商业决策技术课件-07Quantitative-Methods-For-Decision-Maker.ppt

1、Quantitative Methods For Decision Makers3rd EditionChapter 7MARKET RESEARCH AND STATISTICAL INFERENCELearning ObjectivesBy the end of this chapter you should be able to:nUnderstand the difference between a sample and a populationnExplain the principles of a sampling distributionnCalculate and explai

2、n a confidence interval around a sample mean and a sample percentagenCarry out a variety of common hypothesis testsPopulation and sampleTwo key assumptions about the sample:nThe sample is a properly representative subset of the statistical population.nThe data collected is reliable and accurate.Samp

3、ling distributionnA sampling distribution is a distribution of all of the sample means selected from a population.nDifferent samples taken from the same population would generate similar,but not necessarily arithmetically identical,sample means.The central limit theoremIf we take random samples of s

4、ize n from a population,the distribution of sample means will approach that of the Normal probability distribution.This approximation will become closer,the larger is n.nCentral Limit TheoremAs the sample size gets large enough the sampling distribution becomes almost normal regardless of shape of p

5、opulationxThe sampling distributionSample mean valuesMean of all sample meansnIf the data distribution is Normal,then the interval:ncontains about 68%of the values in the populationThe Empirical Rule 1 68%1ncontains about 95%of the values in the populationncontains about 99.7%of the values in the po

6、pulationThe Empirical Rule 2 3 399.7%95%2Characteristics of the sampling distribution Mean SDSample sPopulation Sampling distribution xns/Point and Interval EstimatesnA point estimate is a single number,na confidence interval provides additional information about variabilityPoint EstimateLower Confi

7、dence LimitUpperConfidence LimitWidth of confidence intervalWe can estimate a Population Parameter Point Estimateswith a SampleStatistic(a Point Estimate)MeanProportionpXConfidence IntervalsnHow much uncertainty is associated with a point estimate of a population parameter?nAn interval estimate prov

8、ides more information about a population characteristic than does a point estimatenSuch interval estimates are called confidence intervalsEstimation Process(mean,is unknown)PopulationRandom SampleMean X=50SampleI am 95%confident that is between 40&60.General FormulanThe general formula for all confi

9、dence intervals is:Point Estimate (Critical Value)(Standard Error)Confidence LevelnConfidence LevelnConfidence for which the interval will contain the unknown population parameternA percentage(less than 100%)Finding the Critical Value,ZnConsider a 95%confidence interval:Z=-1.96Z=1.960.9510.02520.025

10、2Point EstimateLower Confidence LimitUpperConfidence LimitZ units:X units:Point Estimate01.96ZCommon Levels of ConfidencenCommonly used confidence levels are 90%,95%,and 99%Confidence LevelConfidence Coefficient,Z value1.281.6451.962.332.583.083.270.800.900.950.980.990.9980.99980%90%95%98%99%99.8%99

11、.9%1Confidence intervals95%CI=1.96xns/nZXCI2/nSZXCI2/ExamplenA sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms.We know from past testing that the population standard deviation is 0.35 ohms.nDetermine a 95%confidence interval for the true mean resistance of the

12、 population.2.4068 1.99320.2068 2.20)11(0.35/1.96 2.20nZ XExamplenA sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms.We know from past testing that the population standard deviation is 0.35 ohms.nSolution:(continued)InterpretationnWe are 95%confident that the t

13、rue mean resistance is between 1.9932 and 2.4068 ohms nAlthough the true mean may or may not be in this interval,95%of intervals formed in this manner will contain the true meanConfidence intervals for proportionsnppZpCI)1(2/nppZpCI)100(2/ExamplenA random sample of 100 people shows that 25 are left-

14、handed.nForm a 95%confidence interval for the true proportion of left-handersExamplenA random sample of 100 people shows that 25 are left-handed.Form a 95%confidence interval for the true proportion of left-handers./1000.25(0.75)1.9625/100p)/np(1Zp0.3349 0.1651 (0.0433)1.96 0.25(continued)Interpreta

15、tionnWe are 95%confident that the percentage of left-handers is between 16.51%and 33.49%.nAlthough the interval from 0.1651 to 0.3349 may or may not contain the true proportion,95%of intervals formed from samples of size 100 in this manner will contain the true proportion.Applications A survey of 61

16、1 office workers investigated telephone answering practices,including how often each office worker was able to answer incoming telephone calls and how often incoming telephone calls went directly to voice mail(USA Today,April 21,2007).A total of 281 office workers indicated that they never need voic

17、e mail and are able to take every telephone call.a.What is the point estimate of the proportion of the population of office workers who are able to take every telephone call?b.What is the 90%confidence interval for the proportion of the population of office workers who are able to take every telepho

18、ne call?What is a Hypothesis?nA hypothesis is a claim(assumption)about a population parameter:npopulation meannpopulation proportionExample:The mean monthly cell phone bill of this city is =$42Example:The proportion of adults in this city with cell phones is =0.68Hypothesis testsSteps:1.Formulate th

19、e null and alternative hypothesis.2.Determine a significance level.3.Identify the rejection area.4.Determine the critical statistical value.5.Calculate the test statistic value.6.Choose between the two hypotheses.The Null Hypothesis,H0nStates the claim or assertion to be testedExample:The average nu

20、mber of TV sets in U.S.Homes is equal to three ()nIs always about a population parameter,not about a sample statistic 3:H03:H03X:H0The Null Hypothesis,H0nBegin with the assumption that the null hypothesis is truenSimilar to the notion of innocent until proven guiltynRefers to the status quonAlways c

21、ontains“=”,“”or“”signnMay or may not be rejected(continued)The Alternative Hypothesis,H1nIs the opposite of the null hypothesisne.g.,The average number of TV sets in U.S.homes is not equal to 3 (H1:3)nChallenges the status quonNever contains the“=”,“”or“”signnMay or may not be provennIs generally th

22、e hypothesis that the researcher is trying to proveFormulate the null and alternative hypothesesnP=63%n=250%55:%55:10HH3 categories of hypothesisxHxHxHxHxHxH:.3:.2:.1101010Lower-tail test:H0:H1:Two-tail test:H0:=H1:/2/2-za-za/2zaza/2Reject H0 if Z ZaReject H0 if Z Za/2 Rejection areaChoose between t

23、he two hypothesesnIf Zcalc Z then we must reject H0nIf Zcalc Z then we cannot reject H0n=0.01 Z=2.33 Zcalc=2.54nReject H0ProportionsnSample proportion in the success category is denoted by pnthe test statistic is a Z value:sizesamplesampleinsuccessesofnumbernXp(continued)n)(1pZExample:Z Test for Pro

24、portion A marketing company claims that it receives 8%responses from its mailing.To test this claim,a random sample of 500 were surveyed with 25 responses.Test at the =0.05 significance level.Z Test for Proportion:Solution =0.05 n=500,p =0.05Reject H0 at =0.05H0:=0.08 H1:0.08Critical Values:1.96Test

25、 Statistic:Decision:Conclusion:z0RejectReject.025.0251.96-2.47There is sufficient evidence to reject the companys claim of 8%response rate.2.47500.08).08(1.08.05n)(1pZ-1.96The test statistic value for a sample mean of a large samplensxZCalcExample A local authority has one of its administrative cent

26、ers located out of town and the majority of employees drive to work.As part of the authoritys green policy various effects have been made to encourage employees to car-share.The authority is trying to assess the impact of these measures on car-sharing.Last year,in a comprehensive exercise,the author

27、ity found that the number of cars parked in the staff car park each day averaged 220.This year based on a sample of 75 days observation,it has calculated that the mean number of cars parked in the staff car park is 205,with a standard deviation of 32.Based on this data,can we comment on the impact o

28、f these measures on car-sharing?What assumptions have you made in assessing this data?The test statistic value for difference between two means of independent populationProfit achieved in region A and B region A region BX 12981.2 17168.8S 17476.3 19688.1n 113 121The test statistic value for differen

29、ce between two means of independent population2221212121)()(nsnsxxZCalcLower-tail test:H0:1 2 0H1:1 2 0Two-tail test:H0:1 2=0H1:1 2 0a aa a/2a a/2a a-za-za/2zaza/2Reject H0 if Z ZaReject H0 if Z Za/2 Hypothesis tests for 1 2 Test on two proportions or percentagesPopulation proportions 212121n1n1)(10

30、0ppZppThe test statistic for p1 p2 is a Z statistic:222111212211nXp ,nXp ,nnpnpnpwhere210:HExample:Two population Proportions Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A?nIn a random sample,36 of 72 men and 31 of 50 w

31、omen indicated they would vote YesnTest at the 0.05 level of significancenThe hypothesis test is:H0:1 2=0 (the two proportions are equal)H1:1 2 0 (there is a significant difference between proportions)nThe sample proportions are:nMen:p1=36/72=.50nWomen:p2=31/50=.62.5491226750723136nnXX2121p The pool

32、ed estimate for the overall proportion is:Example:Two population Proportions(continued)The test statistic is:Example:Two population Proportions(continued).025-1.961.96.025-1.31Decision:Do not reject H0Conclusion:There is not significant evidence of a difference in proportions who will vote yes betwe

33、en men and women.1.31501721.549)(1.5490.62.50n1n1)(1ppz212121PPReject H0Reject H0Critical Values=1.96For =.05Tests on small samplesnConvert sample statistic to a t test statistic(where t is the critical value of the t distribution with n-1 degrees of freedom)The test statistic is:nSXt1-nStudents t D

34、istributiont0t (df=5)t (df=13)t-distributions are bell-shaped and symmetric,but have fatter tails than the normalStandard Normal(t with df=)Note:t Z as n increasest distribution valuesWith comparison to the Z valueConfidence t t t Z Level (10 d.f.)(20 d.f.)(30 d.f.)_ 0.80 1.372 1.325 1.310 1.28 0.90

35、 1.812 1.725 1.697 1.645 0.95 2.228 2.086 2.042 1.96 0.99 3.169 2.845 2.750 2.58Note:t Z as n increasesStudents t DistributionnThe t is a family of distributionsnThe t value depends on degrees of freedom(d.f.)nNumber of observations that are free to vary after sample mean has been calculatedd.f.=n-1

36、If the mean of these three values is 8.0,then X3 must be 9(i.e.,X3 is not free to vary)Degrees of Freedom(df)Here,n=3,so degrees of freedom =n 1=3 1=2(2 values can be any numbers,but the third is not free to vary for a given mean)Idea:Number of observations that are free to vary after sample mean ha

37、s been calculatedExample:Suppose the mean of 3 numbers is 8.0 Let X1=7Let X2=8What is X3?Students t TableUpper Tail Areadf.25.10.0511.000 3.078 6.31420.817 1.886 2.92030.765 1.638 2.353t02.920The body of the table contains t values,not probabilitiesLet:n=3 df=n-1=2 =0.10 /2=0.05/2=0.05Example The av

38、erage cost of a hotel room in New York is said to be$168 per night.A random sample of 25 hotels resulted in X =$172.50 and S=$15.40.Test at the =0.05 level.(Assume the population distribution is normal)H0:=168 H1:168n =0.05n n=25n N30 so use a t statisticn Critical Value:t24=2.0639Example Solution:D

39、o not reject H0:not sufficient evidence that true mean cost is different from$168Reject H0Reject H0/2=.025-t n-1,/2Do not reject H00/2=.025-2.06392.06391.462515.40168172.50nSXt1n1.46H0:=168 H1:168t n-1,/2p-Value Approach to Testingnp-value:Probability of obtaining a test statistic more extreme(or )t

40、han the observed sample value given H0 is truep-Value Approach to TestingnConvert Sample Statistic(e.g.,)to Test Statistic(e.g.,Z statistic)nObtain the p-value from a table or computernCompare the p-value with nIf p-value ,reject H0nIf p-value ,do not reject H0 X(continued)Example:Z Test for Proport

41、ion A marketing company claims that it receives 8%responses from its mailing.To test this claim,a random sample of 500 were surveyed with 25 responses.Test at the =0.05 significance level.Z Test for Proportion:Solution =0.05 n=500,p =0.05Reject H0 at =0.05H0:=0.08 H1:0.08Critical Values:1.96Test Sta

42、tistic:Decision:Conclusion:z0RejectReject.025.0251.96-2.47There is sufficient evidence to reject the companys claim of 8%response rate.2.47500.08).08(1.08.05n)(1pZ-1.96Z Test for Proportion:SolutionnP=P(Z2.47)nP=0.0068+0.0068=0.0136nPthe critical 2 then reject H0.2 TestsnTest calculationsn O E (O-E)

43、(O-E)2/En20 n13n12n12n18n75 Test of association using 2Patient view on appointments systemPatient gender for against no preference totalMale 106 27 20 153Female 97 166 34 297Total 203 193 54 450 Test of association using 2 nThe second use of 2 tests is to test the association of categories,known as

44、a contingency table test.nH0:the two characteristics are independent.nD.o.f=(no of rows-1)(no of columns-1)nIf the calculated 2 the critical 2 then reject H0.Test of association using 2nExpected frequenciesnTable cell O E (O-E)(O-E)2/EnM-F 106 69.02 36.98 19.81nM-A 27nM-NP 20nF-F 97nF-A 166nF-NP 34nTotal 450 nThe expected frequencies for each cell is row total column total/sample total