Statistics 303ANOVA: Comparing Several MeansSlide 3Assumptions for ANOVASlide 5Slide 6Slide 7Slide 8ANOVA TableSlide 10Slide 11Side by Side BoxplotsNormal Quantile PlotsSlide 14Slide 15Slide 16Slide 17Slide 18Slide 19Statistics 303Chapter 12ANalysis Of VArianceANOVA: Comparing Several Means•The statistical methodology for comparing several means is called analysis of variance, or ANOVA.•In this case one variable is categorical.–This variable forms the groups to be compared.•The response variable is numeric.•This methodology is the extension of comparing two means.ANOVA: Comparing Several Means•Examples:–“An investigator is interested in studying the average number of days rats live when fed diets that contain different amounts of fat. Three populations were studied, where rats in population 1 were fed a high-fat diet, rats in population 2 were fed a medium-fat diet, and rats in population 3 were fed a low-fat diet. The variable of interest is ‘Days lived.’” (from Graybill, Iyer and Burdick, Applied Statistics, 1998).–“A state regulatory agency is studying the effects of secondhand smoke in the workplace. All companies in the state that employ more than 15 workers must file a report with the agency that describes the company’s smoking policy. In particular, each company must report whether (1) smoking is allowed (no restrictions), (2) smoking is allowed only in restricted areas, or (3) smoking is banned. In order to determine the effect of secondhand smoke, the state agency needs to measure the nicotine level at the work site. It is not possible to measure the nicotine level for every company that reports to the agency, and so a simple random sample of 25 companies is selected from each category of smoking policy.” (from Graybill, Iyer and Burdick, Applied Statistics, 1998).Assumptions for ANOVA 1. Each of the I population or group distributions is normal. -check with a Normal Quantile Plot (or boxplot) of each group 2. These distributions have identical variances (standard deviations).-check if largest sd is > 2 times smallest sd 3. Each of the I samples is a random sample. 4. Each of the I samples is selected independently of one another.ANOVA: Comparing Several MeansThe null hypothesis (step 1) for comparing several means isI 210:Hwhere I is the number of populations to be comparedThe alternative hypothesis (step 2) isothers) thefromdifferent is means theof oneleast (at equal are theof allnot :HiaANOVA: Comparing Several Means•Step 3: State the significance level•Step 4: Calculate the F-statistic:MSEMSGor Error SquaresMean Group SquaresMean FThis compares the variation between groups (group mean to group mean) to the variation within groups (individual values to group means).This is what gives it the name “Analysis of Variance.”ANOVA: Comparing Several Means•Step 5: Find the P-value–The P-value for an ANOVA F-test is always one-sided.–The P-value is)Pr(21, calculateddfdfFF where df1 = I – 1 (number of groups minus 1) and df2 = N – I (total sample size minus number of groups).F-distribution:P-valueANOVA: Comparing Several Means•Step 6. Reject or fail to reject H0 based on the P-value.–If the P-value is less than or equal to , reject H0.–It the P-value is greater than , fail to reject H0.•Step 7. State your conclusion.–If H0 is rejected, “There is significant statistical evidence that at least one of the population means is different from another.” –If H0 is not rejected, “There is not significant statistical evidence that at least one of the population means is different from another.”ANOVA TableSource df Sum of Squares Mean Square F p-valueGroup(between)I – 1Error(within)N – ITotal N – 1 SSGxxnii2)( SSEsnii2)1( SSTotxxij2)(MSGdfGSSGMSEdfESSEMSTdfTotSSTotcalcFMSEMSG)Pr(calcFF Note: MSE is the pooled sample variance and SSG + SSE = SSTot is the proportion of the total variation explained by the difference in meansSSTotSSGR 2ANOVA: Comparing Several Means•Example: “An experimenter is interested in the effect of sleep deprivation on manual dexterity. Thirty-two (N) subjects are selected and randomly divided into four (I) groups of size 8 (ni). After differing amount of sleep deprivation, all subjects are given a series of tasks to perform, each of which requires a high amount of manual dexterity. A score from 0 (poor performance) to 10 (excellent performance) is obtained for each subject. Test at the = 0.05 level the hypothesis that the degree of sleep deprivation has no effect on manual dexterity.” (from Milton, McTeer, and Corbet, Introduction to Statistics, 1997)ANOVA: Comparing Several Means•Information GivenStddev1 = 0.89316Stddev2 = 0.86603Stddev3 = 0.64507Stddev4 = 0.85206Group I Group II Group III Group IV16 hours 20 hours 24 hours 28 hours8.95 7.7 5.99 3.788.04 5.81 6.79 3.357.72 6.61 6.43 2.456.21 6.07 5.85 4.276.48 8.04 5.78 4.877.81 5.96 7.6 3.147.5 7.3 5.78 3.986.9 7.46 6 2.47Sample size: N = 32Variation between groupsVariation within groupsSide by Side BoxplotsGroupI GroupII GroupII I GroupIV2.003.004.005.006.007.008.009.00Normal Quantile PlotsANOVA: Comparing Several Means•Information GivenError Bars show Mean +/- 1.0 SEDot/Lines show Means16 hours 20 hours 24 hours 28 hoursdeprived4.005.006.007.00dexter7.456.876.283.54Variation Within GroupsAverage Within Group Variation (MSE)ANOVA: Comparing Several Means•Information GivenDot/Lines show Means16 hours 20 hours 24 hours 28 hoursdeprived4.005.006.007.00dexterAverage Between Group Variation (MSG)Variation Between GroupsANOVA: Comparing Several MeansStep 1: The null hypothesis is43210:HStep 2: The alternative hypothesis isequal are theof allnot :HiaStep 3: The significance level is = 0.05ANOVA: Comparing Several Means•Step 4: Calculate the F-statistic:MSEMSGor Error SquareMean Group SquareMean FMSG and MSE are found in the ANOVA table when the analysis is run on the computer:ANOVADEXTER71.928 3 23.976 35.730 .00018.789 28 .67190.716 31Between GroupsWithin GroupsTotalSum ofSquares df Mean Square F Sig.MSGMSE0.67123.976 35.73 ANOVA: Comparing Several Means•Step 5: Find the P-value–The P-value is)Pr(21, calculateddfdfFF where df1 = I – 1 (number of groups minus 1) = 4 – 1 = 3 and df2 = N – I (total sample size minus
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