pasw & hand calculations for...

PASW & Hand Calculationsfor ANOVA

Gravetter & WallnauChapter 13, Problem 26

One possible reason that some birds migrate and others don’t is intelligence. Birds with small brains relative to their body size might not be smart enough to find food during the winter. Birds with larger brains might be creative enough to find food even when the weather turns harsh. The next slide has the relative brain size for individual birds in each sample. Are their significant difference among the groups? Compute η2. If appropriate, calculate Tukey’s HSD to determine which groups are sig. different.

Gravetter & WallnauChapter 13, Problem 26

Non-MigratingShort-Distance

Migrants Long-Distance Migrants

181319121612

61179813

495657

Step 1: H0, H1 & α

H0: µNon-Migrating= µShort-Distance Migrants= µLong-Distance Migrants

H1: not H0

α = .05

Step 2: Critical Value

Skip – use the p value from the ANOVA output

F(2, 15) = 3.68From table of critical F values

dfbetween= # levels of factor – 1 = 3 – 1 = 2

dfwithin = Σ(# of people in a condition – 1) = (6 – 1) + (6 – 1) + (6 – 1) = 15

Step 3: Calculate Fobserved

Load PASW

Create two variablesOne indicates the level of the factor

Name: migrate, Label: Migration Status, Values: 1 = Non-Migrating, 2 = Short-Distance Migrant, 3 = Long-Distance Migrant, Measure: Nominal

One is the DVName: brain, Label: Relative Brain Size, Measure: Scale

Type in the data


Click on Analyze | General Linear Model | Univariate

Move the DV (brain size) into the Dependent Variable box

Move the factor (migrate) into the Fixed Factor(s) box


Click on Post HocMove the factor (migrate) into the Post Hoc Tests for boxSelect Tukey as the desired testClick Continue

Click OptionsSelect Descriptive statistics, Estimates of effect sizeClick Continue

Click OK


F(2, 15) = 19.286, p = .000, MSerror = 6.533, η2

= .720


Non-MigratingShort-Distance

MigrantsLong-Distance

Migrants

181319121612

61179813

495657

N = 18(total number of scores)

G = 180(sum of all scores = 90 + 54 +

36)

ΣX2 = 2150(square each score and sum

the squares)

T = 90 = 18+13+19+12+16+12

M = 15= 90 / 6

SS = 48 = (18 - 15)2 + (13 - 15)2 +

(19 - 15)2 + (12 - 15)2 + (16 -15)2 + (12 - 15)2

T = 54 = 6 + 11 + 7 + 9 + 8 + 13

M = 9= 54 / 6

SS = 34=(6 - 9)2 + (11 - 9)2 + (7 - 9)2 +

(9 - 9)2 + (8 - 9)2 + (13 - 9)2

T = 36= 4 + 9 + 5 + 6 + 5 + 7

M = 6= 36 / 6

SS = 16=(4 - 6)2 + (9 - 6)2 + (5 - 6)2 +

(6 - 6)2 + (5 - 6)2 + (7 - 6)2


35018

1802150

N

GXSS

2

22

total

=

−=

−∑=

98

163448

SSSS menteach treat insidewithin

=++=

∑=

252

1800216486135018

180

6

36

6

54

6

90

N

G

n

TSS

2222

22

between

=−++=

−++=

−∑=

17

118

1Ndf total

=−=−=

15

318

kNdfwithin

=−=−=

2

13

1kdfbetween

=−=−=

1262

252

df

SSMS

between

betweenbetweeen

=

=

=

6.5315

98

df

SSMS

within

withinwithin

=

=

=


19.2866.533

126

MS

MSF

within

between

=

=

=

Step 4: Decide

If p ≤ α then reject H0 otherwise fail to reject H0p = .000, α = .05 therefore, reject H0It is unlikely that all the means in the population are equal

If Fobserved≥ Fcritical then reject H0 otherwise fail to reject H0

Fobserved= 19.286, Fcritical = 3.68 therefore, reject H0

η2

η2 = SSbetween/ SStotal = 252 / 350 = 0.720

TukeyHSD

H0: µNon-Migrating= µShort-Distance Migrant

H1: µNon-Migrating≠ µShort-Distance Migrant

H0: µNon-Migrating= µLong-Distance Migrant

H1: µNon-Migrating≠ µLong-Distance Migrant

H0: µLong-Distance Migrant= µShort-Distance Migrant

H1: µLong-Distance Migrant≠ µShort-Distance Migrant

α = .05

TukeyHSD

Pairs that are reliably different (non-migrating vs. short-distance migrant and non-migrating vs. long-distance migrant)p ≤α

Pair that is not reliably different (short-distance migrant vs. long-distance migrant) p > α

TukeyHSD

835.36

6.5333.675

n

MSHSD treatmentwithin

dfk,α,q

treatmentwithin

=

⋅=

⋅= −

−

Any pair of means that is at least 3.835 apart will be reliably different.

Means are likely different

Insufficient evidence to state that means are likely different

From the table of the Studentized Range Statistic with α= .05, k = 3 (three levels) and dfwithin = 15

Writing

The means and standard deviations are presented in Table 1. The analysis of variance revealed a significant effect of migration style on relative brain size, F(2, 15) = 19.286, p = .000, MSerror = 6.533, η2

= .720. Tukey’s post hoc tests revealed that non-migrating birds had a different relative brain size than short-distance migrants (p = .003) and long-distance migrants (p = .000). There was insufficient evidence to suggest that the long- and short-distance migrants had different brain sizes (p = .138).

Writing

Table 1

Relative Brain Size of Three Types of Birds

Type of Bird M SD

Non-migrating 15.00 3.10

Short-distance migrant 9.00 2.61

Long-distance migrant 6.00 1.79

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