one-way anova independent samples. basic design grouping variable with 2 or more levels continuous...

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

• Yij = + j + eij, or,

• Yij - = j + eij.

• 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, eij

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?

Group Score A 1 2 2 2 3 B 2 3 3 3 4 C 6 7 7 7 8 D 7 8 8 8 9

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

k

sssMSE k

222

21 ....

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.

2meansA snMS

F

• F = MSA / MSE

• If H is true, expect F = error/error = 1.

• If H is false, expect

1

error

treatmenterrorF

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

Deviation Method• SSTOT = (Yij - GM)2

= (1 - 5)2 + (2 - 5)2 +...+ (9 - 5)2 = 138.

• SSA = [nj (Mj - GM)2]

• SSA = n (Mj - GM)2 with equal n’s

= 5[(2 - 5)2 + (3 - 5)2 + (7 - 5)2 + (8 - 5)2] = 130.

• SSE = (Yij - Mj)2

= (1 - 2)2 + (2 - 2)2 + .... + (9 - 8)2 = 8.

Computational Method

= (1 + 4 + 4 +.....+ 81) - [(1 + 2 + 2 +.....+ 9)2] N = 638 - (100)2 20 = 138.

= [102 + 152 + 352 + 402] 5 - (100)2 20 = 130.

SSE = SSTOT – SSA = 138 - 130 = 8.

N

GYSSTOT

22

SST

n

G

NA

j

j

2 2

SST

n

G

NA

j

2 2

Source Table

Source SS df MS F p

Teaching Method 130 3 43.33 86.66 < .001

Error 8 16 0.50

Total 138 19

Magnitude of Effect

• Omega Square is less biased

942.138

1302 Tot

A

SS

SS

927.)1(

2

MSESS

MSEkSS

TOT

A

Magnitude of Effect

• Put a confidence interval on eta-squared.

• http://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 cpc

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:

ji

ji

nnMSE

MMt

11

001. ,416.132.)28()16( pt

• For A versus C and B versus D:

• For B versus C

• For A vs B, and C vs D,

001. ,944.82.)37()16( pt

001. ,180.112.5)16( pt

04. ,236.2)5/15/1(5.1)16( pt

Underlining Means Display

• arrange the means in ascending order

• any two means underlined by the same line are not significantly different from one another

Group A B C D

Mean 2 3 7 8

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

cfw

pc

• 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).

00167.6

01. pc

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

Table 1 Effectiveness of Four Methods of Teaching ANOVA

Method M SD Ancient 2.00A .707

Backwards 3.00A .707 Computer-Based 7.00B .707

Devoted 8.00B .707 Note. Means with the same letter in their superscripts do not differ significantly from one another according to a Bonferroni test with a .01 limit on familywise error rate.

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 = pj Mj =.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.

Semester Mean SD N pj

Spring 89 4.85 .360 34 34/133 = .2556

Fall 88 4.61 .715 31 31/133 = .2331

Fall 87 4.61 .688 36 36/133 = .2707

Spring 87 4.38 .793 32 32/133 = .2406

N

np j

j

.4317.)793(.2406.)688(.2707.)715(.2331.)360(.2556. 22222 jjspMSE