Date post: | 02-Jul-2015 |
Category: |
Documents |
Upload: | iybaro-reyes |
View: | 371 times |
Download: | 4 times |
Friedman two way ANOVA By Rank is a test for comparing three or more related samples
and which makes no assumptions about the underlying distribution of the data. The data is set out in a table comprising n rows and k columns.
The data is ranked horizontally or across the rows and the mean rank for each column is compared.
This test is very useful when the data are ordinal (i.e., ranked)
Friedman Formula2
2
1
12 ( 1)
( 1) 2
k
r j
j
b kR
bk k
2 2
1
123 ( 1)
( 1)
k
r j
j
R b kbk k
EQUATION 1
EQUATION 2
EQUATION 3
Example A water company sought evidence the measures taken to
clean up a river were effective. Biological Oxygen Demand (BOD) at 12 sites on the river were compared before clean up, 1 month later and a year after clean up.
Aqualytic sensor system AL606
Hypothesis Testing Steps
1. Data
Site BOD (biological oxygen demand)
Before After 1month
After 1 year
1 17.4 13.6 13.2
2 15.7 10.1 9.8
3 12.9 9.7 9.7
4 9.8 9.2 9.0
5 13.4 11.1 10.7
6 18.7 20.4 19.6
7 13.9 10.4 10.2
8 11 11.4 11.5
9 5.4 4.9 5.2
10 10.4 8.9 9.2
11 16.4 11.2 11.0
12 5.6 4.8 4.6
Hypothesis Testing Steps
1. Data
Site BOD (biological oxygen demand)
Before After 1month
After 1 year
1 17.4 13.6 13.2
2 15.7 10.1 9.8
3 12.9 10.3 9.7
4 9.8 9.2 9.0
5 13.4 11.1 10.7
6 18.7 20.4 19.6
7 13.9 10.4 10.2
8 11 11.4 11.5
9 5.4 4.9 5.2
10 10.4 8.9 9.2
11 16.4 11.2 11.0
12 5.6 4.8 4.6
Site BOD (biological oxygen demand)
Before After 1month
After 1 year
1 17.4 3 13.6 2 13.2 1
2 15.7 3 10.1 2 9.8 1
3 12.9 3 9.7 1.5 9.7 1.5
4 9.8 3 9.2 2 9.0 1
5 13.4 3 11.1 2 10.7 1
6 18.7 1 20.4 3 19.6 2
7 13.9 3 10.4 2 10.2 1
8 11 1 11.4 2 11.5 3
9 5.4 3 4.9 1 5.2 2
10 10.4 3 8.9 1 9.2 2
11 16.4 3 11.2 2 11.0 1
12 5.6 3 4.8 2 4.6 1
Rj 32 22.5 17.5
Hypothesis Testing Steps
1. Data 2. Assumption
The observations appearing in a given block are independent of the observations appearing in each of the other blocks, and within each block measurement on at least an ordinal scale is achieved.
3. Hypothesis
H0 : The clean up procedure has had no effect on the BOD.HA : The clean up procedure has affected the BOD.
4. Decision Rule: Reject H0 if M > critical value at 5% level of significance
5. Calculation of Test Statistic
Calculating of test statistic……
Friedman’s magic formula!!!!
Where, k = number of columns (treatments)
n = number of rows (blocks)
Rj = sum of the ranks
BOD (biological oxygen demand)
Site Before After 1 month After 1 year
Sum of ranks 32 22.5 17.5
2
(sum of ranks) 1024 506.25 306.25
Number of columns, k 3
Solution: Number of rows, n 12
1836.5 = (1024 + 506.25 + 306.25)
__12__nk(k+1)
0.083 = ___12___12 x 3 x 4
3n(k+1) 144 = 3 x 12 x 4
Test Statistic M 8.43 = 0.083 x 1836.5 - 144
6. Statistical decision
Compare computed M value to critical value at 5% level of significance.
M(computed value) = 8.43
critical value at 5% level of significance is = 6.17
• 7. Conclusion M is > than critical value
Reject the null hypothesis
Alternative hypothesis:
HA : The clean up procedure has affected the BOD.
Critical Values for Friedman’s two way ANOVA by Ranks
k n =0.10 =0.05 =0.1
3 3 6.00 6.00 ---
4 6.00 6.50 8.00
5 5.20 6.40 8.40
6 5.33 7.00 9.00
7 5.43 7.14 8.86
8 5.25 6.25 9.00
9 5.56 6.22 8.67
10 5.00 6.20 9.60
11 4.91 6.54 8.91
12 5.17 6.17 8.67
13 4.77 6.00 9.39
-- 4.61 5.99 9.21
Offline version
Statistic calculator