ANOVA - Primer

Analysis of  Variance 

What is ANOVA

Analysis of Variance

Statistical test for detecting differences in the group means when there is one one parametric dependent variable and one or more independent variable.

We are intrested in determining whther differences exist between the population means.

Types of ANOVA

1 - way and 2-way

1-way:

 - 1 dependent and 1 independent variable

2-way :

 - 2 dependent and 2 or more independent variables

Key terms

Null Hypothesis H0

 - General statement that states that no relationship between 2 measured phonomena or no association amoong groups

Alternative Hypothesis HA

 - Contrary to Null hypothesis, it states that whenever something is happening, a new theory is preferred instead of old

P Value

  - The probability of finding the observed, or more extreme, results when null hypothesis of a study question is True. 

Alpha Value 

  - Creterion for determining whther a test statistic is technically significant.

F - statistics

  - Extent of difference between the means of different trials

Sum of Squares

  - Variation from the mean of different medical trials

Mean

  - Average of all the results from evidences like medical trials.

How we can use ANOVA

Anova determines whther the groups created by the levels of independent  variable are statistically different by  calculatin the whether the means of the different samples are different from the overall mean of the dependent variable

If any of  group means is significantly different from the overall mean, then the null hypothesis is rejected.

F - Statistics

Value you get when you run the ANOVA test to find out if the means bewteen two populations are significantly different.

Between-group variance is large relative to the value within group variance, so F statistic will be larger & > the critical value, therefore significantly different.

ANOVA formula

is made up of numerous parts. The best way to tackle an ANOVA test problem is to organize the formulae inside an ANOVA table. Below are the ANOVA formulae.

Source of Variation	Sum of Squares	Degree of Freedom	Mean Squares	F Value
Between Groups	SSB = Σnj(X̄j– X̄)2	df1 = k – 1	MSB = SSB / (k – 1)	f = MSB / MSE or, F = MST/MSE
Error	SSE = Σnj(X̄- X̄j)2	df2 = N – k	MSE = SSE / (N – k)
Total	SST = SSB + SSE	df3 = N – 1

where,

• F = ANOVA Coefficient
• MSB = Mean of the total of squares between groupings
• MSW = Mean total of squares within groupings
• MSE = Mean sum of squares due to error
• SST = total Sum of squares
• p = Total number of populations
• n = The total number of samples in a population
• SSW = Sum of squares within the groups
• SSB = Sum of squares between the groups
• SSE = Sum of squares due to error
• s = Standard deviation of the samples
• N = Total number of observations

Worked Examples

Example 1: Three different kinds of food are tested on three groups of rats for 5 weeks. The objective is to check the difference in mean weight(in grams) of the rats per week. Apply one-way ANOVA using a 0.05 significance level to the following data:

Food I	Food II	Food III
8	4	11
12	5	8
19	4	7
8	6	13
6	9	7
11	7	9

Solution:

H0: μ1= μ2=μ3

H1: The means are not equal

Since, X̄1 = 5, X̄2 = 9, X̄3 = 10

Total mean = X̄ = 8

SSB = 6(5 – 8)2 + 6(9 – 8)2 + 6(10 – 8)2 = 84

SSE = 68

MSB = SSB/df1 = 42

MSE = SSE/df2 = 4.53

f = MSB/MSE = 42/4.53 = 9.33

Since f > F, the null hypothesis stands rejected.

Example 2: Calculate the ANOVA coefficient for the following data:

Plant	Number	Average span	s
Hibiscus	5	12	2
Marigold	5	16	1
Rose	5	20	4

Solution:

Plant	n	x	s	s2
Hibiscus	5	12	2	4
Marigold	5	16	1	1
Rose	5	20	4	16

p = 3
n = 5
N = 15
x̄ = 16
SST = Σn(x−x̄)2

SST= 5(12 − 16)2 + 5(16 − 16)2 + 11(20 − 16)2 = 160

MST = SST/p-1 = 160/3-1 = 80

SSE = ∑ (n−1) = 4 (4 + 1) + 4(16) = 84

MSE = 7

F = MST/MSE = 80/7
 
F = 11.429

 

Example 3: The following data show the number of worms quarantined from the GI areas of four groups of muskrats in a carbon tetrachloride anthelmintic study. Conduct a two-way ANOVA test.

I	II	III	IV
338	412	124	389
324	387	353	432
268	400	469	255
147	233	222	133
309	212	111	265

Solution:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square
Between the groups	62111.6	8	9078.067
Within the groups	98787.8	16	4567.89
Total	167771.4	24

Since F = MST / MSE

           = 9.4062 / 3.66 = 2.57 

1. Example 4: Three types of fertilizers are used on three groups of plants for 5 weeks. We want to check if there is a difference in the mean growth of each group. Using the data given below apply a one way ANOVA test at 0.05 significant level.

   Fertilizer 1	Fertilizer 2	Fertilizer 3
   6	8	13
   8	12	9
   4	9	11
   5	11	8
   3	6	7
   4	8	12

   Solution:

   $H_0$: ${\mu }_1$ = ${\mu }_2$ = ${\mu }_3$

   $H_1$: The means are not equal

   Fertilizer 1	Fertilizer 2	Fertilizer 3
   6	8	13
   8	12	9
   4	9	11
   5	11	8
   3	6	7
   4	8	12
   ${\overline{X}}_1$ = 5	${\overline{X}}_1$ = 9	${\overline{X}}_1$ = 10

   Total mean, $\overline{X}$ = 8

   $n_1$ = $n_2$ = $n_3$ = 6, k = 3

   SSB = 6(5 - 8)² + 6(9 - 8)² + 6(10 - 8)²

   = 84

   df1 = k - 1 = 2

   Fertilizer 1	(X - 5)2	Fertilizer 2	(X - 9)2	Fertilizer 3	(X - 10)2
   6	1	8	1	13	9
   8	9	12	9	9	1
   4	1	9	0	11	1
   5	0	11	4	8	4
   3	4	6	9	7	9
   4	1	8	1	12	4
   ${\overline{X}}_1$ = 5	Total = 16	${\overline{X}}_1$ = 9	Total = 24	${\overline{X}}_1$ = 10	Total = 28

   SSE = 16 + 24 + 28 = 68

   N = 18

   df2 = N - k = 18 - 3 = 15

   MSB = SSB / df1 = 84 / 2 = 42

   MSE = SSE / df2 = 68 / 15 = 4.53

   ANOVA test statistic, f = MSB / MSE = 42 / 4.53 = 9.33

   Using the f table at $\alpha$ = 0.05 the critical value is given as F(0.05, 2, 15) = 3.68

   As f > F, thus, the null hypothesis is rejected and it can be concluded that there is a difference in the mean growth of the plants.

   Answer: Reject the null hypothesis