One-Way ANOVA Independent Samples

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

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

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?

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 = ( ) / 4 = 0.5

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

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

p F = /.5 = 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

Deviation Method SS TOT =  (Y ij - GM) 2 = (1 - 5) 2 + (2 - 5) (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) (9 - 8) 2 = 8.

Computational Method = ( ) - [( ) 2 ]  N = (100) 2  20 = 138. = [ ]  5 - (100) 2  20 = 130. SSE = SS TOT – SS A = = 8.

Source Table

Magnitude of Effect Omega Square is less biased

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

Magnitude of Effect Enter F and df into NoncF.sav

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

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

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

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.

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

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

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

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

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).

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.

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.

APA-Style Presentation

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.

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