Six Easy Steps for an ANOVA

• 1) State the hypothesis

• 2) Find the F-critical value

• 3) Calculate the F-value

• 4) Decision

• 5) Create the summary table

• 6) Put answer into words

Example

• Want to examine the effects of feedback on self-esteem. Three different conditions -- each have five subjects

• 1) Positive feedback

• 2) Negative feedback

• 3) Control

• Afterward all complete a measure of self-esteem that can range from 0 to 10.

Example:

• Question: Is the type of feedback a person receives significantly (.05) related their self-esteem?

Results

Positive Feedback

Negative Feedback

Control

10 4 3

Step 1: State the Hypothesis

• H1: The three population means are not all equal

• H0: pos = neg = cont

Step 2: Find F-Critical

• Step 2.1• Need to first find dfbetween and dfwithin

• Dfbetween = k - 1 (k = number of groups)

• dfwithin = N - k (N = total number of observations)

• dftotal = N - 1

• Check yourself

• dftotal = Dfbetween + dfwithin

• Dfbetween = 2 (k = number of groups)

• dfwithin = 12 (N = total number of observations)

• dftotal = 14

• Check yourself• 14 = 2 + 12

• Step 2.2

• Look up F-critical using table F on pages 370 - 373.

• F (2,12) = 3.88

Step 3: Calculate the F-value

• Has 4 Sub-Steps

• 3.1) Calculate the needed ingredients

• 3.2) Calculate the SS

• 3.3) Calculate the MS

• 3.4) Calculate the F-value

Step 3.1: Ingredients

Positive Feedback

Negative Feedback

Control

10 4 3 6 3 6

PositiveFeedback

NegativeFeedback

Control

10 4 3

Xp = 40 Xn = 25 Xc = 20

X = 85

PositiveFeedback

NegativeFeedback

Control

8 64 5 25 2 4

7 49 6 36 4 16

9 81 7 49 5 25

10 100 4 16 3 9

6 36 3 9 6 36

Xp = 40 Xn = 25 Xc = 20

X = 85

X2 = 555

X2p = 330 X2

n = 135 X2c = 90

T2 = (X)2 for each groupPositive

FeedbackNegativeFeedback

Control

8 64 5 25 2 4

7 49 6 36 4 25

9 81 7 49 5 25

10 100 4 16 3 9

6 36 3 9 6 36

Xp = 40 Xn = 25 Xc = 20

X = 85

X2 = 555

X2p = 330 X2

n = 135 X2c = 90

T2p = 1600 T2

n = 625 T2c = 400

PositiveFeedback

NegativeFeedback

Control

8 64 5 25 2 4

7 49 6 36 4 25

9 81 7 49 5 25

10 100 4 16 3 9

6 36 3 9 6 36

Xp = 40 Xn = 25 Xc = 20

X = 85

X2 = 555

= 2625

X2p = 330 X2

n = 135 X2c = 90

T2p = 1600 T2

n = 625 T2c = 400

NPositive

Control

8 64 5 25 2 4

7 49 6 36 4 25

9 81 7 49 5 25

10 100 4 16 3 9

6 36 3 9 6 36

Xp = 40 Xn = 25 Xc = 20

X = 85

X2 = 555

= 2625

N = 15

X2p = 330 X2

n = 135 X2c = 90

T2p = 1600 T2

n = 625 T2c = 400

nPositive

Control

8 64 5 25 2 4

7 49 6 36 4 25

9 81 7 49 5 25

10 100 4 16 3 9

6 36 3 9 6 36

Xp = 40 Xn = 25 Xc = 20

X = 85

X2 = 555

= 2625

N = 15

X2p = 330 X2

n = 135 X2c = 90

T2p = 1600 T2

n = 625 T2c = 400

Step 3.2: Calculate SSX = 85

X2 = 555

= 2625

N = 15

n = 5• SStotal

Step 3.2: Calculate SS

• SStotal

1573.33

X = 85

X2 = 555

= 2625

N = 15

• SSWithin

X = 85

X2 = 555

= 2625

N = 15

• SSWithin

5552625

X = 85

X2 = 555

= 2625

N = 15

• SSBetween

X = 85

X2 = 555

= 2625

N = 15

• SSBetween

X = 85

X2 = 555

= 2625

N = 15

• Check!

• SStotal = SSBetween + SSWithin

• Check!

• 73.33 = 43.33 + 30

Step 3.3: Calculate MS

221.67

Calculating this Variance Ratio

Step 3.4: Calculate the F value

2.58.67

Step 4: Decision

• If F value > than F critical– Reject H0, and accept H1

• If F value < or = to F critical– Fail to reject H0

Step 4: Decision

F value = 8.67

F crit = 3.88

Step 5: Create the Summary Table

Source SS df MS F

Between 43.33 2 21.67 8.67*

Within 30.00 12 2.5

Total 73.33 14

Step 6: Put answer into words

• Question: Is the type of feedback a person receives significantly (.05) related their self-esteem?

• The type of feedback a person receives is related to their self-esteem

43.333 2 21.667 8.667 .005

30.000 12 2.500

73.333 14

BetweenGroups

WithinGroups

ESTEEM

Sum ofSquares df

MeanSquare F Sig.

Practice

• You are interested in comparing the performance of three models of cars. Random samples of five owners of each car were used. These owners were asked how many times their car had undergone major repairs in the last 2 years.

Results

VW Beetle

Ford Mustang

Geo Metro

Practice

• Is there a significant (.05) relationship between the model of car and repair records?

Step 1: State the Hypothesis

• H0: V = F = G

• Dfbetween = 2 (k = number of groups)

• dfwithin = 12 (N = total number of observations)

• dftotal = 14

• Check yourself• 14 = 2 + 12

• Step 2.2

• Look up F-critical using table F on pages 370 - 373.

• F (2,12) = 3.88

Step 3.1: Ingredients

X = 60X2 = 304Tj

2 = 1400

• N = 15

• n = 5

Step 3.2: Calculate SSX = 60

X2 = 304

= 1400

N = 15

n = 5• SStotal

• SStotal

X = 60

X2 = 304

= 1400

N = 15

• SSWithin

X = 60

X2 = 304

= 1400

N = 15

• SSWithin

3041400

X = 60

X2 = 304

= 1400

N = 15

• SSBetween

X = 60

X2 = 304

= 1400

N = 15

• SSBetween

X = 60

X2 = 304

= 1400

N = 15

• Check!

• SStotal = SSBetween + SSWithin

• Check!

• 64 = 40 + 24

Calculating this Variance Ratio

Step 4: Decision

F value = 10

F crit = 3.88

Step 5: Create the Summary Table

Source SS df MS F

Between 40 2 20 10*

Within 24 12 2

Total 64 14

Step 6: Put answer into words• Question: Is there a significant (.05) relationship

between the model of car and repair records?

• There is a significant relationship between the type of car a person drives and how often the car is repaired

A way to think about ANOVA

• Make no assumption about Ho

– The populations the data may or may not have equal means

VW Beetle

Ford Mustang

Geo Metro

• The samples can be used to estimate the variance of the population

• Assume that the populations the data are from have the same variance

• It is possible to use the same variances to estimate the variance of the populations

222222 ,, GEOGEOFordFordVWVW SSS

222GEOFordVW

VW Beetle

Ford Mustang

Geo Metro

S2 = .50 S2 = .50 S2 = 5.0

222222 ,, GEOGEOFordFordVWVW SSS

222GEOFordVW

)0.550.50(.2

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

• Assume about Ho is true

– The population mean are not different from each other

• They are three samples from the same population– All have the same variance and the same

VW Beetle

Ford Mustang

Geo Metro

2,1,2,3,2,5,4,3,4,4,9,6,3,7,5

Random A

Random B

Random C

2,1,2,3,2,5,4,3,4,4,9,6,3,7,5

For any population of scores, regardless of form, the sampling distribution of the mean will approach a normal distribution a N (sample size) get larger. Furthermore, the sampling distribution of the mean will have a mean equal to and a standard deviation equal to / N

Central Limit Theorem

For any population of scores, regardless of form, the sampling distribution of the mean will approach a normal distribution a N (sample size) get larger. Furthermore, the sampling distribution of the mean will have a mean equal to and a standard deviation equal to / N

Central Limit Theorem

• Central Limit Theorem (remember)

• The variance of the means drawn from the same population equals the variance of the population divided by the sample size.

)( 22Xe Sn

Can estimate population variance from the sample means with the formula

*This only works if the means are from the same population

Random A

Random B

Random C

2 4 6 S2 = 4.00

)( 22Xe Sn

)00.4(520

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

)00.4(520 *Estimates population variance only if the three means are from the same population

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

*Estimates population variance regardless if the three means are from the same population

What do all of these numbers mean?

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

Why do we call it “sum of squares”?

• SStotal

• SSbetween

• SSwithin

• Sum of squares is the sum the squared deviations about the mean

2)( XX

Why do we use “sum of squares”?

2)( XX

SS are additive

Variances and MS are only additive if df are the same

Another way to think about ANOVA

• Think in “sums of squares”

2..)( XXSS ijtotal

Represents the SS of all observations, regardless of the treatment.

VW Beetle

Ford Mustang

Geo Metro

2 4.00 5 1.00 9 25.00

1 9.00 4 .00 6 4.00

2 4.00 3 1.00 3 1.00

3 1.00 4 .00 7 9.00

2 4.00 4 .00 5 1.00

Overall Mean= 4

64..)( 2 XX ij

2: ijtotal

total Sdf

SSNote

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

64..)( 2 XX ij

15 4.0000 4.571

VAR00001

Valid N(listwise)

N Mean Variance

Descriptive Statistics

2..)( XXnSS jbetween

Represents the SS deviations of the treatment means around the grand mean

Its multiplied by n to give an estimate of the population variance (Central limit theorem)

VW Beetle

Ford Mustang

Geo Metro

Overall Mean= 4

408)5(..)( 2 XXn j

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

408)5(..)( 2 XXn j

2)( jijwithin XXSS

Represents the SS deviations of the observations within each group

VW Beetle

Ford Mustang

Geo Metro

2 0 5 1 9 9

1 1 4 0 6 0

2 0 3 1 3 9

3 1 4 0 7 1

2 0 4 0 5 1

Overall Mean= 4

24)( 2 jijwithin XXSS

40.000 2 20.000 10.000 .003

24.000 12 2.000

64.000 14

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

24)( 2 jijwithin XXSS

Sum of Squares

• SStotal

– The total deviation in the observed scores

• SSbetween

– The total deviation in the scores caused by the grouping variable and error

• SSwithin

– The total deviation in the scores not caused by the grouping variable (error)

Conceptual Understanding

Source SS df MS F

Between -- -- -- --

Within 152 -- - -

Total 182 --

Complete the above table for an ANOVA having 3 levels of the independent variable and n = 20. Test for significant at .05.

Conceptual UnderstandingSource SS df MS F

Between 30 2 15 5.62*

Within 152 57 2.67

Total 182 59

Fcrit = 3.18

Complete the above table for an ANOVA having 3 levels of the independent variable and n = 20. Test for significant at .05.

Fcrit (2, 57) = 3.15

• Distinguish between: Between-group variability and within-group variability

• Between concerns the differences between the mean scores in various groups

• Within concerns the variability of scores within each group

Between and Within Group Variability

Between-group variability

Within-group variability

Between and Within Group Variability

sampling error + effect of variable

sampling error

• Under what circumstance will the F ratio, over the long run, approach 1.00? Under what circumstances will the F ratio be greater than 1.00?

• F ratio will approach 1.00 when the null hypothesis is true

• F ratio will be greater than 1.00 when the null hypothesis is not true

Without computing the SS within, what must its value be? Why?

The SS within is 0. All the scores within a group are the same (i.e., there is NO variability within groups)

Example

• Freshman, Sophomore, Junior, Senior

• Measure Happiness (1-100)

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

• Traditional F test just tells you not all the means are equal

• Does not tell you which means are different from other means

Why not

• Do t-tests for all pairs

• Fresh vs. Sophomore• Fresh vs. Junior• Fresh vs. Senior• Sophomore vs. Junior• Sophomore vs. Senior• Junior vs. Senior

Problem

• What if there were more than four groups?

• Probability of a Type 1 error increases.

• Maximum value = comparisons (.05)

• 6 (.05) = .30

Chapter 12

• A Priori and Post Hoc Comparisons

• Multiple t-tests• Linear Contrasts• Orthogonal Contrasts• Trend Analysis• Bonferroni t• Fisher Least Significance Difference• Studentized Range Statistic• Dunnett’s Test

Multiple t-tests

• Good if you have just a couple of planned comparisons

• Do a normal t-test, but use the other groups to help estimate your error term

• Helps increase you df

Remember

within221

ProofCandy Gender

5.00 1.004.00 1.007.00 1.006.00 1.004.00 1.005.00 1.001.00 2.002.00 2.003.00 2.004.00 2.003.00 2.002.00 2.00

6 5.1667 1.1690 .4773

6 2.5000 1.0488 .4282

GENDER1.00

CANDYN Mean

Std.Deviation

Std. ErrorMean

Group Statistics

.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953

4.159 9.884 .002 2.6667 .6412 1.2358 4.0976

Equalvariancesassumed

Equalvariancesnotassumed

CANDYF Sig.

Levene's Test forEquality of Variances

t dfSig.

(2-tailed)Mean

DifferenceStd. ErrorDifference Lower Upper

95% ConfidenceInterval of the Mean

t-test for Equality of Means

Independent Samples Test

21.333 1 21.333 17.297 .002

12.333 10 1.233

33.667 11

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953

4.159 9.884 .002 2.6667 .6412 1.2358 4.0976

CANDYF Sig.

t dfSig.

(2-tailed)Mean

21.333 1 21.333 17.297 .002

12.333 10 1.233

33.667 11

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

t = 2.667 / .641 = 4.16

.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953

4.159 9.884 .002 2.6667 .6412 1.2358 4.0976

CANDYF Sig.

t dfSig.

(2-tailed)Mean

21.333 1 21.333 17.297 .002

12.333 10 1.233

33.667 11

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

t = 2.667 / .641 = 4.16

within221

.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953

4.159 9.884 .002 2.6667 .6412 1.2358 4.0976

CANDYF Sig.

t dfSig.

(2-tailed)Mean

21.333 1 21.333 17.297 .002

12.333 10 1.233

33.667 11

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

t = 2.667 / .641 = 4.16

6)233.1(2

5.217.5 t

.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953

4.159 9.884 .002 2.6667 .6412 1.2358 4.0976

CANDYF Sig.

t dfSig.

(2-tailed)Mean

21.333 1 21.333 17.297 .002

12.333 10 1.233

33.667 11

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

t = 2.667 / .641 = 4.16

16.4641.

.027 .873 4.159 10 .002 2.6667 .6412 1.2380 4.0953

4.159 9.884 .002 2.6667 .6412 1.2358 4.0976

CANDYF Sig.

t dfSig.

(2-tailed)Mean

21.333 1 21.333 17.297 .002

12.333 10 1.233

33.667 11

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

Also, when F has 1 df between Ft 2tF

Within Variability

• Within variability of all the groups represents “error”

• You can therefore get a better estimate of error by using all of the groups in your ANOVA when computing a t-value

within221

Note: This formula is for equal n

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

Hyp 1: Juniors and Seniors will have different levels of happiness

Hyp 2: Seniors and Freshman will have different levels of happiness

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

within221

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

6)90.100(2

8562 t

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

2397.3

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

2397.3

t crit (20 df) = 2.086

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

2397.3

t crit (20 df) = 2.086

Juniors and seniors do have significantly different levels of happiness

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

within221

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

6)90.100(2

8576 t

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

t crit (20 df) = 2.086

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

t crit (20 df) = 2.086

Freshman and seniors do not have significantly different levels of happiness

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

HAPPY76.0000

9.8793

72.0000

13.3417

62.0000

8.2219

85.0000

7.7717

73.7500

12.6052

Std.Deviation

Report

1636.500 3 545.500 5.406 .007

2018.000 20 100.900

3654.500 23

BetweenGroups

WithinGroups

Sum ofSquares df

MeanSquare F Sig.

Hyp 1: Juniors and Sophomores will have different levels of happiness

Hyp 2: Seniors and Sophomores will have different levels of happiness

PRACTICE!

Practice

• 11.1

• Figure out if 5 days is different than 35 days.

Practice

Source SS df MS F

Between 2100 2 1050 40.13*

Within 392.5 15 26.17

Total 2492.5 17

* p < .05

Six Easy Steps for an ANOVA

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