critical region lies entirely onleft tail of normal curve
- 22. Right tailed :Null HypothesisH0 : μ = μ0Alternative HypothesisH1 : μ > μ0Left tailed :Null HypothesisH0 : μ = μ0Alternative HypothesisH1 : μ < μ0Right tailed
- 23. 7.Critical value : value of sample statisticthat defines regions of acceptance andrejectionCritical value of z for a single tailed ( leftor right ) at a level of significance α is thesame as critical value of z for two tailed testat a level of significance 2α .
- 24. Critical value(Zα )Level of significance1 % 5 % 10 %Two tailed test [Zα ] =2.58[Zα ] = 1.96 [Zα ] = 1.645Right tailedtestZα = 2.33 Zα = 1.645 Zα= 1.28Left tailed test Zα = -2.33Zα = -1.645 Zα = - 1.28
- 25. S.No. Confidence level(1- α )Value of confidencecoefficient Zα ( twotailed test)1 90 % 1.642 95 % 1.963 98 % 2.334 99 % 2.585 Without anyreference toconfidence level3.006α is level of significance which separatesacceptance & rejection level
- 26. 8. Decision :1. if mod .Z < Zαaccept Null Hypothesistest statistic falls in the region ofacceptance2. if mod .Z > Zαreject Null Hypothesis
- 27. Q1.Given a normal distribution with mean 60 &standard deviation 10 , find the probability that xlies between 40 & 74Given μ= 60 , σ =10P ( 40 < x < 74 ) = P ( -2 < z< 1.4 ) = P ( -2< z< 0 ) + P ( 0 < z < 1.4 )= 0.4772+ 0.4192 = 0. 8964
- 28. Q2.In a project estimated time ofcompletion is 35 weeks. Standarddeviation of 3 activities in critical pathsare 4 , 4 & 2 respectively . calculate theprobability of completing the project ina. 30 weeks , b. 40 weeks and c. 42weeks
- 29. Test of significanceMeanNull –there is no significance difference betweensample mean & population mean orThe sample has been drawn from the parentpopulationDeviation from mean xbar- μZ = --------------------------- = --------------Standard Error Standard Errorxbar = sample meanμ = population mean
- 30. 1. Standard Error of mean = σ / √ nWhen population standard deviation is knownσ = standard deviation of the populationn = sample size2. Standard Error of mean = s / √ nWhen sample standard deviation is knowns = standard deviation of the samplen = sample size
- 31. ProprtionNull –there is no significance difference betweensample proportion & population proportion orThe sample has been drawn from a populationwith population proportion PNull hypothesis H0 : P = P0 where P0 isparticular value of PAlternate hypothesis H1 : P ≠ P0 ( i.e. P > P0 orP < P0 )
- 32. P*(1-P)Standard error of proportion (S.E.(p)) = √ ------------nDeviation from proprtion p-PTest statistic Z = --------------------------- = --------Standard Error (p) S.E.(p)
- 33. Q3. a sample of size 400 was drawn andsample mean was 99. test whether thissample could have come from a normalpopulation with mean 100 & standarddeviation 8 at 5 % level of significance
- 34. Ans. Given xbar = 99 , n = 400 μ = 100 , σ = 81.Null hypothesis sample has come from a normalpopulation with mean = 100 & s.d. = 8Null hypothesis H0 : μ = 100Alternate hypothesis H1 : μ ≠ μ0 ( i.e. μ > μ0 or μ <μ0 )Two tail test so out of 5% , 2.5 % on each side ( lefthand & right hand)2.calculation of Test statisticStandard error (s.e.) of xbar = σ / √ n = 8/√ 400 =8/20= 2/5xbar- μ 99-100Test statistic Z = ----------- = -------- = -5/2 =- 2.5S.E. 2/5
- 35. Mod z = 2.53. level of significance 5 % i.e.value of α = .05( hence , level of confidence = 1- α = 1-0.05 = 0.95or 95%)4. Critical value (since it is Two tail test so out of5% , we take two tails on each side i.e.2.5 % oneach side = 0.025 ( left hand & right hand)= read from z –table value corresponding to area= 0.5 -0.025 = 0.4750( 0.4750 on both sides i.e. 2* 0.4750 = 0.95 areawhich means 95% confidence)= value of z corresponding to area is 1.96
- 36. 5. Decision – since mod value of z is more thancritical valueNull Hypothesis is rejected & alternate hypothesisis acceptedSample has not been drawn from a normalpopulation with mean 100 &s.d. 8
- 37. Q4. the mean life time of a sample of 400fluorescent light tube produced by acompany is found to be 1570 hours with astandard deviation of 150 hrs. test thehypothesis that the mean life time of thebulbs produced by the company is 1600 hrsagainst the alternative hypothesis that it isgreater than 1600 hrs at 1 &% level ofsignificance
- 38. Ans. Given xbar = 1570 , n = 400 μ = 1600, standard deviation of sample mean s= 150= 8Null hypothesis : mean life time of bulbs is1600 hrsi.e. Null hypothesis H0 : μ = 1600Alternate hypothesis H1 : μ > 1600i.e. it is a case of right tailed test2.calculation of Test statisticStandard error (s.e.) of xbar = s / √ n150/√ 400 = 150/20= 7.5
- 39. xbar- m 1570-1600Z = ----------- = -------------- = -30/7.5= - 4S.E. 7.5Mod z = 43. level of significance 1 % i.e.value of α = .01( hence , level of confidence = 1- α = 1-0.01 =0.99 or 99%)
- 40. 4. Critical value (since it is right tail test so out of 1%,we take 1 % only one side= 0.01 on right hand)= read from z –table value corresponding to area =0.5 -0.01 = 0.4900 ( 0.49 on one side together with+0.5 makes total 0.99 which means 99% confidence)= value of z ( corresponding to area 0 .49 is 2.335. Decision – since mod value of z is more thancritical valueNull Hypothesis is rejected & alternate hypothesis isacceptedHence mean life time of bulbs is greater than 1600hrs
- 41. Q5.in a sample of 400 burners there were 12whose internal diameters were not withintolerance . Is this sufficient to conclude thatmanufacturing process is turning out more than2 % defective burners. Take α = .05
- 42. Given P= 0.002 Q= 1-P =1-0.02 =0. 98& p= 12/400 = 0.03Null hypothesis H0 : P = process is under controlP ≤ 0.02Alternate hypothesis H1 : P > 0.02Left tail testCalculation of Standard error of proportionP*(1-P) 0.02*0.98(S.E.(p)) = √ ------------ = √--------------n 400
- 43. Deviation 0.03-0.02 0.001Z = ----------- = ------------ = -------- = 1.429S.E.(p) √ (0.02*0.98) / 400 0.0073. level of significance 5 % i.e.value of α = .05( hence , level of confidence = 1- α = 1-0.01 = 0.95or 95%)
- 44. 4. Critical value (since it is left tail test so out of5% , we take full 5 % only one side = 0.05 on lefthand) = read from z –table value corresponding toarea = 0.5 -0.05 = 0.4500 ( 0.45 on one sidetogether with +0.5 makes total 0.95 which means95% confidence) = value of z ( corresponding toarea 0 .45 is 1.6455. Decision – since mod value of z is less thancritical valueNull Hypothesis is acceptedHence process is not out of control
- 45. Q6. a manufacturer claimed that at least 95 %of the equipment which he supplied isconforming to specifications. A examination ofsample of 200 pieces of equipment revealedthat 18 were faulty. Test his claim at level ofsignificance i.) 0.05 ii.) 0.01
- 46. Given P= 0.95 Q= 1-P =1-0.95 =0. 05n= 200p= 18/200 = - (200-18) / 200 = 182/200 =0.91Null hypothesis H0 : P = process is undercontrol P = 0.95Alternate hypothesis H1 : P < 0.95Left tail test
- 47. P*(1-P) 0.95*0.05S.E.(p) = √ ------------ = √---------n2000.91-0.95 -0.04Z = ------------ = --------- = -2.6√ (0.02*0.98) / 200 0.0154
- 48. 3a. level of significance 5 % i.e.value of α = .05( hence , level of confidence = 1- α = 1-0.05 = 0.95or 95%)4a. Critical value (since it is right tail test so out of5% , full 5 % on one side = 0.05 on right hand) =read from z –table value corresponding to area = 0.5-0.05 = 0.4500 ( 0.45 on one side together with +0.5makes total 0.95 which means 95% confidence) =value of z ( corresponding to area 0 .45 is 1.645)5a. Decision – since mod value of z is more thancritical valueNull Hypothesis is rejectedManufacturer’s claim is rejected at 5 % level ofsignificance
- 49. 3b. level of significance 1 % i.e.value of α = .01( hence , level of confidence = 1- α = 1-0.01 =0.99 or 99%)4b. Critical value (since it is right tail test so outof 1% , we take 1 % only one side = 0.01 on righthand) = read from z –table value correspondingto area = 0.5 -0.01 = 0.4900 ( 0.49 on one sidetogether with +0.5 makes total 0.99 which means99% confidence) = value of z ( correspondingto area 0 .49 is 2.33
- 50. 5b. Decision – since mod value of z ismore than critical valueNull Hypothesis is rejectedManufacturer’s claim is rejected at 1 %level of significance

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