Chapter 5: ANOVA
1. A using the formulas for the One Way Repeated Measures ANOVA below,
Calculate the F statistic for the data.
N
1
2
3
4
5
6
Group1
8
7
8
6
9
7
Step 1: Calculate sum of all scores squared
∑X2 
Answer:
Total Squares= 660.
Step 2: Calculate the correction factor

∑(X) 2 / N
G= Grand total of all of the scores added together
N= number of scores in the table
Answer:
Correction Factor =10816/18
=600.8889
Step 3: Calculate the Model Sum of Squares


SSModel ∑C2 / Ncorrection - factor
DoFModel= k-1
Where:
C= Total score per condition (i.e. column sum)
N= Amount of participants in the experiment (6)
k= Amount of conditions (in this case, groups)
DoFModel= Model Degrees of Freedom
Answer:
C - Group 1= 45; C Group 2= 36; C Group 3= 23
SSModel= ((2025+1296+529) / 6)- 600.889
= ((2025+1296+529) / 6)- 600.889
Group2
6
7
5
5
8
5
Goup3
5
3
3
5
4
3
= ((3850) / 6)- 600.889
= 641.6667-600.889
= 40.77778
DoFModel= 3-1=2
Step 4: Calculate the Error Sum of Squares

SSError ∑X2 - ∑ 𝑐 2 / N - ∑ 𝑝2 / K + correction - Factor
DoFError= (k-1)(N-1)
Where:
P= total score per participant (i.e. row sum)
N= Number of participants in the experiment
k= number of conditions in the experiment
DoFError= Error Degrees of Freedom
Answer:
P- Participant (row) 1= 19; Participant (row) 2=17; Participant (row) 3=16
Participant (row) 4=16; Participant (row) 5=21; Participant (row) 6=15
SSerror = 660- 641.6667-((361+289+256+256+441+225)/3)+600.889
SSerror = 660- 641.6667-(1828/3)+600.889
SSerror = 660- 641.6667-609.3333+600.889
SSerror = 9.889
DoFError= 2x5=10
Step 5: Calculate Mean Square Model & Mean Square Error

MSModel SSModel lDoFmod el

MSError SSErrorDoFError
Answer:
MS model= 40.77778/2 = 20.38889
MS error= 9.889/10 = 0.9889
Step 6: Calculate the F Ratio
F MSModel / MSError

Answer:
F=20.38889
Chapter6: Chi-Square
Let X be the number of defects in printed circuit boards. A random sample of n = 60
Printed circuit boards are taken and the number of defects recorded. The results were as
follows:
Number Of Defects
0
1
2
3
Observed Frequency
32
15
9
4
Does the assumption of a Poisson distribution seem appropriate as a model for these data?
The null hypothesis is HO : X ------- Poisson
The alternative hypothesis is H1 : X ------------ does not follow a Poisson distribution.
Calculating Poisson distribution Probabilities:
μ = (32 × 0) + (15 × 1) + (9 × 2) + (4 × 3)/60 = 0.75
p0 = P(X = 0) = 𝑒 −0.75(0.750 ) / 0! = 0.472
E= 0.472*60= 28.32
p1 = P(X = 1) =𝑒 −0.75(0.751 ) / 1! =0.354
E1 = 0.354 × 60 = 21.24
p2 = P(X = 2) =𝑒 −0.75(0.752 ) / 2! = 0.133
E2 = 0.133 × 60 = 7.98
p3 = P(X _ 3) = 1 − (p0 + p1 + p2) = 0.041 E3 = 0.041 × 60 = 2.46
Calculating 𝑥 2
Number of Defects
0
1
2 or more
Observed Frequency
32
15
9
Expected Frequency
28.32
21.24
10.44
𝑥 2 = ∑(𝑂 − 𝐸)2 /𝐸
(32 − 28.32)2/28.32 +(15 − 21.24)2/21.24 +(13 − 10.44)2/10.44 = 2.94
Df= K – P – 1
df= 3-1-1=1
P-value from the table = 0.1
0.1>0.05
Decision: Reject H0