dr. sinn, psyc301, the joy of 1-way anova1 unit 3 outline day 1: introduce f return tests (20) power...

Dr. Sinn, PSYC301, The joy of 1-way ANOVA 2

Lecture Overview on ANOVA

Reviewhypothesis testing; inferential statistics

z-test, t-test, independent & dependent t-test

New StuffPower – Ability to reject Ho

ANOVA• Analysis of Variance

• Done with 3 or more groups

• Playground Exercise

• Complete SPSS Example


Power

Review: Hypothesis Testing ErrorsWrongly rejecting Ho: Chance of Type I error: α

Wrongly retaining Ho: Chance of Type II error: β

PowerOpposite of β

Power = 1- β

Ability to reject Ho (when Ho should be rejected).

Researchers want Power! • Want ability to reject Ho; Show you were right to suspect a

difference.

• Want to show IV affects your DV.


Error Areas

α area (where we reject the Ho, and we shouldn’t)

• beyond tcritical

• under Ho

β area (where we retain the Ho, and we shouldn’t)

• inside tcritical

• under Ha

Ho: μ=55

tctc

αα β

Ha: μ>55

tc


Increasing Power

#1: Increase Treatment: Increase difference between groups (μ’s)

β

Reality: μ=57

tc

βtc

H0: μ=55

H0: μ=55

Reality: μ=72

tctc

ββ

#2: Decrease Sampling Error: Decrease differences within groups.


Examples of increasing power

#1 Increase Treatment Effect(Increase BG differences)

Rat study

0,3,or 6 mg

0,10,or 20 mg

Therapy study

10 therapy sessions

1 therapy session

#2 Decrease Sampling Error(Decrease WG differences)

Rat study

Different strains of rats

Same strain of rat

Rats allowed to eat freely

Rats all unfed for 24 hours

Therapy study

Diff. types of Therapy

Same type of Therapy

Rat Study: IV:Caffeine Level DV:Amt. Food Found

Therapy Study: IV:Therapy (drug, talk, drug+talk, or control) DV: Improvement]


1-Way ANOVA

ANOVAAnalysis of Variance1-way means 1 Independent Variable (IV)

Purpose: ANOVA allows hypothesis testing with 3+ sample means

• Imagine study on interventions to help frosh make friends• Three IV levels: Standard courses, interactive courses,

clustered courses.

WG

BG

MS

MSF

ANOVA uses F-test

Strategy: Compare variability within group to variability between groups.

F is ratio between two values:


ANOVA Playground(Download from Website)

1


Matching Exercise

D r . S i n n , P S Y C 3 0 1 , T h e j o y o f 1 - w a y A N O V A 1

M a t c h i n g V a r i a n c e t o O u t p u t ( & * & )

S t a n d a r d C o u r s e s

I n t e r a c t i v e C o u r s e s

C l u s t e r e d C o u r s e s

0 2 92 1 0 1 27 4 82 6 1 26 9 64 3 50 8 1 1

m e a n 3 . 0 6 . 0 9 . 0

s t d . d e v . 2 . 8 3 . 1 2 . 8

N u m b . o f " g o o d f r i e n d s " b y e n d o f y e a r

1 D r . S i n n , P S Y C 3 0 1 , T h e j o y o f 1 - w a y A N O V A 5


6 3 . 0 0 7 . 5

8 . 4 4

F =

W G

B G

WG

BG

MS

MSF =

A






0 7 72 2 67 2 42 6 16 2 24 0 00 4 2

m e a n 3 . 0 3 . 3 3 . 1

s t d . d e v . 2 . 8 2 . 5 2 . 6


2 D r . S i n n , P S Y C 3 0 1 , T h e j o y o f 1 - w a y A N O V A 6


0 . 1 4 0 . 0

6 . 9 0

F =

W G

B G

WG

BG

MS

MSF =

B






2 5 82 5 82 6 83 6 94 6 1 04 7 1 04 7 1 0

m e a n 3 . 0 6 . 0 9 . 0

s t d . d e v . 1 . 0 0 . 8 1 . 0


3



6 3 . 0 0 7 0 . 9

0 . 8 9

F =

W G

B G

WG

BG

MS

MSF =

C






3 4 44 5 55 5 55 6 76 6 77 6 87 9 1 0

m e a n 5 . 3 5 . 9 6 . 6

s t d . d e v . 1 . 5 1 . 6 2 . 1


4



2 . 9 0 1 . 0

3 . 0 0

F =

W G

B G

WG

BG

MS

MSF =

D


Draw Conclusions from Playground

What does a large F mean?

What two things will make F large?


Partitioning Variance

Partitionfancy word for “divide up”ANOVA partitions variance (MS means variance)

Types of varianceTotal variance = MSWG + MSBG

MSWG= sampling error (background noise)

MSBG = sampling error + treatment (includes effect of Independent Variable)

error

errortreatment

MS

MSF

WG

BG

If just error F tends toward 1.0

If treatment effect F gets larger


Example of 1-way ANOVA

Studying effect of caffeine on productivity

Does caffeine help or hurt?

IV: Level of Caffeine: 0, 10, 20 mg

DV: Number of Food Pellets Found

0 mg 10 mg 20 mg

23142

12312

444455

Number of Food

Pellets Found


SPSS Data Entry

DV

IV

Label levels of IV so output is easier to read.


SPSS AnalysisGo to Analyze, Compare Means, & select One-way ANOVA

Put IV here.

Put DV here.


SPSS Analysis, Part #2

Conducts “after the fact” test to compare all pairs of sample means.

Select this to get descriptive statistics like sample means & standard deviations.

Gives you a line graph of the sample means

Alpha level still set to .05, just like it was with t-tests.


SPSS OutputDescriptives

FOOD_FND

5 2.40 1.140 .510 1 4

5 1.80 .837 .374 1 3

6 4.33 .516 .211 4 5

16 2.94 1.389 .347 1 5

0 mg

10 mg

20 mg

Total

N MeanStd.

DeviationStd.Error Min Max

ANOVA

FOOD_FND

19.604 2 9.802 13.65 .001

9.333 13 .718

28.937 15

Between Groups

Within Groups

Total

Sum ofSquares df

MeanSquare F Sig.

WG

BG

MS

MSF

Source of Variation Table

Sample means from 3 groups, plus mean amount of food found overall.


Where does F come from?

MSWG = SSWG/dfWG = Sum of Squares / degrees of freedom

MSBG = SSBG/dfBG = Sum of Squares / degrees of freedom

Degrees of freedomdfWG: NT – K (Total # of subjects - # of groups)

dfBG: K-1 (# of groups – 1)

dfTOTAL: NT – 1 (Total # of subjects – 1)

Expectations:If I give you df and SS, you can calculate F

You don’t have to get any SS by hand.


SPSS Output –Post Hoc Test

FOOD_FND

Student-Newman-Keulsa,b

5 1.80

5 2.40

6 4.33

.270 1.0

GROUP10 mg

0 mg

20 mg

Sig.

N 1 2

Subset foralpha = .05

Means for groups in homogeneous subsets are displayed.

GROUP

20 mg10 mg0 mg

Mea

n of

FO

OD

_FN

D

4.5

4.0

3.5

3.0

2.5

2.0

1.5

No Sig. Diff. Between 0 & 10mg

Rats at 20 mg found significantly more food than rats on 0 or 10 mg of caffeine.


SPSS Output– Practical Significance

η2 (“eta squared”)Effect size statistic – indicates % of variance explained

Measures impact of IV on DV

We can explain 68% of the variance in how much food a rat finds if we know the level of caffeine.

ANOVA

FOOD_FND

19.604 2 9.802 13.65 .001

9.333 13 .718

28.937 15

Between Groups

Within Groups

Total

Sum ofSquares df

MeanSquare F Sig.

6775.937.28

604.192 Total

BG

SS

SS


Hypothesis Testing Steps

1. Comparison: cf. three sample means.

2. Hypothesis: Ho: μ1= μ2 = μ3 Ha: Not all μ’s equal

3. Set-up: α= .05 , dfbg= K-1= 2, dfwg= NT-K = 16-3=13, Fcrit = 3.80

4. Fobt = 13.653

5. Reject Ho.

The hypothesis was largely supported. Rats found sig. more food on 20mg of caffeine (M=4.33) than on 0mg (M=2.40) or 10mg (M=1.80), F(2,13) = 13.653, p <=.05. Caffeine has a large effect on food finding behavior, accounting for about 68% of the variance, η2 = .6775.


F-table df Between Groups

df Within Groups


Lab #8: 1-way ANOVA

TV Problem: The hypothesis was supported.

Light TV users provided more community service

(M = 6.13) than did moderate users (M = 4.00),

who provided more than heavy users (M = 1.75),

F(2,21) = 15.963, p ≤ .05. TV accounts for about

60% of the variance in community service, η2

= .6032.


Follow-up Questions

Q1: Variance within group? MSwg = 2.399

Q2: Variance between groups? MSbg=38.292

Q3: Replacing heavy scores with 4,5,4,5,6,5,4,3 would decrease the difference between groups because the heavy users would then difference less from the other groups.

Q4: Decreasing between group differences (decreasing treatment) would decrease F.


Problem #2: Post Hoc Explanation

Student-Newman-Keulsa

5 1.20

5 1.60

5 2.40 2.40

5 3.40

.077 .068

commute45 min commute

60 min commute

30 min commute

0 min commute

Sig.

N 1 2

Subset for alpha = .05



Problem #2: Post Hoc Explanation


5 1.20

5 1.60

5 2.40 2.40

5 3.40

.077 .068

commute45 min commute

60 min commute

30 min commute

0 min commute

Sig.

N 1 2

Subset for alpha = .05



Problem #2:

The hypothesis was supported. People commuting

0 minutes participated significantly more (M=3.4

hours) than people commuting 45 (M=1.2) or 60

minutes (M=1.6), F (3,16) = 7.256, p≤.05.

Commuting accounted for a large amount of

variance in community involvement, η2 = .5764.


Follow-up Questions

Q1: Variance within group? MSwg = .650

Q2: Variance between groups? MSbg=4.717

Q3: Replacing 30 minute commuting scores with 1,4,1,4,3 would increase the within group variability.

Q4: Increasing sampling error would decrease F.


Review Partitioning

Study: Does alcohol affect reaction time? Identify the treatment

effect in this case.

Explain how sampling error might arise.

No Alcohol

2 Beers 4 Beers

10 15 20

20 25 15

15 30 30

10 20 40

14 23 26 Sample Means

μna=?? μ2b=?? μ4b=?? Population Means

Dr. Sinn, PSYC301, The joy of 1-way ANOVA

30

One-Way ANOVA

Part 2!!


Review Partitioning

Study: Does alcohol affect reaction time?

What accounts for variability within groupswithin groups?

What accounts for variability between groups?

What’s the Formula for F?

No Alcohol

2 Beers 4 Beers

10 15 20

20 25 15

15 30 30

10 20 40


Review Partitioning

Study: Does alcohol affect reaction time?

If the alcohol content of the beers is not held constant, what happens to F?

a. increasesb. decreases

c. neither

No Alcohol

2 Beers 4 Beers

10 15 20

20 25 15

15 30 30

10 20 40

If the alcohol content of the beers is not held constant, what happens?

a. error increases

b. error decreases

c. treatment effect increases

d. treatment effect decreases





3. Set-up: α= .05 , dfbg=K-1=3-1=2, dfwg=NT-K=12-3=9, Fcrit = 4.26

• now do one-way ANOVA on SPSS


SPSS Output - Charts

ANOVA

ALCOHOL

329.167 2 164.583 2.633 .126

562.500 9 62.500

891.667 11

Between Groups

Within Groups

Total

Sum ofSquares df

MeanSquare F Sig.

Descriptives

ALCOHOL

4 13.75 4.787 2.394

4 22.50 6.455 3.227

4 26.25 11.087 5.543

12 20.83 9.003 2.599

no alcohol

2 beers

4 beers

Total

N Mean Std. Deviation Std. Error

ALCOHOL


4 13.75

4 22.50

4 26.25

.118

GROUPno alcohol

2 beers

4 beers

Sig.

N 1

Subset foralpha = .05


Uses Harmonic Mean Sample Size = 4.000.a.


SPSS Output - Graphs

GROUP

4 beers2 beersno alcohol

Mea

n of

ALC

OH

OL

28

26

24

22

20

18

16

1412 444N =

GROUP

4 beers2 beersno alcoholM

ean

+-

2 S

E A

LCO

HO

L

40

30

20

10

0





3. Set-up: α= .05 , dfbg=K-1=3-1=2, dfwg=NT-K=12-3=9, Fcrit = 4.26

4. Fobt = 2.633

5. Retain Ho.

The hypothesis was not supported. The reaction times following no alcohol (M=13.75), two beers (M=22.50), and four beers (M=26.25) did not differ significantly, F(2,9) = 2.633, n.s..


Numb. of Words Recalled: Dataset A

Bet. Group Varib: L M H

MSbg: _______

With. Group Varib: L M H

MSwg: _______

4 8 12

5 9 10

4 9 11

5 8 12

WG

BG

MS

MSF


Numb. of Words Recalled: Dataset B


MSbg: _______


MSwg: _______

8 4 10

9 5 12

9 5 11

8 4 12

WG

BG

MS

MSF


Numb. of Words Recalled: Dataset C


MSbg: _______


MSwg: _______

7 3 9

10 6 13

7 6 10

10 3 13

WG

BG

MS

MSF


Numb. of Words Recalled: Dataset D


MSbg: _______


MSwg: _______

7 6 7

10 8 7

7 6 12

10 10 12

WG

BG

MS

MSF


Numb. of Words Recalled: Dataset E


MSbg: _______


MSwg: _______

7 6 7

10 8 7

7 6 12

10 10 12

7 6 7

10 8 7

7 6 12

10 10 12

WG

BG

MS

MSF


Numb. of Words Recalled: Dataset F


MSbg: _______


MSwg: _______

WG

BG

MS

MSF

dr. sinn, psyc301, the joy of 1-way anova1 unit 3 outline day 1: introduce f return tests (20) power...

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