Psych 230 Psychological Measurement and Statistics Pedro Wolf November 18, 2009

Psych 230

Psychological Measurement and Statistics

Pedro Wolf

November 18, 2009

This Time….

• Analysis of Variance (ANOVA)

• Concepts of variability

• Why bother with ANOVA?

• Conducting a test

Statistical Testing

1. Decide which test to use2. State the hypotheses (H0 and HA)

3. Calculate the obtained value4. Calculate the critical value (size of )5. Make our conclusion

Statistical Testing

1. Decide which test to use2. State the hypotheses (H0 and HA)

3. Calculate the obtained value4. Calculate the critical value (size of )5. Make our conclusion6. Conduct post-hoc tests

Analysis of Variance (ANOVA)

Analysis of Variance

• In this statistical test, we are interested in seeing if

there are significant differences between more

than two groups

• In an experiment involving only two conditions of

the independent variable, you can use either a t-

test or the ANOVA

• We will look at the variance within each group and

compare that to the variance found between the

groups

• Remember: Variance is the degree to which scores

are dispersed in our data

• Remember: H0 is that all the group means in our

experiment are the same

– any difference between them is due to random chance

• H1 is that there is a difference between our group

– a difference that is so unlikely to have happened by chance

that we conclude it is due to the independent variable

• There is always variability in our data

• This variability can be due to two factors:

1. The independent variable• Systematic factors

2. Error variance• Random factors

• So, to draw a conclusion about whether the

independent variable makes a difference to the

dependent variable, we need to know what kind

of variance there is in our data

– variance due to the independent variable (systematic)

– variance due to random factors (error)

• To assess this, we need to look at the variance both

within each of our experimental conditions and

also at the variance between each of our

experimental conditions

• Assume H0 is true - there is no effect of our

independent variable

• What type of variance might we expect to see?

• Example: we want to see if people differ in their shoe

size by where they sit in the class

– three groups of students: front row, middle row and back

– expect to find significant differences in shoe size?

• Front row: vary in their shoe size: 6,8,5,4,7

• Middle row: vary in their shoe size : 9,9,4,6,7

• Back row: vary in their shoe size : 4,6,12,10,8

Analysis of VarianceBack Middle Front

5 4 12

4 6 10

6 7 8 Mean=7

5 4 12

4 6 10

6 7 8 Mean=7

Total variance

5 4 12

4 6 10

6 7 8 Mean=7

Variance within the groups

Total variance

Variance between the groups

Back Middle Front

5 4 12

4 6 10

6 7 8 Mean=7

Total variance

5 4 12

4 6 10

6 7 8 Mean=7

Total Variance =

Variance between + Variance within

Total variance

• So, when H0 is actually true, we should expect the

same amount of variance both within each group

and between the groups

• If we divide Variancebetween by Variancewithin, we

should get?

• If H0 is true, this ratio should be close to 1

• How about if H1 is actually true?

• In this case, we know the independent variable is

having some effect

• So, we should expect more variance between each

group than there is within each group

• Example: we want to see if people differ in their

attendance by where they sit in the class

– front row, middle row and back row

– expect to find significant differences in attendance?

• Front row: vary in their attendance: 6,5,7,6,6

• Middle row: vary in their attendance: 7,8,6,7,7

• Back row: vary in their attendance: 8,8,7,9,8

Back Middle Front

6 7 8 Mean=7

Total Variance =

Variance between + Variance within

Total variance

• When HA is true, we have more variance between

each group than there is within each group

• If we divide Variancebetween by Variancewithin, we should

• If HA is true, this ratio should be more than 1

– the F-ratio

• A one-way ANOVA is performed when only one

independent variable is tested in the experiment

• Example: we are interested in the differences

between freshmen, sophomores and juniors on

tests of socialization

– dependent variable: socialization scores

– independent variable: class standing

• A two-way ANOVA is performed when two

independent variables are tested in the experiment

• Example: we are interested in the differences

between male and female freshmen, sophomores

and juniors on tests of socialization

– dependent variable: socialization scores

– independent variable 1: class standing

– independent variable 2: gender

• When an independent variable is studied using independent samples in all conditions, it is called a between-subjects factor

• A between-subjects factor involves using the formulas for a between-subjects ANOVA

• When a factor is studied using related (dependent) samples in all levels, it is called a within-subjects factor

• This involves a set of formulas called a within-subjects ANOVA

ANOVA - assumptions

1. All conditions contain independent samples

2. The dependent scores are normally distributed, interval or ratio score

3. The variances of the populations are homogeneous

ANOVA - why bother?

• We want to see if there are differences between our three groups:– Freshmen

– Sophomores

– Juniors

• Why not just do a bunch of t-tests?– Freshmen vs. Sophomores– Freshmen vs. Juniors– Sophomores vs. Juniors

ANOVA - experiment-wise error

• The overall probability of making a Type I error somewhere in an experiment is call the experiment-wise error rate

• When we use a t-test to compare only two means in an experiment, the experiment-wise error rate equals

ANOVA - experiment-wise error

• When there are more than two means in an experiment, the multiple t-tests result in an experiment-wise error rate that is much larger than the we have selected– Freshmen vs. Sophomores: =0.05– Freshmen vs. Juniors: =0.05– Sophomores vs. Juniors: =0.05– experiment-wise error rate = 0.05+0.05+0.05=approx 0.15

• Using the ANOVA allows us to compare the means from all levels of the factor and keep the experiment-wise error rate equal to

Conducting an ANOVA

1. Decide which test to use

• Are we comparing a sample to a population?– Yes: Z-test if we know the population standard deviation– Yes: One-sample T-test if we do not know the population std dev– No: Keep looking

• Are we looking for the difference between samples?– Yes: How many samples are we comparing?

• Two: Use the Two-sample T-test– Are the samples independent or related?

» Independent: Use Independent Samples T-test» Related: Use Related Samples T-test

• More than Two: Use Anova test

2. State the Hypotheses

• H0 : 1 = 2 = ……. = k

– there is no difference in the means

• HA : not all s are equal – there is a difference between some of the means

• Only conduct two-tailed tests using ANOVA

3. Calculate the obtained value (Fobt)

• The statistic for the ANOVA is F

• When Fobt is significant, this indicates only that somewhere among the means at least two of them differ significantly

• It does not indicate which groups differ significantly

• When the F-test is significant, we perform post hoc comparisons (step 6)

• Remember, we are trying to compare the between group variance to the within group variance

• We use the mean squares to calculate this

• The mean square within groups is an estimate of the variability in scores as measured by differences within the conditions

• The mean square between groups is an estimate of the differences in scores that occurs between the levels in a factor

• The F-ratio is therefore the mean square between groups divided by the mean square within groups

bnobt MS

• When H0 is true, Fobt should be close to 1

• When HA is true, Fobt should be greater than 1

bnobt MS

• The ANOVA table:

Source Sum of Squares

df Mean Squares

Between SSbn dfbn MSbn Fobt

Within SSwn dfwn MSwn

Total SStot dftot

df Mean Squares

Total SStot dftot

bnobt MS

df Mean Squares

Total SStot dftot

bnobt MS

bnbn df

wnwn df

df Mean Squares

Total SStot dftot

bnobt MS

bnbn df

wnwn df

dftot= N - 1

dfwn= N - k

dfbn= k - 1

df Mean Squares

Total SStot dftot

bnobt MS

bnbn df

wnwn df

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bntotwn SSSSSS

columnin

)columnin(

tottot

6 7 8 Mean=7

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751 dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

bntotwn SSSSSS

columnin

)columnin(

tottot

df Mean Squares

Total SStot dftot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

columnin

)columnin(

df Mean Squares

Total SStot dftot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

columnin

)columnin(

df Mean Squares

Total SStot dftot

= [(302/5) + (352/5) + (402/5)] - (1052/15)

= [(900/5) + (1225/5) + 1600/5)] - (11025/15)

= [180 + 245 + 320] - (735)

= 745 - 735 = 10

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

columnin

)columnin(

df Mean Squares

Between 10 dfbn MSbn Fobt

Total SStot dftot

= [(302/5) + (352/5) + (402/5)] - (1052/15)

= [(900/5) + (1225/5) + 1600/5)] - (11025/15)

= [180 + 245 + 320] - (735)

= 745 - 735 = 10

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total SStot dftot

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total SStot dftot

= (751) - (1052/15)

= 751 - 735

= 751 - 735 = 16

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total 16 dftot

= (751) - (1052/15)

= 751 - 735

= 751 - 735 = 16

tottot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total 16 dftot

= 16 - 10

bntotwn SSSSSS

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 dfwn MSwn

Total 16 dftot

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 dfwn MSwn

Total 16 dftot

= 3 - 1

dfbn= k - 1

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 MSbn Fobt

Within 6 dfwn MSwn

Total 16 dftot

= 3 - 1

dfbn= k - 1

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 dfwn MSwn

Total 16 dftot

= 15 - 3

dfwn= N - k

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 12 MSwn

Total 16 dftot

= 15 - 3

dfwn= N - k

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 12 MSwn

Total 16 dftot

= 15 - 1

dftot= N - 1

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 12 MSwn

Total 16 14

= 15 - 1

dftot= N - 1

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 12 MSwn

Total 16 14

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 6 12 MSwn

Total 16 14

= 10 / 2

bnbn df

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 Fobt

Within 6 12 MSwn

Total 16 14

= 10 / 2

bnbn df

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 Fobt

Within 6 12 MSwn

Total 16 14

= 6 / 12

wnwn df

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 Fobt

Within 6 12 0.5

Total 16 14

= 6 / 12

wnwn df

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 Fobt

Within 6 12 0.5

Total 16 14

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 Fobt

Within 6 12 0.5

Total 16 14

= 5 / 0.5

bnobt MS

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 10

Within 6 12 0.5

Total 16 14

= 5 / 0.5

bnobt MS

6 7 8 Mean=7

∑X=30 ∑X=35 ∑X=40 ∑Xtot=105

∑X2=182 ∑X2=247 ∑X2=322 ∑X2tot=751

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 10 2 5 10

Within 6 12 0.5

Total 16 14

4. Calculate the critical value

• Assume =0.05• Always a two-tail test with ANOVA

• dfbetween = k - 1• dfwithin = N - k

• k is the number of levels, or groups• N is the total number of subjects

df betweendf within 1 2 3 4 5 12 0.05 4.75 3.88 3.49 3.26 3.11

Fcrit and Fobt

Fcrit= 3.88

Fobt= 10

5. Make our Conclusion

• Fcrit = 3.88• Fobt = 10

• If Fobt is outside the rejection region, we retain H0

• If Fobt is inside the rejection region, we reject H0 and accept HA

• We conclude that there is a significant difference between some of our groups– but which groups?

6. Conduct post-hoc tests

• To determine which means differ from each other significantly, we conduct post-hoc tests

• Post hoc comparisons are like t-tests

– we compare all possible pairs of means from a factor, one pair at a time

• There are many possible post-hoc tests to use:

– Tukey’s HSD

– Scheffe

– Bonferroni

• When the n’s in all levels of the factor are equal, we can use Tukey’s HSD test

– “honestly significant difference”

• Calculate the difference between each pair of means

• Compare each difference between the means to the HSD

• If the absolute difference between two means is greater than the HSD, then these means differ significantly

• Calculating the HSD:

where qk is found using the appropriate table

MSqHSD k

HSD = (qk) [√(MSwn/n)]

qk: Look up Table L pg. 562 - Need:

– k (number of means being compared)

– dfwn

= (3.77) [√(0.5/5)]

= (3.77) [√(0.1)]

= (3.77) (0.316)

= 1.19

MSqHSD k

• HSD = 1.19

• Difference between means:

• ‘Back’ (mean=6) and ‘Middle’ (mean=7) : 1

• ‘Middle’ (mean=7) and ‘Front’ (mean=8): 1

• ‘Back’ (mean=6) and ‘Front’ (mean=8): 2

• Because 2 is greater than 1.19, there is a significant difference between the ‘front’ and ‘back’ groups in terms of their attendance

ANOVA - example

A clinical psychologist has noted that autistic children seem to respond to treatment better if they are in a familiar environment. To evaluate the influence of environment, the psychologist selects a group of 18 children who are currently in treatment and randomly divides them into three groups. One group continues to receive treatment in the clinic as usual. For the second group, treatment sessions are conducted entirely in the child’s home. The third group gets half of the treatment in the clinic and half at home. After six weeks the data look as follows (high scores are good):

– Clinic: 14, 13, 14, 13, 10, 15

– Home: 10, 12, 11, 13, 11, 10

– Both: 11, 14, 14, 15, 13, 15

Do the data indicate any significant differences between the three settings? Use =0.05

• More than Two: Use Anova test

• H0 : 1 = 2 = ……. = k

3. Calculate the obtained value (Fobt)Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229 dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

df Mean Squares

Total SStot dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

df Mean Squares

Total SStot dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

columnin

)columnin(

= [(802/6) + (672/6) + (822/6)] - (2292/18)

= [(6400/6) + (4489/6) + 6724/6)] - (52441/18)

= [1066.67 + 748.17 + 1120.67] - (2913.39)

= 2935.51 - 2913.39 = 22.12

df Mean Squares

Between 22.11 dfbn MSbn Fobt

Total SStot dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

columnin

)columnin(

= [(802/6) + (672/6) + (822/6)] - (2292/18)

= [(6400/6) + (4489/6) + 6724/6)] - (52441/18)

= [1066.67 + 748.17 + 1120.67] - (2913.39)

= 2935.51 - 2913.39 = 22.11

df Mean Squares

Total SStot dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= (2969) - (2292/18)

= 2969 - 2913.39

= 55.61

tottot

df Mean Squares

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= (2969) - (2292/18)

= 2969 - 2913.39

= 55.61

tottot

df Mean Squares

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 55.61 - 22.12

= 33.49

bntotwn SSSSSS

df Mean Squares

Within 33.49 dfwn MSwn

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 55.61 - 22.12

= 33.49

bntotwn SSSSSS

df Mean Squares

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 3 - 1

dfbn= k - 1

df Mean Squares

Between 22.11 2 MSbn Fobt

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 3 - 1

dfbn= k - 1

df Mean Squares

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 18 - 3

dfwn= N - k

df Mean Squares

Within 33.49 15 MSwn

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 18 - 3

dfwn= N - k

df Mean Squares

Total 55.61 dftot

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 18 - 1

dftot= N - 1

df Mean Squares

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 18 - 1

dftot= N - 1

df Mean Squares

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 22.12 / 2

= 11.06

bnbn df

df Mean Squares

Between 22.11 2 11.055 Fobt

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 22.12 / 2

= 11.06

bnbn df

df Mean Squares

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 33.49 / 15

= 2.23

wnwn df

df Mean Squares

Within 33.49 15 2.23

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 33.49 / 15

= 2.23

wnwn df

df Mean Squares

Within 33.49 15 2.23

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 11.06 / 2.23

= 4.96

bnobt MS

df Mean Squares

Between 22.11 2 11.055 4.957

Within 33.49 15 2.23

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

= 11.06 / 2.23

= 4.96

bnobt MS

df Mean Squares

Between 22.11 2 11.05 4.96

Within 33.49 15 2.23

Total 55.61 17

Clinic Home Both

14 10 11

13 12 14

14 11 14

14 13 15

10 11 13

15 10 15

13.33 11.17 13.67 Mean=12.72

∑X=80 ∑X=67 ∑X=82 ∑Xtot=229

∑X2=1082 ∑X2=755 ∑X2=1322 ∑X2tot=2969

n1=6 n2=6 n3=6 N = 18k=3

• dfbetween = k - 1 = 3 - 1 = 2• dfwithin = N - k = 18 - 3 = 15

• Fcrit for 2 and 15 degrees of freedom and = 0.05 is 3.68

Fcrit and Fobt

Fcrit= 3.68

Fobt= 4.96

• Fcrit = 3.68• Fobt = 4.96

• Fobt is inside the rejection region, we reject H0 and accept HA

• We conclude that there is a significant difference between some of our groups– which groups?

qk: Look up Table L - Need:

– dfwn

= (3.675) [√(2.233/6)]

= (3.675) [√(0.372)]

= (3.675) (0.61)

= 2.242

MSqHSD k

• HSD = 2.242

• Clinic - Home = 2.166

• Clinic - Both = -0.334

• Home - Both = -2.5

• Therefore, there is a significant difference in scores between treatment at home, and treatment at both home and at the clinic

ANOVA - your turn

Problem

A psychologist is interested in the effects of room temperature on learning. 15 subjects are recruited for an experiment to answer this questions - 5 subjects undergo a learning task in a room with a temperature of 50 degrees, another group of 5 learn the same material in room with a temperature of 70 degrees and the final group learn in a temperature of 90 degrees. The data are below, with higher numbers indicating greater learning. Does room temperature affect rates of learning? Use =0.05

– 50 degrees: 0, 1, 3, 1, 0

– 70 degrees : 4, 3, 6, 3, 4

– 90 degrees : 1, 2, 2, 0, 0

Analysis of Variance50° 70° 90°

dftot= N - 1

dfwn= N - k

dfbn= k - 1

bnbn df

wnwn df

bnobt MS

bntotwn SSSSSS

columnin

)columnin(

tottot

• More than Two: Use Anova test– No: Keep looking

• Are we looking for the relationship between variables?– Yes: Use the Correlation test

• H0 : 1 = 2 = ……. = k

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total SStot dftot

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total SStot dftot

columnin

)columnin(

= [(52/5) + (202/5) + (52/5)] - (302/15)

= [(25/5) + (400/5) + 25/5)] - (900/15)

= [5 + 80 + 5] - (60)

= 90 - 60 = 30

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total SStot dftot

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

columnin

)columnin(

= [(52/5) + (202/5) + (52/5)] - (302/15)

= [(25/5) + (400/5) + 25/5)] - (900/15)

= [5 + 80 + 5] - (60)

= 90 - 60 = 30

df Mean Squares

Total SStot dftot

= (106) - (302/15)

= 106 - 60

tottot

)(50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total 46 dftot

= (106) - (302/15)

= 106 - 60

tottot

)(50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Total 46 dftot

= 46 - 30

bntotwn SSSSSS 50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 dfwn MSwn

Total 46 dftot

= 46 - 30

bntotwn SSSSSS 50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 dfwn MSwn

Total 46 dftot

= 3 - 1

dfbn= k - 150° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 dfwn MSwn

Total 46 dftot

= 3 - 1

dfbn= k - 150° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 dfwn MSwn

Total 46 dftot

= 15 - 3

dfwn= N - k50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 12 MSwn

Total 46 dftot

= 15 - 3

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

dftot= N - 1

df Mean Squares

Within 16 12 MSwn

Total 46 dftot

= 15 - 1

dftot= N - 150° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 12 MSwn

Total 46 14

= 15 - 1

dftot= N - 150° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 12 MSwn

Total 46 14

= 30 / 2

bnbn df

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 30 2 15 Fobt

Within 16 12 MSwn

Total 46 14

bnbn df

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

= 30 / 2

df Mean Squares

Within 16 12 MSwn

Total 46 14

= 16 / 12

= 1.33

wnwn df

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Within 16 12 1.33

Total 46 14

wnwn df

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

= 16 / 12

= 1.33

df Mean Squares

Within 16 12 1.33

Total 46 14

= 15 / 1.33

= 11.28

bnobt MS

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 30 2 15 11.28

Within 16 12 1.33

Total 46 14

= 15 / 1.33

= 11.28

bnobt MS

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

df Mean Squares

Between 30 2 15 11.28

Within 16 12 1.33

Total 46 14

50° 70° 90°

1 4 1 Mean=2

∑X=5 ∑X=20 ∑X=5 ∑Xtot=30

∑X2=11 ∑X2=86 ∑X2=9 ∑X2tot=106

n1=5 n2=5 n3=5 N = 15k=3

• dfbetween = k - 1 = 3 - 1 = 2• dfwithin = N - k = 15 - 3 = 12

• Fcrit for 2 and 15 degrees of freedom and = 0.05 is 3.88

Fcrit and Fobt

Fcrit= 3.88

Fobt= 11.28

• Fcrit = 3.88• Fobt = 11.28

• Fobt is inside the rejection region, we reject H0 and accept HA

• We conclude that there is a significant difference between some of our groups– which groups?

qk: Look up Table L - Need:

– dfwn

= (3.77) [√(1.33/5)]

= (3.77) [√(0.266)]

= (3.77) (0.52)

= 1.96

MSqHSD k

• HSD = 1.96

• 50° group - 70° group = -3

• 50° group - 90° group = 0

• 70° group - 90° group = 3

• Therefore, there is a significant increase in learning when the temperature is 70° as compared to either 50° or 90°

Psych 230 Psychological Measurement and Statistics Pedro Wolf November 18, 2009

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