Analysis of Variance ANOVA (F test): It is the method of testing and comparing three or more means. It is an extension of t test for two independent samples. ASSUMPTIONS: The populations from which the samples were obtained must be normally or approximately normally distributed. The samples must be independent of each other. The variances of the population must be equal. SOURCES OF VARIABILITY: Total variation among scores Within group Variation Between group Variation Sample size need not to be equal Its always right tail

2. ANALYSIS OF VARIANCE
(ANOVA)
Shakir Rahman
BScN, MScN, MSc Applied Psychology, PhD Nursing (Candidate)
University of Minnesota USA.
Principal & Assistant Professor
Ayub International College of Nursing & AHS Peshawar
Visiting Faculty
Swabi College of Nursing & Health Sciences Swabi
Nowshera College of Nursing & Health Sciences Nowshera
1

3. EXAMPLE 1
The mean serum creatinine level is measured in 13 white womenwho
received a newly proposed antibiotic was 1.2 mg/dl and the
standard deviation was 0.6 mg/dl.
Another sample of 12 black women who have received anold
antibiotic have mean serum creatinine level of 1.0 mg/dl with
standard deviation of 0.4 mg/dl.
Using a significance level of 0.05, test if the meanserum
creatinine level in white women is different than theblack
women. 2

4. t test for two
independent samples
4

5. EXAMPLE 2
5
The mean serum creatinine level is measured in 13 white women
who received a newly proposed antibiotic was 1.2 mg/dl and
the standard deviation was 0.6 mg/ dl. Another sample of 12
black women who have received an old antibiotic have mean
serum creatinine level of 1.0 mg/dl with standard deviation of
0.4 mg/dl. The third sample of 14 Asian women who have
received placebo have mean serum creatinine level of 0.8
mg/dl with standard deviation of 0.2 mg/dl.
Using a significance level of 0.05, test if the mean
serum creatinine level is different among three groups.

6. Analysis of Variance
ANOVA (F test)
6

7. ANOVA
7
 It is the method of testing and comparing three
or more means.
 It is an extension of t test for two independent
samples.

8. ASSUMPTIONS
8
 The populations from which the samples were
obtained must be normally or approximately
normally distributed.
 The samples must be independent of each other.
 The variances of the population must be equal.

9. SOURCES OF VARIABILITY
Total variation among scores
Within group
Variation
Between group
Variation
9

10. STEP 1
10
Null Hypothesis (Ho)
Ho: µ1=µ2=µ3
Alternate Hypothesis (Ha)
Ha: At least one mean is different fromthe
others
Note:ANOV
Aonly shows that a difference exist among the three
means. It doesn’t reveal where the difference lies

11. STEP 2
11
Critical value:
d.f.N (Numerator) = k – 1
d.f.D (Denominator) = N –k
K = number of groups
N = sum of sample size of the groups
(N=n1+n2+n3+…. all samples)

12.  Sample size need not to be equal
 Its always right tail
12

13. STEP 3 (PART A)
Grand mean
∑X
X GM =
N
∑X= adding all eachvalue
N= total sample size

14. Between GroupVariance
S2
B = ∑ni(Xi – X GM)2
K - 1
XGM = Calculated in part B
STEP 3 (PART B)
14

15. ∑ (ni – 1) s2
i
S2
w =
∑ (ni – 1)
n= total sample size of eachgroup
STEP 3 (PART C)
15

16. STEP 3 (PART D
15 )
S2
B (between group variance)
F =
S2
w (Within group Variance)

17. Step 4
Reject Ho if Fcal is > F tab
Step 5
since Fcal is greater than Ftab we reject Ho….
OR
since Fcal is less than Ftab we fail to reject Ho….
17

18. EXAMPLE
18
A researcher three different
techniques to
wishes to compare
lower the blood pressure of
individual diagnosed with high blood pressure. The
subjects are randomly assigned to three groups;
the first group takes medication, the second group
does exercise, and the third group uses diet to
recorded after four weeks to test
control BP
. Reduction in each person’s BP is
the claim that
there is no difference among the mean reduction of
significance level.
BP in these groups. Use a 0.05
The data are following:

19. EXAMPLE
Medication Exercise Diet
10 6 5
12 8 9
9 3 12
15 0 8
13 2 4
e
Within group B tween group variation
variation 19

20. STEP 1
20
Ho: µ1=µ2=µ3
Ha: At least one mean is different from theothers

21. STEP 2
21
Critical value:
d.f.N = k – 1 = 3-1 = 2
d.f.D = N –k = 15 – 3 = 12
Critical value is 3.89 [F(tab)] obtained from the
table with α=0.05

22. 22

23. Medication (n1 = 5) Exercise (n2 = 5) Diet (n3 = 5)
10 6 5
12 8 9
9 3 12
15 0 8
13 2 4
X = 11.8 X = 3.8 X = 7.6
23
S2
1 = 5.7 S2
2 = 10.2 S2
3 = 10.3

24. STEP 3 (PART A)
Grand mean
_
∑X 10+12+9+……+4
X GM = =
N 15
116
=
15
_
X GM = 7.73

25. STEP 3 (PART B
25 )
∑ni(Xi – X GM) 2
S2
B =
k – 1
= 5 (11.8 – 7.73) 2 + 5 (3.8 – 7.73)2 + 5
(7.6 – 7.73)2
3-1 d.f.N = k -1

26. Medication (n1 = 5) Exercise (n2 = 5) Diet (n3 = 5)
10 6 5
12 8 9
9 3 12
15 0 8
13 2 4
_
X = 11.8
_
X = 3.8
_
X = 7.6
S2
1 = 5.7 S2
2 = 10.2 S2
3 = 10.3

27. ∑ (ni – 1)s2
i
S2
w =
∑ (ni –1)
27
(5-1) (5.7) + (5-1) (10.2) + (5-1)(10.3)
(5-1) + (5-1) +(5-1)
= 104.80
S2
W = 104.80 / 12 = 8.73
STEP 3 (PART C)

28. S2
B (between group variance)
F =
S2
w (Within group Variance)
80.07
= = 9.17
8.73
STEP 3 (PART D)

29. STEP 4
F (tab) = F alpha, df1, df2
= F 0.05, 2, 12 = 3.89
9.17>3.89 (Graph)
Reject the null hypothesis
Step 5:
There is sufficient evidence to reject the
null hypothesis and conclude that at least
one mean is different from the others.

30. CONCLUSION
30
 Since F cal (9.17) is greater than F tab so we
reject the null hypothesis at 5% level of
significance and conclude that at least one of the
means is different.

31. Acknowledgements
Dr Tazeen Saeed Ali
RM, RM, BScN, MSc ( Epidemiology & Biostatistics), Ph.D.
(Medical Sciences), Post Doctorate (Health Policy & Planning)
Associate Dean School of Nursing & Midwifery
The Aga Khan University Karachi.
Kiran Ramzan Ali Lalani
BScN, MSc Epidemiology & Biostatistics (Candidate)
Registered Nurse (NICU)
Aga Khan University Hospital

32. REFERENCE
 Bluman, G. A. (2008). Elementary Statistics, Astepby
step approach (7th ed.) McGraw Hill.