Bài giảng Statistical Techniques in Business and Economics - Chapter 12 Analysis of Variance

Chapter 12Analysis of Variance1.Discuss the general idea of analysis of variance.2.List the characteristics of the F distribution.Chapter GoalsWhen you have completed this chapter, you will be able to:Organize data into a one-way and a two-way ANOVA table.3.Conduct a test of hypothesis to determine whether the variances of two populations are equal. 4.and...5.Define the terms treatments and blocks.6.Conduct a test of hypothesis to determine whether three or more treatment means are equal.Chapter Goals7.Develop multiple tests for difference between each pair of treatment means. Characteristics of the F-Distribution There is a “family of F-Distributions:Each member of the family is determined by two parameters: the numerator degrees of freedom, and the denominator degrees of freedomF cannot be negative, and it is a continuous distributionThe F distribution is positively skewedIts values range from 0 to  as F  , the curve approaches the X-axis Test for Equal VariancesFor the two tailed test, the test statistic is given by:The null hypothesis is rejected if the computed value of the test statistic is greater than the critical valuewhereand are the sample variances for the two samples21s22sColin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent. QuestionThe mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the .05 significance level, can Colin conclude that there is more variation in the internet stocks?SolveDo not reject H0Reject H0 and accept H1State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4Step 5 Hypothesis Testing Compute the value of the test statistic and make a decisionRecall Hypothesis Test State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4 = 0.05The test statistic is the F distributionState the decision ruleStep 4Compute the test statistic and make a decisionStep 5Reject H0 if F > 3.68 The df are 9 in the numerator and 7 in the denominator. Do not reject the null hypothesis; there is insufficient evidence to show more variation in the internet stocks. Conclusion: = 1.2416F 2221ss==22)5.3()9.3(220:U IHssÊU221:IHs>sThis this technique is called analysis of variance or ANOVAThe F distribution is also used for testing whether two or more sample means came from the same or equal populationsUnderlying Assumptions for ANOVAthe populations have equal standard deviationsthe samples are randomly selected and are independentthe sampled populations follow the normal distributionANOVA requires the following conditionsthat The Null Hypothesis (H0) is that the population means are the same The Alternative Hypothesis (H1) is that at least one of the means is differentANOVA Procedure The Test Statistic is the F distribution The Decision rule is to reject H0 if F(computed) is greater than F(table) with numerator and denominator dfAnalysis of VarianceTerminologyTotal Variationis the sum of the squared differences between each observation and the overall meanRandom Variationis the sum of the squared differences between each observation and its treatment meanTreatment Variation is the sum of the squared differences between each treatment mean and the overall meanAnalysis of VarianceSSESSTF()kn-=()k-1Procedure If there are a total of n observations the denominator degrees of freedom is n - k The test statistic is computed by: If there k populations being sampled, the numerator degrees of freedom is k – 1 SS Total is the total sum of squaresAnalysis of VarianceProcedurenXX22)(TotalSSS-S= SST is the treatment sum of squares Analysis of VarianceProcedure()nXnTSSTcc22S-ữữứửỗỗốổS= TC is the column total, nc is the number of observations in each column, X the sum of all the observations, and n the total number of observationsSSE is the sum of squares errorAnalysis of VarianceProcedureSST - totalSS SSE=Analysis of VarianceProcedureEasy Meals Restaurants specialize in meals for senior citizens. Katy Smith, President, recently developed a new meat loaf dinner. Before making it a part of the regular menu she decides to test it in several of her restaurants. She would like to know if there is a difference in the mean number of dinners sold per day at the Aynor, Loris, and Lander restaurants. Use the .05 significance level. Example Aynor Loris Lander 13 10 18 12 12 16 14 13 17 12 11 17 17Tc 51 46 85nc 4 4 5Analysis of VarianceProcedureExamplecontinuedAnalysis of VarianceProcedureExamplecontinued SS Total (is the total sum of squares) = 8613= 2634 -)( TotalSS22S-S=nXX(182)2Analysis of VarianceProcedureExamplecontinued()nXnTSSTcc22S-ữữứửỗỗốổS=SST is the treatment sum of squares()()()= 76.2513)182(5 85446451222 2-ữữứửỗỗốổ++=SSE is the sum of squares errorAnalysis of VarianceProcedureExamplecontinuedSSE = SS Total - SST 86 – 76.25= 9.75 Hypothesis Test State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4 = 0.05The test statistic is the F distributionState the decision ruleStep 4Compute the test statistic and make a decisionStep 5Reject H0 if F > 4.10 The df are 2 in the numerator and 10 in the denominator. = 39.101:H0:H1m2m==3mTreatment means are not all equalSSESSTF()kn-=()k-1 109.75 2=76.25Decision The decision is to reject the null hypothesis The treatment means are not the same The mean number of meals sold at the three locations is not the sameAnalysis of VarianceProcedureExamplecontinuedConclusion:Analysis of VarianceSource DF SS MS F PFactor 2 76.250 38.125 39.10 0.000 Error 10 9.750 0.975 Total 12 86.000Individual 95% CIs For Mean Based on Pooled St.DevLevel N Mean St.Dev ---------+---------+---------+-------Aynor 4 12.750 0.957 (---*---) Loris 4 11.500 1.291 (---*---) Lander 5 17.000 0.707 (---*---) ---------+---------+---------+-------Pooled St.Dev = 0.987 12.5 15.0 17.5ANOVA Tablefrom the Minitab system Analysis of Variance in ExcelSeeUsingClick on DATA ANALYSISSeeClick on ToolsHighlight ANOVA: SINGLE FACTORClick OKUsingSeeSeeUsing INPUT NEEDSA1:C6SeeClick on OKSeeInput the sample data in Columns A, B, C.UsingSS TotalSSTSSEF testInferences About Treatment MeansWhen we reject the null hypothesis that the means are equal, we may want to know which treatment means differ One of the simplest procedures is through the use of confidence intervals Inferences About Treatment MeansConfidence IntervalConfidence Interval for the Difference Between Two Meanswhere t is obtained from the t table with degrees of freedom (n - k).MSE = [SSE/(n - k)]()XX12-t±MSEnn1211+ổốỗửứữDevelop a 95% confidence interval for the difference in the mean number of meat loaf dinners sold in Lander and Aynor. Can Katy conclude that there is a difference between the two restaurants?Confidence Interval for the Difference Between Two MeansExampleSolveMSE()XX12-t±MSEnn1211+ổốỗửứữ(17-12.75) 2.228±.9751415+ổốỗửứữ..425148±ị (2.77,5.73)Confidence Interval for the Difference Between Two MeansBecause zero is not in the interval, we conclude that this pair of means differsThe mean number of meals sold in Aynor is different from LanderConfidence Interval for the Difference Between Two MeansExamplecontinuedFor the two-factor ANOVA we test whether there is a significant difference between the treatment effect and whether there is a difference in the blocking effect!Let Br be the block totals (r for rows)Let SSB represent the sum of squares for the blocks ANOVATwo-FactorSSBBkXnr=ộởờựỷỳ-SS22()The Bieber Manufacturing Co. operates 24 hours a day, five days a week. The workers rotate shifts each week. Todd Bieber, the owner, is interested in whether there is a difference in the number of units produced when the employees work on various shifts. A sample of five workers is selected and their output recorded on each shift. At the .05 significance level, can we conclude there is a difference in the mean production by shift and in the mean production by employee?ANOVATwo-FactorExampleANOVATwo-FactorEmployeeDayOutputEveningOutputNightOutputMcCartney312535Neary332633Schoen282430Thompson302928Wagner282627Examplecontinued Hypothesis Test State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4 = 0.05The test statistic is the F distributionState the decision ruleStep 4Compute the test statistic and make a decisionStep 5Reject H0 if F > 4.46. The df are 2 and 8 1:H0:H1m2m==3mNot all means are equal=()())1)(1(1---bkSSEkSSTFDifference between various shifts?SolveCompute the various sum of squares: SS(total) = 139.73 SST = 62.53 SSB = 33.73 SSE = 43.47 df(block) = 4, df(treatment) = 2 df(error)=8ANOVATwo-FactorExamplecontinuedUsing to get these results Since 5.754 > 4.46, H0 is rejected. ()()()()151343.471353.62---=ANOVATwo-FactorExamplecontinuedStep 5Conclusion: There is a difference in the mean number of units produced on the different shifts.=()())1)(1(1---bkSSEkSSTF= 5.754 Hypothesis Test State the null and alternate hypothesesStep 1Select the level of significanceStep 2Identify the test statisticStep 3State the decision ruleStep 4 = 0.05The test statistic is the F distributionState the decision ruleStep 4Compute the test statistic and make a decisionStep 51:H0:H1m2m==3mNot all means are equal=()())1)(1(1---bkSSEkSSTFDifference between various shifts?SolveReject H0 if F > 3.84 The df are 4 and 8 ANOVATwo-FactorExamplecontinuedStep 5=()())1)(1(1---bkSSEkSSTF Since 1.55 < 3.84, H0 is not rejected. = 1.55()()4243.47433.73=Conclusion: There is no significant difference in the mean number of units produced by the various employees.Units versus Worker, ShiftAnalysis of Variance for Units Source DF SS MS F PWorker 4 33.73 8.43 1.55 0.276Shift 2 62.53 31.27 5.75 0.028 Error 8 43.47 5.43Total 14 139.73from the Minitab systemANOVATwo-FactorUsingSeeHighlight ANOVA: TWO FACTOR WITHOUT REPLICATIONClick OKSelect INPUT DATASS TotalSSTSSESSBFtestFcriticalUsingSince F(test) < F(critical), there is not sufficient evidence to reject H0Conclusion:There is no significant difference in the average number of units produced by the different employees.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 12