IntroductionApplications, Definations and NotationConfidence IntervalsComputationz-Confidence Intervalst-Confidence IntervalsSample Size DeterminationAn ExampleComparing Two PopulationsComparing Means of Two Independent PopulationsComparing Means of Two Dependent GroupsConfidence Interval for ProportionConfidence Interval for Population ProportionSample Size DeterminationChapter 7Confidence IntervalsJ.C. WangGoal and ObjectivesGoal: to learn confidence intervalsObjectives:ITo understand that each interval has two end-points (lowerand upper bound) and Interpret the confidence intervalITo compute the confidence interval: a point estimate ± themargin of errorITo determine the sample sizeOutlineIntroductionApplications, Definations and NotationConfidence IntervalsComputationz-Confidence Intervalst-Confidence IntervalsSample Size DeterminationAn ExampleComparing Two PopulationsComparing Means of Two Independent PopulationsComparing Means of Two Dependent GroupsConfidence Interval for ProportionConfidence Interval for Population ProportionSample Size DeterminationApplications of Estimation in BusinessexamplesIStore inventory valueIManufacture processIDistribution processIDrug deliveryIAuditorDefinitionsISample statistic: a value computed from the sample (i.e.,from data).IPoint estimate (pt.est): a single sample statistic thatestimates the population parameter such as the mean orproportion.IInterval estimate of the true population parameter takesinto account the sampling distribution of the point estimatewhere we have an upper bound and a lower bound.Notationsto be discussed and used laterICI — confidence intervalICVal — critical valueIME — margin of errorISE — standard errorISD — standard deviationIpt.est. — point estimateIZα/2— normal distribution critical value (use invnorm)Itn−1— students t distribution critical value with n − 1degrees of freedom (use math solver or the invT)Computation ofconfidence intervalsIpt.est ± MEIWhere the point estimate estimates population mean µ (byx) or population proportion p (byˆp)ImarginOfError = criticalValue × standardErrorIIn other words, ME = (CVal)(SE).iClicker Question 7.1 pre-lectureiClicker Question 7.1 pre-lectureStandard ErrorIMost of the time we will not have the SD of populationmean, but we can compute sample SE of the mean:SEx=s√nIAlso, we will not have the SD of population proportion, butwe can compute sample proportion SE:SEˆp=rˆp(1 −ˆp)nCritical ValueIz for normal distributionIt for students t-distributionIThe students t-distribution has n − 1 degrees of freedom,df = n − 1z-Critical ValueINotation: zα/2= upper (100 × α/2)th standard normalpercentileIThat is: P(Z > zα/2) = α/2 ⇐⇒ P(Z ≤ zα/2) = 1 − α/2So, zα/2= invNorm(1 − α/2)IExample 95% confidence interval will give 2.5% in each tailof the bell-shaped curve; therefore, the z-CVal,zcv= z.025= invNorm(1−.025) = invNorm(.975) = 1.96.z-Critical Valuecontinued1 −− ααzαα2αα2αα2area to the left of zαα2 is 1 −−αα2t-Critical Valueusing TI calculators1. math −→ solver −→ tcdf(L, U, D) − A/T , whereIL = tcv(to be solved)IU = 9999 (i.e., U = ∞)ID = df = n − 1IA = α (error rate)IT = number of tails = 2 for c.i.2. or use invT(1 − α/2, df )Cereal Box Packaging ExampleConsider a cereal packaging plant in Battle Creek that isconcerned with putting 368 gram of cereal into a box.IWhat are the costs associated with putting too much cerealin a box?IWhat are the costs associated with putting too little cerealin a box?ILet’s construct a 95% confidence interval.Cereal Box Packaging ExamplecontinuedISuppose sample size n = 25ISuppose sample average x = 365 gramsISuppose SD is a process SD; therefore, σ = 15 gramsISuppose we want a 95% confidence intervalITherefore, the critical value is zcv= 1.96Cereal Box Packaging Examplecontinued, margin of errorIRecall ME = CVal × SEIThe critical value (CVal) for 95% CI means that the areaunder the curve of one tail is 5% ÷ 2 or 0.025; therefore,zcv= invNorm(1 − .025) = invNorm(.975) = 1.96ISE =σ√n=15√25= 3IME = 1.96 × 3 = 5.88Cereal Box Packaging Examplecontinued, confidence intervalISince the confidence interval is the pt.est ± MEICI = 365 ± 5.88 = (359.12, 370.88).ITherefore, we are 95% confident that the population meanis between 359 and 371.ISince 368, the value that is printed on the box indicates themanufacturing process is working properly (is within theinterval), there is no reason to conclude that anything iswrong with the process.z-Confidence Interval Using TI CalculatorsexampleLet’s use TI calculator:IDo this: STAT → TESTS → Zinterval → STATS↓ σ:15 ↓ x:365 ↓ n:25 ↓ C-Level:.95 ↓ CALCULATEIREADOUT:Zinterval(359.12, 370.88)x = 365n = 25Since 368, the target of the package, is within the interval;production should continue.Note on z-Confidence IntervalsIThe value of z selected for constructing such a confidenceinterval is called the critical value for the distribution.IThere are different critical values for each level ofconfidence (or confidence level, CL), 1 − α, where α =significance level, SL (or error rate).IFrequently Used zcv:SL CL 2-tailed CVal10% 90% 1.6455% 95% 1.961% 99% 2.58Note: There is a trade off between the width of the confidenceinterval and the level of confidence.Problem When SD is UnknownWe have been dealing with N(µ, σ) where σ (population orprocess SD) is known. What happens when standard deviation(σ) is not from a population or process SD? Is this requirementrigid? Can we compute standard deviation from the sample?Let us review some history first.History of the Student t DistributionWilliam Gosset, an employee of Guinness Breweries in Ireland,had a preoccupation with making statistical inferences aboutthe mean when SD was unknown. Since the employees of thecompany were not allowed to publish their scientific work undertheir own name. He chose the pseudonym “Student.”Therefore, his contribution is still known as Student’st-Distribution.Comparing Standard Normal Curvewith t curves−3 −2 −1 0 1 2 3 40.00.10.20.30.4Comparison of Standard Normalwith t CurvesxdensityN(0,1)t1t5t10t-Confidence Interval for the Meansummer II quiz exampleConstruct a 95% CI for the mean score for Summer II QuizData of 14 studentsGiven: 95% CL,x = 25, s = 10.777, n = 14,SE =10.777√14= 2.8803CVal = tα/2,n−1= t.025,13= invT(1 − .025, 13) = 2.1604ME = 2.1604 × 2.8803 = 6.2225pt.est = 2595%CI = pt.est ± ME =