Bài giảng Statistical Techniques in Business and Economics - Chapter 8 Sampling Methods & Central Limit Theorem

lChapter 8pgnimaSMethods&Central Limit TheoremWhen you have completed this chapter, you will be able to:Chapter GoalsDescribe methods for selecting a sample. Define and construct a sampling distribution of the sample mean.1.2.3. Explain under what conditions sampling is the proper way to learn something about a population.Explain the central limit theorem.4.Use the central limit theorem to find probabilities of selecting possible sample means from a specified population. 5.We use sample information to make decisions or inferences about the population.Two KEY steps:SamplingChoice of a proper method for selecting sample data &2. Proper analysis of the sample data (more later)KEY 1.SamplingSamplingKEY 1.If the proper method for selecting the sample is NOT MADE the SAMPLE will not be truly representative of the TOTAL Population! and wrong conclusions can be drawn! of the physical impossibility of checking all items in the population, and, also, it would be too time-consuming the studying of all the items in a population would NOT be cost effectivethe sample results are usually adequate the destructive nature of certain testsWhy Sample the Population?BecauseSamplingwith ReplacementEach data unit in the population is allowed to appear in the sample more than onceEach data unit in the population is allowed to appear in the sample no more than onceEach data unit in the population has a known likelihood of being included in the sampleNon-Probability SamplingDoes not involve random selection; inclusion of an item is based on convenienceProbability Samplingwithout ReplacementTechniquesMethodsSimple RandomSystematic RandomStratified RandomClusterSampling...each item(person) in the population has an equal chance of being includeditems(people) of the population are arranged in some order. A random starting point is selected, and then every kth member of the population is selected for the samplea population is first divided into subgroups, called strata, and a sample is selected from each strataa population is first divided into primary units, and samples are selected from each unitSamplingTerminology is the difference between a sample statistic and its corresponding population parameter is a probability distribution consisting of all possible sample means of a given sample size selected from a population“Sampling error”“Sampling distribution of the sample mean”ExampleSamplingThe law firm of Hoya and Associates has five partners. At their weekly partners meeting each reported the number of hours they billed their clients last week:ExamplePartnerHoursDunn22Hardy26Kiers30Malinowski26Tillman22If two partners are selected randomlyhow many different samples are possible?SamplingPartnerHoursDunn22Hardy26Kiers30Malinowski26Tillman22Objects5taken 2 at a timeUsing 5C2 for a Total of 10 Samples!If two partners are selected randomlyhow many different samples are possible?SamplingPartnerHoursDunn22Hardy26Kiers30Malinowski26Tillman22Objects55C2 =5!=2!= 10 Samples(5 – 2!)If two partners are selected randomlyhow many different samples are possible?SamplingPartnersSamples of 2Mean1&21&31&41&52&32&42&53&43&54&5(22+26)/2 =(22+30)/2 =(22+26)/2 =242624(22+22)/2 =(26+30)/2 =(26+26)/2 =(26+22)/2 =(30+26)/2 =(30+22)/2 =(26+22)/2 =22282624282624Sample MeanFrequencyRelative frequencyProbabilitySamplingOrganize the sample means into a Sampling DistributionExample continuedMean242624222826242826242224262814321/104/103/102/1010 SamplesSample MeanFrequency10= 22(1)+ 24(4)+ 26(3) + 28(2)SamplingExample continued222426281432Compute the mean of the sample means. Compare it with the population mean = 25.2mXSamplingExample continued52226302622++++=mThe population mean is also the same as the sample means25.2 hours!NotePartnerHoursDunn22Hardy26Kiers30Malinowski26Tillman22 = 25.2Sampling The sampling distribution of the means of all possible samples of size ngenerated from the population will be approximately normally distributed!Central Limit TheoremSampling Distributions:µVariance2 /nMean (µx )/ nStandard Deviation  (standard error of the mean)X sample meansample standard deviationsample variancesample proportionA point estimate is one value ( a single point) that is used to estimate a population parameterPoint EstimatesExamplesSamplingMorePoint EstimatesSamplingPopulation follows the normal distributionThe sampling distribution of the sample means also follows the normal distributionProbability of a sample mean falling within a particular region, use:Z =nsXm-Population does NOT follow the normal distributionIf the sample is of at least 30 observations, the sample WILL follow the normal distributionProbability of a sample mean falling within a particular region, use:Z =nXm-sCentral Limit TheoremSamplingChart 8 – 6 Results for Several Populations251139Generating Random Numbers in ExcelSeeUsingClick on ToolsClick on DATA ANALYSISSeeHighlight RANDOM NUMBER GENERATIONClick OKSeeUsingSeeUsing INPUT NEEDS1200100$A:$ASeeClick on OKSeeUsingSeeIf you want whole numbers, use the FUNCTION WIZARD (fx) to ROUND to the nearest integer.UsingClick onSeeHighlight Math & TrigScroll Down find RoundUsingSeeHighlight, and click OKUsingSeeClick on OK INPUT REQUIRED VALUESA10UsingSee66 CLICK on B1 and DRAG to Fill COLUMN B SeeSelecting Simple Random Sample in ExcelUsingInput Data in Column ASelectSeeScroll toSamplingClick OKSelectUsing INPUT REQUIRED VALUES$A:$A10$B:$BClick on OKUsingSince this is random number generation, you will get different numbers each time you do thisSuppose it takes an average of 330 minutes for taxpayers to prepare, copy, and mail an income tax return form. Using the Sampling Distribution of the Sample Mean= 12.6A consumer watchdog agency selects a random sample of 40 taxpayers and finds the standard deviation of the time needed is 80 minutesWhat is the standard error of the mean?Data /nFormula = 80 / 40What is the likelihood the sample mean is greater than 320 minutes?Using the Sampling Distribution of the Sample MeanSuppose it takes an average of 330 minutes for taxpayers to prepare, copy, and mail an income tax return form. A consumer watchdog agency selects a random sample of 40 taxpayers and finds the standard deviation of the time needed is 80 minutes. DataAnswerUsing the Sampling Distribution of the Sample MeanWhat is the likelihood the sample mean is greater than 320 minutes? * average of 330 minutes *random sample of 40 * standard deviation is 80 minutesDatansXzm-=Formula 4080330320-== 0.791330 320 a1Using the Sampling Distribution of the Sample MeanWhat is the likelihood the sample mean is greater than 320 minutes? * average of 330 minutes *random sample of 40 * standard deviation is 80 minutesDataLook up 0.79 in Table2a1 =0.2852Required Area =0.2852 + .5 = 0.7852330 320 a1 Sampling Distribution of ProportionThe normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n.Use when np and n(1- p ) are both greater than 5!np=)1 (p np-=Mean and Varianceof a Binomial Probability Distribution 2 smmFormula 2sFormula A multinational company claims that 55% of its employees are bilingual. To verify this claim, a statistician selected a sample of 60 employees of the company using simple random sampling and found 48% to be bilingual. np = A60(.55) = 33n(1- p ) = 60(.45) = 27The sample size is big enough to use the normal approximation with a mean of .55 and a standard deviation of (.55)(.45)/60 = 0.064 Sampling Distribution of ProportionBased on this information, what can we say about the company’s claim?sXzm-=1Z = (0.48 -0.55) / 0.064Z = -1.09Look up 1.09 in Table2a1 =0.3621Required Area = .5 – 0.3621 = 0.1379 or 14% Sampling Distribution of Proportion continuedAFormula .55 .48 a1sXzm-=1Z = (0.48 -0.55) / 0.064Z = -1.09Look up 1.09 in Table2a =0.3621Required Area = .5 – 0.3621 = 0.1379 or 14%There is approximately a 14% chance that the company’s claim is true, based on this sample. Sampling Distribution of ProportionConclusion continuedAFormula Suppose the mean selling price of a litre of gasoline in Canada is $.659. Further, assume the distribution is positively skewed, with a standard deviation of $0.08. What is the probability of selecting a sample of 35 gasoline stations and finding the sample mean within $.03 of the population mean? Sampling Distribution of Mean Sampling Distribution of MeannszXm-=13508.0$659$.629$.-=22.-2=nszXm-=23508.0$659$.689$.-=2.22 = mean selling price is $.659 SD of $0.08Sample of 35 gasoline stations Probability of sample mean within $.03? Data1Find the z-scores for.659 +/- .03Stepi.e. 0.629 and .689.629 .689We would expect about 97% of the sample means to be within $0.03 of the population mean.a1 = .4868a2 = .4868 Sampling Distribution of Mean mean selling price is $.659 SD of $0.08Sample of 35 gasoline stations Probability of sample mean within $.03? DataFind areas from table2StepRequired A = .9736z= -2.221z= 2.22 2Test your learning www.mcgrawhill.ca/college/lindClick onOnline Learning Centrefor quizzesextra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat dataand much more!This completes Chapter 8