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One-Way ANOVA Independent Samples

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Basic Design Grouping variable with 2 or more levels Continuous dependent/criterion variable H : 1 = 2 =... = k Assumptions –Homogeneity of variance –Normality in each population

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The Model Y ij = + j + e ij, or, Y ij - = j + e ij. The difference between the grand mean ( ) and the DV score of subject number i in group number j is equal to the effect of being in treatment group number j, j, plus error, e ij

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Four Methods of Teaching ANOVA Do these four samples differ enough from each other to reject the null hypothesis that type of instruction has no effect on mean test performance?

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Error Variance Use the sample data to estimate the amount of error variance in the scores. This assumes that you have equal sample sizes. For our data, MSE = (.5 +.5 +.5 +.5) / 4 = 0.5

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Among Groups Variance Assumes equal sample sizes VAR(2,3,7,8) = 26 / 3 MSA = 5 26 / 3 = 43.33 If H is true, this also estimates error variance. If H is false, this estimates error plus treatment variance.

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F F = MSA / MSE If H is true, expect F = error/error = 1. If H is false, expect

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p F = 43.33 /.5 = 86.66. total df in the k samples is N - 1 = 19 treatment df is k – 1 = 3 error df is k(n - 1) = N - k = 16 Using the F tables in our text book, p <.01. One-tailed test of nondirectional hypothesis

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Deviation Method SS TOT = (Y ij - GM) 2 = (1 - 5) 2 + (2 - 5) 2 +...+ (9 - 5) 2 = 138. SS A = [n j (M j - GM) 2 ] SS A = n (M j - GM) 2 with equal n’s = 5[(2 - 5) 2 + (3 - 5) 2 + (7 - 5) 2 + (8 - 5) 2 ] = 130. SSE = (Y ij - M j ) 2 = (1 - 2) 2 + (2 - 2) 2 +.... + (9 - 8) 2 = 8.

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Computational Method = (1 + 4 + 4 +.....+ 81) - [(1 + 2 + 2 +.....+ 9) 2 ] N = 638 - (100) 2 20 = 138. = [10 2 + 15 2 + 35 2 + 40 2 ] 5 - (100) 2 20 = 130. SSE = SS TOT – SS A = 138 - 130 = 8.

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Source Table

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Magnitude of Effect Omega Square is less biased

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Magnitude of Effect Put a confidence interval on eta-squared. http://core.ecu.edu/psyc/wuenschk/SPSS/ SPSS-Programs.htmhttp://core.ecu.edu/psyc/wuenschk/SPSS/ SPSS-Programs.htm

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Magnitude of Effect Enter F and df into NoncF.sav

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Magnitude of Effect Run syntax file NoncF3.sps. CI runs from.837 to.960.

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Pairwise Comparisons and Familywise Error fw is the alpha familywise, the conditional probability of making one or more Type I errors in a family of c comparisons. pc is the alpha per comparison, the criterion used on each individual comparison. Bonferroni: fw c pc

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c = 6, pc =.01 alpha familywise might be as high as 6(.01) =.06. What can we do to lower familywise error?

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Fisher’s Procedure Also called the “Protected Test” or “Fisher’s LSD.” Do ANOVA first. If ANOVA not significant, stop. If ANOVA is significant, make pairwise comparisons with t. For k = 3, this will hold familywise error at the nominal level, but not with k > 3.

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Computing t Assuming homogeneity of variance, use the pooled error term from the ANOVA: For A versus D:

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For A versus C and B versus D: For B versus C For A vs B, and C vs D,

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Underlining Means Display arrange the means in ascending order any two means underlined by the same line are not significantly different from one another GroupABCD Mean2378

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The Bonferroni Procedure Does NOT require that ANOVA be conducted or, if conducted, that ANOVA be significant. Compute an adjusted criterion of significance to keep familywise error at desired level

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For our data, Compare each p with the adjusted criterion. For these data, we get same results as with Fisher’s procedure. In general, this procedure is very conservative (robs us of power).

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Ryan-Einot-Gabriel-Welsch Test Does not require a significant ANOVA. Holds familywise error at the stated level. Has more power than other techniques which also adequately control familywise error. SPSS will do it for you.

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Which Test Should I Use? If k = 3, use Fisher’s Procedure If k > 3, use REGWQ Remember, ANOVA does not have to be significant to use REGWQ.

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APA-Style Presentation

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Teaching method significantly affected test scores, F(3, 16) = 86.66, MSE = 0.50, p <.001, 2 =.942, CI.95 =.837,.960. Pairwise comparisons were made with Bonferroni tests, holding familywise error rate at a maximum of.01. As shown in Table 1, the computer-based and devoted methods produced significantly better student performance than did the ancient and backwards methods.

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Computing ANOVA From Group Means and Variances with Unequal Sample Sizes GM = p j M j =.2556(4.85) +.2331(4.61) +.2707(4.61) +.2406(4.38) = 4.616. Among Groups SS = 34(4.85 ‑ 4.616) 2 + 31(4.61 ‑ 4.616) 2 + 36(4.61 ‑ 4.616) 2 + 32(4.38 ‑ 4.616) 2 = 3.646. With 3 df, MSA = 1.215, and F(3, 129) = 2.814, p =.042.

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