chapter 17 comparing multiple population means: one-factor anova

Chapter 17

Comparing Multiple Population Means: One-factor ANOVA

What if we have more than 2 conditions/groups?

Interest - the effects of 3 drugs on depression - Prozac, Zoloft, and Elavil

Select 24 people with depression, randomly assign (blindly) to one of four conditions: 1) Prozac, 2) Zoloft, 3) Elavil, and 4) Placebo

After 1 month of drug therapy, we measure depression

Research Design and Data

Prozac Zoloft Elavil Placebo10 14 19 21 8 12 15 2715 18 14 2012 16 16 23 9 13 18 15 6 17 20 22

Multiple t-tests?Differences between drugs?

Prozac vs. Zoloft Prozac vs. Elavil

Prozac vs. Placebo Zoloft vs. Elavil

Zoloft vs Placebo Elavil vs. Placebo6 separate t-tests

Probability Theory (Revisited)The probability of making a correct

decision when the null is false is 1 - α (generally .95)

Each test is independentThe probability of making the correct

decision across all 6 tests is the product of those probabilities

or, (.95)(.95)(.95)(.95)(.95)(.95) = .735

Type 1 error & multiple t-testsThus, the probability of a type 1 error is

not α, but 1 - (1 - α)C, where C is the number of comparisons

Or, in the present case 1 - .735 = .265

t statistic as a ratio

obtained difference t = ———————————————— difference expected by chance (“error”)

Easy – Pool Variance

Hmmm…

Differences in the t test

M1 – M2 or MD

Can we subtract multiple means from one another?

M1 – M2 – M3 – M4 = ????

M4 – M1 – M2 – M3 = ????Is there another statistic that tells us how much things differ from one another?

What statistic describes how scores differ from one another?

Variance

How do a set a means differ from one another? Answer – variance between means/groups

t statistic as a ratio

obtained difference t = ———————————————— difference expected by chance (“error”)

variance between means/groups t = ———————————————— pooled variance

F statistic between-groups variance estimateF = —————————————— within-groups variance estimate

Mean-square Treatment (MST or MSB) s2B

F = ———————————————— = —

Mean-square Error (MSE or MSW) s2W

ANOVAAnalysis of Variance, or ANOVA,

allows us to compare multiple group means, without compromising α

And, even though an ANOVA uses variances and the F statistic, it helps test hypotheses about means

F statisticBetween-groups variance (MST or MSB) is

based on the variability between the groupsWithin-groups variance (MSE or MSW) is a

measure of the variability within the groups– if there is no difference between these 2 measure of

variability (due to no differences between groups), F will be close to 1

– if there is greater variability between-groups (due to differences between groups), F will be greater than 1

Between-groups variance (MST, MSB or s2

B) k groups

where Mi is the mean of the ith group, and MG is the grand mean (the mean of all scores)

Within-groups variance(MSE, MSW, or s2

W) k groups

SST (Sums of Squares Total)The sums of squares total can be used

either as a check, or to calculate SSW

An ANOVA TableThe results of an ANOVA are often

presented in a table:

Source SS df MS FBetweenWithinTotal

An ANOVA TableThe results of an ANOVA are often

presented in a table:

Source SS df MS FBetween 180 2 90.0 36.00Within 30 12 2.5Total 210 14

Procedure for Completing an ANOVA1. Arrange Data by Group2. Compute for each group (k

groups):Σx

Σx2

M

SS(x)

n

Procedure for Completing an ANOVA 3. Compute the grand mean ( MG), by

adding all the scores and dividing by NMG = Σx/N

4. Compute SSB = Σ ni( Mi - MG)2 5. Compute SSW

SSW = SS(x1) + SS(x2) + ···+ SS(xk) 6. Compute SST = Σx2 - (Σx)2/N

Procedure for Completing an ANOVA7. Compute df

dfB = k - 1

dfW = N - k

dfT = N -18. Fill in ANOVA table 9. Compute MS (SS/df)10. Compute F = MSB/MSW

1. ANOVA Calculations

Prozac Zoloft Elavil Placebo10 14 19 21 8 12 15 2715 18 14 2012 16 16 23 9 13 18 15 6 17 20 22


Prozac (Group 1)

10 ΣX1 = 50

8 ΣX12 = 650

15 M1 = 10

12 SS(X1) = 50

9 n1 = 6 6


Zoloft (Group 2)

14 ΣX2 = 90

12 ΣX22 = 1378

18 M2 = 15

16 SS(X2) = 28

13 n2 = 617


Elavil (Group 3)

19 ΣX3 = 102

15 ΣX32 = 1762

14 M3 = 17

16 SS(X3) = 28

18 n3 = 620


Placebo (Group 4)

21 ΣX4 = 128

27 ΣX42 = 2808

20 M4 = 21.33

23 SS(X4) = 77.33

15 n4 = 622


MG = Σx/N

= (ΣX1+ ΣX2 + ΣX3 + ΣX4)/ (n1+n2+n3+n4) = (60+90+102+128)/(6+6+6+6) = 380/24 = 15.83


SSB = Σ ni ( Mi - XG)2

= 6(10 - 15.83)2 + 6(15 - 15.83)2 + 6(17 - 15.83)2 + 6(21.33 - 15.83)2

= 6(34.03) + 6(.69) + 6(1.36) + 6(30.25) = 204.18 + 4.14 + 8.16 + 181.5 = 398.00


SSW = SS(X1) + SS(X2) + ···+ SS(Xk) = 50 + 28 + 28 + 77.33 = 183.33


SST = ΣX2 - (ΣX)2/N = (650 + 1378 + 1762 + 2808) - (60 + 90 + 102 + 128)2/24 = 6598 - 144400/24

= 6598 - 6016.67 = 581.33

Check

SST = SSB + SSW581.33 = 398 + 183.33 581.33 = 581.33


dfB = k -1 = 4 -1 = 3

dfW = N - k = 24 - 4 = 20

dfT = N - 1 = 23


Source SS df MS FBetween 398.00 3Within 183.33 20Total 581.33 23


Source SS df MS FBetween 398.00 3 132.67Within 183.33 20 9.17 Total 581.33 23


Source SS df MS FBetween 398.00 3 132.67 14.47Within 183.33 20 9.17 Total 581.33 23

Hypothesis test of Anti-depressants1. State and Check Assumptions

– About the population Normally distributed? - don’t know Homogeneity of variance – we’ll check

– About the sample Independent Random sample? – yes Independent samples

– About the sample Interval level

Hypothesis test of Anti-depressants

2. HypothesesHO : μProzac = μZoloft = μElavil = μPlacebo

HA : the null is wrong

That’s an Odd HA

You might think that the alternative hypothesis should look like this:HA : μProzac ≠ μZoloft ≠ μElavil ≠ μPlacebo

Accepting this alternative indicates that all of the means are unequal, which is not what ANOVA determines

What does ANOVA determine?That at least one of the means is

different than at least one other meanSince, that is a difficult statement to

write, we say“the null is wrong”

Hypothesis test of Anti-depressants

3. Choose test statistic– 4 groups

independent samples

One-factor ANOVA

Hypothesis test of Anti-depressants 4. Set Significance Level

α = .05

Critical Value

Non-directional Hypothesis with

dfB = k – 1 and dfW = N – k

dfB = 3 and dfW = 21

From Table D

Fcrit = 3.07, so we reject HO if

F ≥ 3.07

Hypothesis test of Anti-depressants5. Compute Statistic

Source SS df MS FBetween 398.00 3 132.67 14.47Within 183.33 20 9.17 Total 581.33 23

Hypothesis test of Anti-depressants6. Draw Conclusions

– because our F falls within the rejection region, we reject the HO, and

– conclude that at least one medicine is better than at least one other medicine in treating depression

Violations of AssumptionsAs with t-tests, ANOVA is fairly

ROBUST to violations of normality and homogeneity of variance, but

IF there are severe violations of these assumptions,

Use a Kruskal-Wallis H test (a non-parametric alternative)

Procedure for completing a Kruskal-Wallis H 1. Arrange data in columns, 1 group per

column, skipping columns between groups 2. Rank all the scores, assigning the lowest

rank (1) to the lowest score (put ranks in the column next to the raw scores)

3. Sum the ranks in each column (ΣTj) 4. Square the sum of the ranks of each

column (ΣTj)2

Procedure for completing a Kruskal-Wallis H test

5. Compute SSB

6. Compute H

Procedure for completing a Kruskal-Wallis H test

6. Compute df = k - 17. H is distributed as a χ2

– Look up critical value in χ2 (chi-square) table with appropriate df

Dependent Samples(more than 2 conditions)Experiments are often conducted

comparing more than 2 conditions– ANOVA– Kruskal-Wallis H

Samples are often related - “dependent samples” (within-subjects, repeated measures, etc.)

Dependent Samples ANOVA

SS(T) = SS(B) + SS(Bl) + SS(E)Calculate SS(T), SS(B), and SS(Bl)

SS(E) = SS(T) - SS(B) - SS(Bl)

Why “Blocks”?A dependent samples ANOVA is

sometimes referred to as a “Randomized-Block” design

Each group of related measurements, either within-subject, or with matching, is a “Block” of measurements

SS(Bl)Sum of Squares Blocks - the sum of

the squared deviations of each block mean from the grand mean

SS(Bl) = Σk( Mi - MG )2, or

SS(Bl) = ΣBl2/k - N( MG2), where

Bl = sum of the scores in a block

Procedure for Completing A dependent samples ANOVA 1. Arrange data where columns are

conditions, rows are blocks (subjects or matched-subjects)

2. Compute for each column (conditions)n

ΣXΣX2

MSS(X)s2

Procedure for Completing A dependent samples ANOVA3. Total the scores in the rows in a

new column to the right (Block Totals)4. Square the block totals in the next

column5. Compute the grand mean ( MG),

by adding all the scores and dividing by N

MG = ΣX/N

6. Compute SS(B) = Σ ni ( Mi - MG)2

Procedure for Completing A dependent samples ANOVA

7. Compute SS(T) = ΣX2 - NMG2

8. Compute SS(Bl) = ΣBl2/k – NMG2

9. Compute SS(E) = SS(T) - SS(B) - SS(Bl)

10. Compute dfdfB = k - 1

dfBl = n - 1

dfE = (N - k) - (n - 1)

dfT = N -1

Procedure for Completing A dependent samples ANOVA

11. Fill in ANOVA table 12. Compute MS (SS/df)13. Compute F = MSB/MSE

Dependent Samples ANOVA tableSource SS df MS FBetween

Blocks Error

Total

Example A researcher is interested in the effects of

three new sleep-aids, Sleep E-Z, Zonked, and NockOut

He selects 5 subjects and they take each of the 3 new drugs in a random order

The number of hours slept per night on each of the new sleep-aids is recorded

Data

Subject Sleep E-Z ZonkedNockOut1 6 5 82 5 6 7

3 6 6 9

4 7 7 6

5 4 5 8

Hypothesis Test – Sleep aids 1. State and Check Assumptions

– Population Normally Distributed – not sure, assume for time being H of V – not sure, but we’ll check sample variances

– Sample Dependent samples Random assignment

– Data Interval/Ratio

Hypothesis Test – Sleep aids2. State Null and Alternative Hypotheses

HO : μ1 = μ2 = μ3 (the population means are equal)

HA : HO is wrong (at least one of the means differs, can’t say “μ1 ≠ μ2 ≠ μ3” because this means “all the means differ from one another”)

Hypothesis Test – Sleep aids3. Choose Test Statistic

– Parameter of interest – means– Number of Groups – 3– One factor (or IV being manipulated)– Dependent Samples

One-factor ANOVA for Dependent Samples (F)

Hypothesis Test – Sleep aids4. Set Significance Level

α = .05

F = MSB/MSE, dfB = k – 1, dfE = (N – k) – (n – 1), where N = total number of obs, k = number of groups/conditions, n = number of subs/blocks

dfB = 3 –1 = 2, dfE = (15 – 3) – ( 5 – 1)

Fcrit(2, 8) = 4.46

If our F ≥ 4.46, we Reject HO

Hypothesis Test – Sleep aids5. Compute test Statistic

Computations

Sub S E-Z Z NO 1 6 5 8 2 5 6 73 6 6 94 7 7 65 4 5 8

Sub S E-Z Z NO 1 6 5 8 2 5 6 73 6 6 94 7 7 65 4 5 8n 5 5 5ΣX 28 29 38ΣX2 162 171 294M 5.6 5.8 7.6SS(X) 5.2 2.8 5.2s2 1.3 0.7 1.3 – H of V Otay!

Sub S E-Z Z NO Bl 1 6 5 8 192 5 6 7 183 6 6 9 214 7 7 6 205 4 5 8 17n 5 5 5ΣX 28 29 38ΣX2 162 171 294M 5.6 5.8 7.6SS(X) 5.2 2.8 5.2s2 1.3 0.7 1.3

Sub S E-Z Z NO Bl Bl2

1 6 5 8 19361

2 5 6 7 18324

3 6 6 9 21441

4 7 7 6 20400

5 4 5 8 17289

N 5 5 5 ΣBl2 = 1815

ΣX 28 29 38ΣX2 162 171 294M 5.6 5.8 7.6SS(X) 5.2 2.8 5.2s2 1.3 0.7 1.3

Computations

MG = ΣX/N = (28 + 29 + 38) / (15)

= 6.333

Computations

SS(B) = Σ ni ( Mi - MG)2

= 5(5.6 - 6.33)2 + 5(5.8 - 6.33)2 +

5(7.6 - 6.33)2

= 12.133

Computations

SS(T) = Σ X2 - (Σ X)2/N= (162 + 171 + 294) - (28 + 29 + 38)2/15

= 627 - 601.66= 25.33

Computations

SS(Bl) = ΣBl2/k - N( MG2)

= (361 + 324 + 441 + 400 + 289)/3 -

15(6.33)2

= 1815/3 - 601.66= 3.33

Computations

SS(E) = SS(T) - SS(B) - SS(Bl)= 25.33 - 12.13 - 3.33= 9.87

Computations

dfB = k - 1 = 3 - 1 = 2

dfBl = n - 1 = 5 - 1 = 4

dfE = (N - k) - (n - 1) = (15 - 3) - (5 - 1) = 12 - 4 = 8

dfT = N -1 = 15 -1 = 14

Computations

Source SS df MS FBetween 12.13 2

Blocks 3.33 4Error 9.87 8

Total 25.33 14

Computations

Source SS df MS FBetween 12.13 2 6.07

Blocks 3.33 4 .83Error 9.87 8 1.23

Total 25.33 14

Computations

Source SS df MS FBetween 12.13 2 6.07 4.93

Blocks 3.33 4 .83Error 9.87 8 1.23

Total 25.33 14

Hypothesis Test6. Draw Conclusions

– Since our F > 4.46, we Reject HO, accept HA

– And conclude that the at least one of the medications resulted in more sleep than the others

Dependent samples ANOVAWhat if we violate one of the

assumptions?Friedman test

– means (or distribution) are of interest– more than 2 groups/conditions– dependent samples– concerns about normality, homogeneity of

variance, etc.

Friedman Fr

1. Arrange data in columns, 1 group/condition per column, (conditions = columns = k)2. Place correlated measures (matched, repeated, etc.) across conditions in the same rows (n rows)3. Rank the scores in each row from 1 to k, assigning the lowest rank (1) to the lowest score (put ranks in the column next to the raw scores)

Friedman (continued)

4. Sum the ranks of each column (ΣTk)5. Compute the mean of the Ts, T

6. Compute S

Friedman (continued)

7. Compute the Friedman test statistic Fr

8. Compute df = k-19. Look up critical value in Χ2 table or use

Excel to find p

chapter 17 comparing multiple population means: one-factor anova

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