Module 10 – examples

 

One Way Anova Table

 

Variation

Degree of Freedom

Sum of Squares

Mean Square

F

Explained by treatments

(between columns)

c-1

SST = # rows x [(column avg-Grand mean)2 + (column avg –Grand mean)2}

 

MST=SST/c-1

MST/MSE

Error or unexplained

(within columns)

(r-1)c

SSE = (treatment - column avg)2 + (treatment – column avg)2

Or (SS-SST)

 

MSE = SSE/(r-1)c

 

 

Total

Rc-1

SS    = (treatment – Grand mean)2 + (treatment – Grand mean)2

Or (SST + SSE)

 

 

 

 

Two Way Anova Table

 

Variation

Degree of Freedom

Sum of Squares

Mean Square

F

Explained by treatments

(between columns)

c-1

SST =# rows x[ (column avg-Grand mean)2 + (column avg –Grand mean)2 ]

MST=SST/c-1

MST/MSE

Explained by blocks

(between rows)

r-1

SSB =# columns x[ (row avg - Grand mean)2 + (row avg – Grand mean)2]

 

MSB = SSB / r –1

 

MSB/MSE

Error or unexplained

(within columns)

(r-1)(c-1)

SSE = SS-SST-SSB

 

MSE = SSE/(r-1)(c-1)

 

 

Total

Rc-1

SS    = (treatment – Grand mean)2 + (treatment – Grand mean)2

 

 

 

 

Critical F Test

F table (c-1 , error of unexplained within columns) 5% significance level

 

Column                                                           Across Top (c-1)

error of unexplained within columns            5%answer

 

Example:

Hypothesis:  Average examination score for each package comes from populations with the same mean.  It will therefore indicate whether the packages make a difference to eventual examination results.

 

 

A

B

C

Average

School 1

64

61

55

60

School 2

69

63

63

65

School 3

72

68

70

70

School 4

58

65

60

61

School 5

64

60

59

61

School 6

63

61

59

61

Average

65

63

61

Grand mean = 63

 

SST =  # rows x[(column avg-Grand mean)2 + (column avg –Grand mean)2]

 

SST = 6 x [ (65-63)2 +63-63)2 + (61-63)2]

SST = 6 x [ (2)2 + (0)2 + (-2)2]

SST = 6 x [4+4]

SST = 6 x 8

SST = 48

 

SSB = # columns x[( (row avg - Grand mean)2 + (row avg – Grand mean)2]

 

SSB = 3 x [ (60-63)2 +(65-63)2 + (70-63)2 + (61-63)2 + (61-63)2 + (61-63)2]

SSB = 3 x [ (-3)2 + (2)2 +(7)2 + (-2) 2 + (-2)22 + (-2)2]

SSB = 3 x [9+4+49+4+4+4]

SSB = 3 x 74

SSB = 222

 

SS = (treatment – Grand mean)2 + (treatment – Grand mean)2

 

 

A

 

 

B

 

 

C

 

School 1

(64-63)2=

1

 

(61-63)2=

4

 

(55-63)2=

64

School 2

(69-63)2=

36

 

(63-63)2=

0

 

(63-63)2=

    0

School 3

(72-63)2=

81

 

(68-63)2=

25

 

(70-63)2=

49

School 4

(58-63)2=

25

 

(65-63)2=

4

 

(60-63)2=

9

School 5

(64-63)2=

1

 

(60-63)2=

9

 

(59-63)2=

16

School 6

(63-63)2=

0

 

(61-63)2=

4

 

(59-63)2=

16

 

 

144

 

 

46

 

 

154

 

SS = 144 + 46 + 154

SS =  344

 

Or for 1 way Anova

SS = SST + SSE

 

SS = 48 + 296

SS = 344

 

SSE = (treatment - column avg)2 + (treatment – column avg)2

 

 

A

 

 

B

 

 

C

 

School 1

(64-65)2=

1

 

(61-63)2=

4

 

(55-61)2=

36

School 2

(69-65)2=

16

 

(63-63)2=

0

 

(63-61)2=

4

School 3

(72-65)2=

49

 

(68-63)2=

25

 

(70-61)2=

81

School 4

(58-65)2=

49

 

(65-63)2=

4

 

(60-61)2=

1

School 5

(64-65)2=

1

 

(60-63)2=

9

 

(59-61)2=

4

School 6

(63-65)2=

4

 

(61-63)2=

4

 

(59-61)2=

4

 

 

120

 

 

46

 

 

130

 

SSE = 120 + 46 + 130 = 296

 

Or for 1 way Anova

SSE = SS – SST

 

For 2 way Anova

SSE = SS - SST – SSB

 

 

One Way Anova Table compares difference between treatment columns

Hypothesis:  Average examination score for each package comes from populations with the same mean.  It will therefore indicate whether the packages make a difference to eventual examination results.

 

Critical F Test

F table (2,15) 5% significance level

Column           Across Top 2

15                                3.59   

 

Variation

Degree of Freedom

Sum of Squares

Mean Square

F

Explained by treatments

(between columns)

c-1

3-1=2

SST = =# rows x[ (column avg-Grand mean)2 + (column avg –Grand mean)2 ]SST =    48

MST=SST/c-1

MST = 48/2= 24

MST/MSE

24/19.73 = 1.2164

Error or unexplained

(within columns)

(r-1)c 

(6-1)3=15

SSE = (treatment - column avg)2 + (treatment – column avg)2

Or (SS-SST)

SSE = 296

MSE = SSE/(r-1)c

MSE = 296/15=19.73

 

Total

rc-1   

(6 x 3)-1=17

SS    = (treatment – Grand mean)2 + (treatment – Grand mena)2

Or (SST + SSE)

SS     = 344

 

 

 

Value 1.22 is much smaller.  The hypothesis is accepted.  There is no significant difference in the examination results.  The packages do not appear to make a difference to examination results.

 

Two Way Anova Table compares difference between treatment columns and blocks (rows)

Above the fact  that treatment group came form populations with the same mean was accepted.  However, the experiment was carried out in several schools.  It could well be a large part of the variation between the groups was accounted for by difference between  schools. 

 

A two way anova table will neutralize and remove from calculation example difference between schools

 

Hypothesis:  Average examination score for each package comes from populations with different mean so we will remove the school parameter from the equation the same mean.  It will therefore indicate whether the packages make a difference to eventual examination results.

 

Critical F Test

F table (2,10) 5% significance level

Column           Across Top 2

10                                4.10   

 

 

 

Variation

Degree of Freedom

Sum of Squares

Mean Square

F

Explained by treatments

(between columns)

c -1

3 -1=2

SST = # rows x[ (column avg-Grand mean)2 + (column avg –Grand mean)2 ]SST = 48

MST=SST/c-1

MST = 48/2= 24

MST/MSE

24/7.4 = 3.2432

Explained by blocks

(between rows)

r- 1

6 –1= 5

SSB =# columns x[ (row avg - Grand mean)2 + (row avg – Grand mean)2]

SSB = 222

MSB = SSB / r –1

MSB =   222 /5 = 44.4

MSB / MSE

44.4 / 7.4 = 6.0

Error or unexplained

(within columns)

(r-1)(c-1) 

(6-1)(3-1)=10

SSE = SS-SST-SSB

SSE = 74

MSE = SSE/(r-1)(c-1) 

MSE = 74/10= 7.4

 

Total

rc-1   

(6 x 3)-1=17

SS    = (treatment – Grand mean)2 + (treatment – Grand mean)2

SS    = 344

 

 

 

Value 3.24 is less than 4.10 .  The hypothesis is accepted.  There is no significant difference in the examination results.  The packages do not appear to make a difference to examination results