ANOVA Nursing Research nursing Statistics SPSS

1. UNIT : XV
ANALYSIS OF
VARIANCE
(ANOVA)
1
Mrs. D. Melba Sahaya Sweety RN,RM
PhD Nursing , MSc Nursing (Pediatric Nursing), BSc Nursing
Associate Professor
Department of Pediatric Nursing
Enam Nursing College, Savar,
Bangladesh.

2. INTRODUCTION
 ANOVA is a statistical method that stands for analysis of
variance. ANOVA was developed by Ronald Fisher in 1920
and is the extension of the t and the z test. This test is also
called the Fisher analysis of variance or F Test, ANOVA is
a hypothesis test to check whether the mean or average of
three or more populations are equal or not.
 The goal of the ANOVA test is to check for variability
within the groups as well as the variability among the
groups.
 It works when the Y variable is continuous and the X
variable is discrete..
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3. ASSUMPTION
• Homogeneity: Equal variances and normal
distribution within the groups under comparison
• Continuous Scale of Measurement : Dependent
variable Measured on Interval scale or continuous.
• Independence : Samples are independently and
randomly drawn from the population.
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4. TYPES OF ANOVA
One Way
ANOVA
N – Way
Multivariate
ANOVA or
MANOVA
Two Way
ANOVA
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5. TYPES OF ANOVA
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One Way
ANOVA
• One-way ANOVA refers to a type of ANOVA test where
there will be only one independent variable. The test
compares means of groups, generally three or more groups,
to analyze the variance.
Assumptions:
• Compare mean of independent groups
• Preferable when there are three or more data
• One independent variable and dependent variable
• Independent variable should have three or more levels
• Use F- distribution for calculation.

6. TYPES OF ANOVA
A two-way ANOVA (“analysis of variance”) is used to determine whether or not there is a
statistically significant difference between the means of three or more independent
groups that have been split on two variables (sometimes called “factors”).
Assumptions:
Variables: Dependent variable should be continuous , two independent variables should
be in categorical, independent groups.
Independence – that each sample has been drawn independently of the other samples
Variance Equality – That the variance of data in the different groups should be the
same
Normality – That each sample is taken from a normally distributed population
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Two Way
ANOVA

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• It applies to multiple independent
variables that affect the dependent variable.
It is more effective than Analysis of
Variance as one can use it to observe
multiple dependent variables
simultaneously.
TYPES OF ANOVA
N – Way Multivariate
ANOVA or MANOVA

8. DIFERRENCE BETWEEN ONE WAY AND
TWO WAY ANOVA
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Basis of comparison One-way ANOVA Two-way ANOVA
Number of
Independent
Variables
A one-way ANOVA only involves one
factor or independent variable.
A two-way ANOVA involves
two independent variables and
one dependent variable.
Number of Groups Of
Sample
There is only one independent variable
that has multiple groups. For example,
‘Age’ as an independent variable may
have multiple groups:
12-18 years,19-26 years
27-35 years, Above 36 years
Compares multiple groups of
two factors. For example, in
addition to ‘Age’, ‘Gender’ may
have the following multiple
groups: Male, Female
Number Of
Observations
The number of observations (sample
size) need not be the same in each
group.
The number of observations
(sample size) needs to be the
same in each group.

9. DIFERRENCE BETWEEN ONE WAY AND TWO
WAY ANOVA
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Basis of comparison One-way ANOVA Two-way ANOVA
Description
One-way ANOVA is a hypothesis
test that allows one to make
comparisons between the means of
three or more groups of data.
Two-way ANOVA is a hypothesis test
that allows one to make comparisons
between the means of three or more
groups of data, where two
independent variables are considered.
Effect
It accesses only one variable at a
given time.
It accesses two variables at the same
time. It therefore also shows whether
there is any interaction effect between
the independent variables.
Design of experiments Need to satisfy only two
principles.
All three principles needs to be
satisfied.

10. DIFERRENCE BETWEEN ONE
WAY AND TWO WAY ANOVA
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Basis of comparison One-way ANOVA Two-way ANOVA
Example How does tea consumption affect
weight?
Independent variable: Tea
(Green tea, Black tea, Milk tea
No tea)
Dependent variable: Weight
(Below 50 kg, 51-70 kg, 71-80
kg, Above 80 kg)
How do tea consumption and
exercise intensity affect weight?
Independent variable 1: Tea
(Green tea, Black tea, Milk tea
No tea)
Independent variable 2: Exercise
intensity
(High intensity, Medium intensity
Low intensity, No exercise)
Dependent variable: Weight
(Below 50 kg, 51-70 kg, 71-80 kg
Above 80 kg )

11. STEPS OF ONE WAY ANOVA
1, Formulation of Hypothesis
2, Calculate Grand Total (GT) = = ∑ x1+ ∑ x2+ ∑ x3+……+ ∑ xn
3, Correction Factor (CF) =
4, Total Sum of Squares (TSS) = ∑x1
2 + ∑x2
2 + ∑x3
2 +… ∑xn
2 – CF
5, Sum of squares between Samples (SSS)= ∑x1
2 + ∑x2
2 + ∑x3
2 + ... ∑xn
2 – CF
n1 n2 n3 n4
6, Sum of square within the Sample = TSS – SSS
7, Degrees of Freedom for between and within the samples
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GT 2
n

12. A, Df for between the samples = Number of groups – 1
B, Df for within the samples = Number of samples in all the group – Number of Groups
C, Total Df = Number of Samples in all the group - 1
8, Calculate the mean of sum of squares
A, Mean of sum of squares between the Samples = Sum of squares between the samples
Df for between the samples
B, Mean of sum of squares within the samples = Sum of squares within the samples
Df for within the samples
9, Calculate the F – ratio = Mean of sum of squares between the Samples
Mean of sum of squares within the samples
10, Find the F table value 12
STEPS OF ONE WAY ANOVA

13. ONE WAY ANOVA
• Example: 1, To Assess the significant of possible variations in performance in a
aptitude test between the colleges of a city. A common test was given to a number of
students from 5 classes of 4 colleges the results are given below make an analysis of
variance.
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A B C D
8 12 18 13
10 11 12 9
12 9 16 12
8 14 6 16
7 4 8 15

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ONE WAY ANOVA
Solution:
Hypothesis Testing:
Null Hypothesis: There will be no significant of possible
variations in performance in a aptitude test between the
colleges of a city H0: µ1 = µ2 = µ3 = µ4
Alternative Hypothesis: There will be a significant of
possible variations in performance in a aptitude test
between the colleges of a city H1: µ1 ≠ µ2 ≠ µ3 ≠ µ4
Level of Significance : 5 per cent level of significance

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15
ONE WAY ANOVA
Test Statistic :
A B C D
x1 x1
2 x2 x2
2 x3 x3
2 x4 x4
2
8 64 12 144 18 324 13 169
10 100 11 121 12 144 9 81
12 144 9 81 16 256 12 144
8 64 14 196 6 36 16 256
7 49 4 16 8 64 15 225
45 421 50 558 60 824 65 875
1, Grand Total (GT) :
= ∑ x1+ ∑ x2+ ∑ x3+ ∑ x4
= 45 + 50 + 60 + 65
= 220

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16
ONE WAY ANOVA
2, Correction Factor (CF) = ∑ x1+ ∑ x2+ ∑ x3+ ∑ x4
2
n1+n2+n3+n4
= (45+50+60+65) 2
5 + 5 + 5 + 5
= (220) 2
20
= 48400
20
CF = 2420
Or GT 2
n

17. 3,Total Sum of Squares (TSS) = ∑x1
2 + ∑x2
2 + ∑x3
2 + ∑x4
2 – CF
= 421+ 558+824+875 - 2420
= 2678 – 2420
= 258
4, Sum of squares between Samples (SSS)= ∑x1
2 + ∑x2
2 + ∑x3
2 + ∑x4
2 – CF
n1 n2 n3 n4
= (45)2 + (50)2 + (60) 2 + (65) 2 - 2420
5 5 5 5
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ONE WAY ANOVA

18. Sum of squared between Samples (SSS ) = 2025 + 2500 + 3600 + 4225 - 2420
5 5 5 5
= 405 + 500 + 720 + 845 – 2420
= 2470 – 2420
= 50
5, Sum of square within the Sample = TSS - SSS
= 258 – 50
= 208
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ONE WAY ANOVA

19. 6, Degrees of Freedom for between and within the samples
A, Df for between the samples = Number of groups – 1
= 4 – 1
= 3
B, Df for within the samples = Number of samples in all the group – Number of
Groups
= 20 – 4
= 16
C, Total Df = Number of Samples in all the group - 1
= 20 – 1
= 19
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ONE WAY ANOVA

20. 7, Calculate the mean of sum of squares
A, Mean of sum of squares between the Samples = Sum of squares between the samples
Df for between the samples
= 50/ 3
= 16.66
B, Mean of sum of squares within the samples = Sum of squares within the samples
Df for within the samples
= 208 / 16
= 13
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ONE WAY ANOVA

21. Square
of
Variance
Sum of
squares
Degree of
freedom
Mean
Square
F -
ratio
Between
the
Samples
50 3 16.66
1.28
Within
the
Samples
208 16 13
Total 258 19
21
ONE WAY ANOVA
8, Calculate the F – ratio =
Mean of sum of squares
between the Samples
Mean of sum of squares within
the samples
= 16.66
13
= 1.28
9, Find the F table value =
Greater Df / Smaller Df
=F(16, 3)
= 3.2389

22. • Inference
Fcal < Ftab
Thus we accept null hypothesis
There will be no significant of possible variations in
performance in a aptitude test between the colleges of a city
H0: µ1 = µ2 = µ3 = µ4
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ONE WAY ANOVA

23. ONE WAY ANOVA
• Example: 2, To Assess the significant of possible variations in
whiteness reading made in 15 cloths with four types of detergents
A,B,C are given below
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A B C
8 7 12
10 5 9
7 10 13
14 9 12
11 9 14

24. ONE WAY ANOVA
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Solution:
Hypothesis Testing:
Null Hypothesis: There will be no significant of
possible variations in whiteness reading of the clothes
by the detergents H0: µA = µB = µC = µD
Alternative Hypothesis: There will be significant of
possible variations in whiteness reading of the clothes
by the detergents H1: µA ≠ µB ≠ µC ≠ µD
Level of Significance : 5 per cent level of
significance

25. ONE WAY ANOVA
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Test Statistic :
A B C
x1 x1
2 x2 x2
2 x3 x3
2
8 64 7 49 12 144
10 100 5 25 9 81
7 49 10 100 13 169
14 196 9 81 12 144
11 121 9 81 14 196
∑ x1 = 50 ∑x1
2 = 530 ∑ x2 = 40 ∑x2
2 = 336 ∑ x3 = 60 ∑x3
2 = 734

26. ONE WAY ANOVA
26
Test Statistic :
1, Grand Total (GT) = ∑ x1+ ∑ x2+ ∑ x3
= 50+40+60
= 150
2, Correction Factor (CF) =
= 150 2
15
= 22500
15
= 1500
GT 2
n

27. 3,Total Sum of Squares (TSS) = ∑x1
2 + ∑x2
2 + ∑x3
2 – CF
= 530+ 336+ 734 - 1500
= 1600 – 1500
= 100
4, Sum of squares between Samples (SSS)= ∑x1
2 + ∑x2
2 + ∑x3
2 – CF
n1 n2 n3
= (50)2 + (40)2 + (60) 2 - 1500
5 5 5
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ONE WAY ANOVA

28. Sum of squared between Samples (SSS ) = 2500 + 1600 + 3600 - 1500
5 5 5
= 500 + 320 + 720 – 1500
= 1540 – 1500
= 40
5, Sum of square within the Sample = TSS - SSS
= 100 – 40
= 60
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ONE WAY ANOVA

29. 6, Degrees of Freedom for between and within the samples
A, Df for between the samples = Number of groups – 1
= 3 – 1
= 2
B, Df for within the samples = Number of samples in all the group – Number of
Groups
= 15 – 3
= 12
C, Total Df = Number of Samples in all the group - 1
= 15 – 1
= 14
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ONE WAY ANOVA

30. 7, Calculate the mean of sum of squares
A, Mean of sum of squares between the Samples = Sum of squares between the samples
Df for between the samples
= 40/ 2
= 20
B, Mean of sum of squares within the samples = Sum of squares within the samples
Df for within the samples
= 60 / 12
= 5
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ONE WAY ANOVA

31. Square
of
Variance
Sum of
squares
Degree of
freedom
Mean
Square
F -
ratio
Between
the
Samples
40 2 20
4
Within
the
Samples
60 13 5
Total 100 14
31
ONE WAY ANOVA
8, Calculate the F – ratio =
Mean of sum of squares
between the Samples
Mean of sum of squares within
the samples
= 20
5
= 4
9, Find the F table value =
Greater Df / Smaller Df
=F(12, 2)
= 3.49

32. ONE WAY ANOVA
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• Inference
Fcal > Ftab
Thus we reject null hypothesis
There will be significant of possible variations in
whiteness reading of the clothes by the
detergents H1: µA ≠ µB ≠ µC ≠ µD

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