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Comparing between Groups (Parametric)

For comparison between groups, we use one factor ANOVA. ANOVA, short for "Analysis of Variance", is a statistical

approach to determine whether there are significant differences between two or more independent groups. If this is

performed for two groups, the results is identical to those in t-test. ANOVA can be used if your research data is interval

or ratio, and normally distributed.

Example :

In a clinical trial, a new drug is tested for effectiveness towards lowering blood pressure level. There are three different

compositional mix of a new drug ingredients (X1, X2 and X3). Thirty patients were chosen and ten were randomly

allocated for each prescription. The percentages of reduction on blood pressure level were recorded after certain period

of time and were as shown below. Is there an effect of drug composition towards the percentage of blood pressure

reduction?

X1 X2 X3

15 21 28

20 25 30

14 29 32

13 18 28

18 26 26

16 22 30

13 26 25

12 24 36

18 28 20

11 21 25

Analysis:

Ho : There is no significant difference on blood pressure reduction due to the different drug composition

H1 : There is a significant difference on blood pressure reduction due to the different drug composition

X1 X12 X2 X2

2 X3 X3

2

15 225 21 441 28 784 ΣX = 670

20 400 25 625 30 900 (ΣX)2 = 448900

14 196 29 841 32 1024 ΣX2 = 16210

13 169 18 324 28 784

18 324 26 676 26 676

16 256 22 484 30 900

13 169 26 676 25 625

12 144 24 576 36 1296

18 324 28 784 20 400

11 121 21 441 25 625

Mean ��� = 15 ��� = 24 �� = 28

Total T1 = 150 ΣX12 = 2328 T2 = 240 ΣX2

2 = 5868 T3 = 280 ΣX32 = 8014

Squared Total T12 = 22500 T2

2 = 57600 T32 = 78400

T = total of scores in a group

No. of conditions/groups, k = 3

No. of scores per group/condition, n = 10

Total no. of data, N = 30

bg= between group, wg = within group

2

Degrees of freedom:

dftotal = N - 1 = 29

dfbg = k - 1 = 2

dfwg = dftotal - dfbg = 29 - 2 = 27

Sum of squares:

�� � �� = � �� − �∑ ���� = 16210 − 44890030 = 1246.67

���� = ∑ �� − �∑ ���

� = 22500 + 57600 + 7840010 − 44890030 = 886.67

��"� = �� � �� − ���� = 1246.67 − 886.67 = 360

Mean squares:

#��� = $$%&'(%& = ))*.*+� = 443.33 ; #�"� = $$,&'(,& = *-�+ = 13.33

Variance Ratio (F):

. = #���#�"� = 443.3313.33 = 33.26 ANOVA Summary Souce of variation Degrees of

Freedom,

df

Sums of

Squares,

SS

Mean

Square

Variance

Ratio, F

Probability

Between Groups

(bg)

2 886.67 443.33 33.26 P < 0.01

Within Groups (wg) 27 360.00 13.33

Total 29 1246.67

For p<0.01, from table, F(2,27) = 3.35.

33.26 > 3.35 (significant) � null hypothesis rejected. Different drug composition does have significant effects towards

the percentages of blood pressure reduction.

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ANOVA Post-Hoc Analysis

Post-Hoc is carried out only if Ho is rejected after performing the F-test (ANOVA). If the F-test fails to reject the null

hypothesis post-hoc procedure should not be used.

(1) Least Significant Difference (LSD)

LSD method calculates difference between pairs of means and the difference calculated is compared to a calculated t-

statistic, at certain confidence level.

Procedures:

1. Calculate the LSD between two means (say, A and B) using the following formula. Note the t-test considered is two

tailed. 0�12,4 = 56/�,'(,&8#�"� 9 1 2 + 1 4:

2. if |��2 − ��4| ≥ 0�12,4, then we reject null hypothesis of Ho: µA = µB.

Example:

From last example,

X1 X2 X3 ��� = 15 ��� = 24 �� = 28

ANOVA Summary Souce of variation Degrees of

Freedom,

df

Sums of

Squares,

SS

Mean

Square

Variance

Ratio, F

Probability

Between Groups

(bg)

2 886.67 443.33 33.26 P < 0.01

Within Groups (wg) 27 360.00 13.33

Total 29 1246.67

As null hypothesis is rejected, we use LSD to find out whether difference between two particular groups are significant.

Consider two groups A and B, LSD between the two groups is defined as:

0�12,4 = 56/�,'(,&8#�"� 9 1 2 + 1 4:

if |��2 − ��4| ≥ 0�12,4, then we reject null hypothesis of Ho: µA = µB.

In this example, since n1 = n2 = n3 = 10, at α = 0.05, t0.05/2,27 = 2.052. Thus, 0�1�,� = 2.052=13.33 × ? ��- + ��-@ = 3.35

Thus, we can perform the following further LSD tests:

|��� − ���| = 9 �≥ 17.41� significant, reject null hypothesis of Ho: µ1 = µ2.

|��� − ��| = 13 �≥ 17.41� significant, reject null hypothesis of Ho: µ1 = µ3.

|��� − ��| = 4 �≥ 17.41� significant, reject null hypothesis of Ho: µ2 = µ3.

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(2) Tukey Test

The Tukey HSD (honestly significant difference) test allows us to compare each pair of conditions to see if their

difference is significant. To use this test, we need to calculate using the q-statistics. The statistic q is called Studentized

range statistic.

A = 'B((.�C "CCD �DE "� FC�DGG '.CHH�H �( IC 'B((.�C "CCD �DE "� FC�DG = J�KLJ�M=NO,&P

, Thus Q�1 = A=R$,&D

Procedures:

1. Calculate the HSD using formula Q�1 = A=R$,&D .

2. if the difference between any two means, say from A and B, exceeds HSD value e.g. |��2 − ��4| ≥ Q�1 , then we

reject null hypothesis of Ho: µA = µB.

3. Normally, for equal number of subjects in each group the n value is easy to be determined. For groups with different

subject size, we can take the smallest sample size n. A more sophisticated way to handle this is by using the

following formula:

= S9 TPTU TPVU⋯U TPX: , with k = number of groups/conditions.

Example:

From previous example,

X1 X2 X3 ��� = 15 ��� = 24 �� = 28

ANOVA Summary Souce of variation Degrees of

Freedom,

df

Sums of

Squares,

SS

Mean

Square

Variance

Ratio, F

Probability

Between Groups

(bg)

2 886.67 443.33 33.26 P < 0.01

Within Groups (wg) 27 360.00 13.33

Total 29 1246.67

As null hypothesis is rejected, we may use Tukey test to find out whether difference between two particular groups are

significant. The q-value requires degree of freedom, df and number of conditions, k. For HSD, we use dfwg = 27 and k = 3

(number of groups/conditions) and n=10 (number of scores in each group). For this, q = 3.51 from table.

Q�1 = A8#�"� = 3.51 × 813.3310 = 4.04

Tukey Tests:

|��� − ���| = 9 �≥ 4.04� significant, reject null hypothesis of Ho: µ1 = µ2.

|��� − ��| = 13 �≥ 4.04� significant, reject null hypothesis of Ho: µ1 = µ3.

|��� − ��| = 4 �≤ 4.04� not significant, accept null hypothesis of Ho: µ2 = µ3. (However, it is near to rejection value).

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