Simple Mixed Design - Analysis of Variance Designs - page 367

B 2
( a )( n ) −

T 2
( a )( b )( n )

SS B =

(80) 2 (53) 2

(3)(5)

(133) 2
(3)(2)(5)

+

= −

= . − . =

613 93 589 63 24.30 (13.3)

AB 2
n

A 2
( b )( n ) −

B 2
( a )( n ) +

T 2
( a )( b )( n )

SS A × B = −

(30) 2 (5) 2 (18) 2 (14) 2

+ +···+ +

=

5

(35) 2 (66) 2 (32) 2 (80) 2 (53) 2

(3)(5)

(133) 2
(3)(2)(5)

+ + +

− − +

(2)(5)

= 725 . 00 − 660 . 50 − 613 . 93 + 589 . 63 = 40.20 (13.4)

AB 2
n

AS 2
b

A 2
( b )( n )

− − Y 2 SS B × S / A = +

(5) 2 + (6) 2 +···+ (3) 2 + (3) 2 − (30) 2 + (5) 2 +···+ (18) 2 + (14) 2

=

5

(5) 2 (8) 2 (5) 2 (6) 2 (35) 2 (66) 2 (32) 2

+ +···+ + + +

− +

2 (2)(5)

= . − . − . + . =

743 00 725 00 675 50 660 50 3.00 (13.5)

T 2
( a )( b )( n )

Y 2 SS T = −

(133) 2
(3)(2)(5)

(5) 2 (6) 2 (3) 2 (3) 2

= + +···+ + −

= 743 . 00 − 589 . 63 = 153.37 . (13.6)

13.4.2 CALCULATING DEGREES OF FREEDOM
Below are the formulas for the degrees of freedom associated with each sum
of squares and the simple computations involved based on our numerical
example:

df A = a − 1 = 3 − 1 = 2

df S / A = ( a )( n − 1) = (3)(5 − 1) = 12

df B = b − 1 = 2 − 1 = 1

df A × B = ( a − 1)( b − 1) = (2)(1) = 2

df B × S / A = ( a )( b − 1)( n − 1) = (3)(1)(4) = 12

df T = ( a )( b )( n ) − 1 = (3)(2)(5) − 1 = 29 .