Bài giảng Statistical Techniques in Business and Economics - Chapter 11 Statistical Inference

Chapter 11Statistical Inference: 1.Explain the the difference between dependent and independent samples.2.Conduct a test of hypothesis and obtain a confidence interval estimate for the difference between Chapter GoalsWhen you have completed this chapter, you will be able to: two population means using matched pair sample. two population proportions. two population means using independent samples.State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4Step 5 Hypothesis Testing Do not reject H0Reject H0 and accept H1Compute the value of the test statistic and make a decisionRecallIf both samples contain at least 30 observations we use the z distribution as the test statistic.Comparing Two PopulationsWe wish to know whether the the distribution of the differences in sample means has a mean of 0.Comparing Two PopulationsThe samples are from independent populations.The formula for computing the value of z is: No assumptions about the of the populations are required.shapez2221n2sn1s+=1X-2XTwo colleges are located in York Region. The local paper recently reported that the mean starting salary of a graduate from College 1 is $38,000 with a standard deviation of $6,000 for a sample of 40 graduates. QuestionThe same article reported the mean starting salary of a graduate from College 2 is $35,000 with a standard deviation of $7,000 for a sample of 35 graduates. At the .01 significance level can we conclude the mean salary of College 1 is more? Solve Hypothesis Test State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4Compute the test statistic and make a decisionStep 5H0: à1 = à2 H1: à1 > à2  = 0.01Because both samples are more than 30, the test statistic is ZReject H0 if z > 2.33 Do not reject the null hypothesis; insufficient evidence. Conclusion:22212121nsnsXXz+-= = 1.98$38,000 - $35,00035)000 ,7($40000),6($22+=zState the decision ruleStep 4The p-value is:alternatep(z > 1.98) == .0239.4761.5000 -Look up in TableSince p of .0239 is greater than the chosen alpha of 0.01, there is insufficient evidence to reject H0 1.980Small Sample Tests of MeansIf one or more of the samples contain less than 30 observations we use the t distribution as the test statistic.Required Assumptions:Both populations must follow the normal distribution.The populations must have equal standard deviations.The samples are from independent populations. Pool the sample variances Determine the value of T from the formulaFinding the value of the test statisticTwo steps needed:Small Sample Tests of Meansữữứửỗỗốổ+-=2112n1ns1XTp2X2)1()1(21222211-+-+-nnsnsn2=sp A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. Question A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the .05 significance level can the EPA conclude that the mpg is higher on the imported cars? Solve Hypothesis Test State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4Compute the test statistic and make a decisionStep 5H0: àd = à1 H1: àd 2.365There are 7 d.f.Solution continuesH1: àD = 01 TDnsD=/ DD2 Hypothesis Test Compute the test statistic and make a decisionStep 5CityHertz ($)Avis ($)Halifax$42$40Quebec City$56$52Kingston4543Toronto4848Winnipeg3732Calgary4548Saskatoon4139Vancouver465024416205-3 2-440259 416Solution continuesTDnsD=/ Hypothesis Test DD224205-3 2-440259 416nDDS=80.8=00.1=-1-nnS2 D=sDS2 D= 3.1623 7818-882-=81623.300.1=T = 0.894Compute the test statistic and make a decisionStep 5Because 0.894 1.645Solution continuesH1: Pu > Pm 2121ˆnnXXp++=H0: pu pm = Hypothesis Test Compute the test statistic and make a decisionStep 5ỳỷựờởộ+211nn-)ˆ1(ˆpp-21ˆˆpp=z2121ˆnnXXp++=.1036=250+ 30022 + 35=250)1036.1(1036.300)1036.1(.1036-+-2502230035-== 1.10 Conclusion:= .5000 - .3643= .1457The p-value is: P(z > 1.10) Hypothesis Test The null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers.Conclusion:Test 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 11