ANOVA makes assumptions about error for significance tests What are the assumptions What might happen why would it be a problem if the assumption of normality equality of error independence of error turned out to be false ID: 1047188
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1. Introduction to ANOVA
2. QuestionsANOVA makes assumptions about error for significance tests. What are the assumptions?What might happen (why would it be a problem) if the assumption of {normality, equality of error, independence of error} turned out to be false?What is an expected mean square? Why is it important?Why do we use the F test to decide whether means are equal in ANOVA?
3. Questions (2)Correctly interpret ANOVA summary tables.Find correct values of critical F from tabled values for a given test.Suppose someone has worked out that a one-way ANOVA with 6 levels has a power of .80 for the overall F test. What does this mean?Describe (make up) a concrete example of a one-way ANOVA where it makes sense to use an overall F test. Explain why ANOVA (not t, chi-square or something else) is the best method for the analysis.
4. ANOVA AssumptionsRecall we can partition total SS into between (treatment) and within (error) SS. No assumptions needed.To conduct tests about population effects, have to make assumptions:Within cells (treatments) error is normal.Homogeneity of error variance.Independent errors.
5. AssumptionsNormality – sampling distribution of means,variances; not bad if N is large; e.g. reaction timeHomogeneity – pooled estimate of population value. Where are means different? Assumed equal error for each. E.g., ceiling effects in training.Independence – sampling distribution again; e.g., cheating on exam, nesting (schools, labs)
6. Mean Square Between GroupsMean square = SS/df = = variance estimate.MS between =E(MS between) =If there is no treatment effect, MS between = error variance.If there is a treatment effect, MS between is bigger than error variance. (J treatments)
7. Mean Square Within GroupsMS within =E(MS within) =Expected mean square for error is . Expected mean square for treatment is same plus treatment effect: .When there is no treatment effect, between and within estimate same thing. (N is total sample size and J is number of groups.)
8. ReviewANOVA makes assumptions about error for significance tests. What are the assumptions?What might happen (why would it be a problem) if the assumption of {normality, equality of error, independence of error} turned out to be false?What is an expected mean square? Why is it important?
9. The F Test (1)SupposeThe null is equivalent to: If the null is true, then for all jfor some jThe ratio of the two variance estimates will be distributed as F with J-1 and N-J degrees of freedom. Note that the p-value will be a statement of conditional probability (if the null is true, then…)
10. The F Test (2)This is a big deal because we can use variance estimates to test the hypothesis that any number of population means are equal. Equality of means is same as testing population treatment effect(s).For a treatment effect to be detected, F must be larger than 1. F is one-tailed in the tables which show upper tail values of F given the two df.
11. F Table – Critical ValuesNumerator df: dfB dfW123455 5% 1%6.6116.35.7913.35.4112.15.1911.45.0511.010 5% 1%4.9610.04.107.563.716.553.485.993.335.6412 5% 1%4.759.333.896.943.495.953.265.413.115.0614 5% 1%4.608.863.746.513.345.563.115.042.964.70
12. ReviewWhy do we use the F test to decide whether means are equal in ANOVA?Suppose we have an ANOVA design with 3 cells and 5 people per cell. What is the critical value of F at alpha = .05?
13. Calculating F – 1 Way ANOVASums of squares (squared deviations from the mean) tell the story of variance. The simple ANOVA designs have 3 sums of squares.The total sum of squares comes from the distance of all the scores from the grand mean. This is the total; it’s all you have.The within-group or within-cell sum of squares comes from the distance of the observations to the cell means. This indicates error.The between-cells or between-groups sum of squares tells of the distance of the cell means from the grand mean. This indicates IV effects.
14. Computational Example: Caffeine on Test ScoresG1: ControlG2: Mild G3: JoltTest Scores758070778272798474818676838878Means798474SDs (N-1)3.163.163.16
15. G1757916Control77794M=7979790SD=3.1681794837916G280791M=8482799SD=3.16847925867949887981G3707981M=74727949SD=3.167479257679978791Sum370Total Sum of Squares
16. G1757916Control77794M=7979790SD=3.1681794837916G2808416M=8482844SD=3.168484086844888416G3707416M=7472744SD=3.167474076744787416Sum120Within Sum of Squares
17. G179790Control79790M=7979790SD=3.167979079790G2847925M=84847925SD=3.16847925847925847925G3747925M=74747925SD=3.16747925747925747925Sum250Between Sum of Squares
18. SourceSSdfMSFBetween Groups250J-1=3-1=2SS/df250/2=125 =MSBF = MSB/MSW = 125/10=12.5Within Groups120N-J=15-3=12120/12 = 10 = MSWTotal370N-1=15-1=14ANOVA Source (Summary) Table
19. ANOVA SummaryCalculate SS (total, between, within)Each SS has associated df to calculate MSF is ratio of MSb to MSwCompare obtained F (12.5) to critical value (3.89). Significant if obtained F is larger than critical.One-tailed test makes sense for F.
20. R code simple ANOVADownload the .R file Caffeine and .csv file from Canvas. Run the problem you saw in the slide show.Compare your results to what you see in the slides.
21. ReviewSuppose we have 4 groups and 10 people per group. We find that SSB = 60 and SSW = 40. Construct an ANOVA summary table and test for significance of the overall effect.
22. ANOVA Descriptive Stats Because SStot = SSb+SSw we can figure proportion of total variance due to treatment.Proportion of total variance due to treatment is:R2= SSb/SStot. Varies from 0 (no effect) to 1 (no error).Sample value is biased (too large).
23. Estimating PowerPower for what? For one-way ANOVA, power usually means for the overall F, i.e., at least 1 group mean is different from the others.Howell uses noncentral F for sample size calculation.Where k is the number of treatment goups; n is sample size per group. Variance of error is MSE in the population (variance of DV within cells). Mu(j) are treatment means; mu is grand mean.
24. SAS Power calculationSAS will compute sample size requirements for a given scenario.You input the expected means and a common (within cell) standard deviation, (along with alpha and desired power) and it will tell you the sample size you need.
25. SAS Inputrun;*********************************************************** Power computation example from Howell, 2010, p. 350.* Note the standard deviation is the square root of the* provided MSE: sqrt(240.35) = ~ 15.5.**********************************************************;proc power ; onewayanova groupmeans = 34 | 50.8 | 60.33 | 48.5 | 38.1 stddev = 15.5 alpha = 0.05 npergroup = . power = .8;
26. SAS OutputThe POWER Procedure Overall F Test for One-Way ANOVA Fixed Scenario Elements Method Exact Alpha 0.05 Group Means 34 50.8 60.33 48.5 38.1 Standard Deviation 15.5 Nominal Power 0.8 Computed N Per Group Actual N Per Power Group 0.831 8
27. ReviewSuppose someone has worked out that a one-way ANOVA with 6 levels has a power of .80 for the overall F test. What does this mean?Describe (make up) a concrete example of a one-way ANOVA where it makes sense to use an overall F test. Explain why ANOVA (not t, chi-square or something else) is the best method for the analysis.