The document outlines the steps to perform the Wilcoxon Signed Rank Test to compare two related samples: 1) Obtain the differences between paired values in two samples and rank the absolute differences. 2) Assign ranks to positive and negative differences and calculate the sum of ranks. 3) Compare the smaller sum of ranks (T) to critical values to determine if the null hypothesis that the samples are identical can be rejected.

2. a. State the null hypothesis
Ho: The two samples are identical
b. State the alternative hypothesis
Ha: The two samples are not identical
c. Obtain the differences between
paired values in the two
samples/treatments.

3. d. Rank the absolute values of these
differences, giving the smallest
difference rank 1 and assigning
average ranks to tied differences.
e. Assign to each rank the sign of the
observed differences.
f. Get the sum of the positive ranks
and the sum of the negative ranks.

4. g. Find the smaller of the absolute
values of the two sums of the ranks
and denote it by T.
h. Test Ho by comparing T with the
critical values of T for the Wilcoxon
Signed Rank Test found in the
appendix.
i. If T is less than or equal to Tcritical for
the chosen level of significance, α,
then the null hypothesis is rejected.
Otherwise the null hypothesis is not

5. A research team subjected to a sensory evaluation test a camote rops
extract (CTE) beverage by comparing it to a commercially available green tea
drink (GTD) in terms of acceptability of the product. The drinks were ranked
from 1 (lower rank) to 10 (highest rank). The following data were obtained.
Evaluator
CTE
GTD
Evaluator
CTE
GTD
1
7
9
9
8
9
2
2
3
10
1
2
3
5
2
11
3
2
4
4
1
12
6
3
5
6
5
13
9
7
6
1
1
14
3
5
7
5
4
15
4
1
8
6
4
16
6
5
Test the Ho which states that CTE and GTD are identical in terms of
acceptability.

6. •Ho: CTE and GTD are identical.
•HA: CTE and GTD are not identical.
•Get the differences (D) between paired values. Rank the absolute values
of D.
Evaluator
1
2
3
4
5
6
7
8
R= -36
D
-2
-1
3
3
1
0
1
2
R= 100
Rank
10.5
-5
14.5
14.5
5
1
5
10.5
Evaluator
9
10
11
12
13
14
15
16
D
-1
-1
1
3
2
-2
3
1
T= 36
•Test the null hypothesis by comparing the value of T= 36 with
Tcritical= 30 at α= 0.05.
•Decision: Since T > Tcritical, the null hypothesis is not rejected. It
means that the two treatments are identical.
Rank
-5
-5
5
14.5
10.5
-10.5
14.5
5