Finding Critical Values Assume that we have two treatments (A and B) that produce quantitative results, and we have only two observations for treatment A and two observations for treatment B. We cannot use the Wilcoxon signed ranks test given in this section because both sample sizes do not exceed 10.

Rank				Rank Sum of Treatment A
1	2	3	4
A	A	B	B	3

a. Complete the accompanying table by listing the five rows corresponding to the other five possible outcomes, and enter the corresponding rank sums for treatment A.

b. List the possible values of R and their corresponding probabilities. (Assume that the rows of the table from part (a) are equally likely.)

c. Is it possible, at the 0.10 significance level, to reject the null hypothesis that there is no difference between treatments A and B? Explain.

Four observations (two of treatment A and two of treatment B) are to be arranged to compute the ranks, the sum of the ranks of treatment A, and the corresponding probability values of the sums.

a.

There area total of four observations (two of treatment A and two of treatment B).

Ranks are assigned to observations such that the lowest value gets rank 1, the next lowest value gets rank 2, and so on.

The two observations of treatment A and the two observations of treatment B can be arranged in the following six ways so that all possible ranks of all observations are considered.

The six possible ways are:

The different sums of the ranks corresponding to treatment A are computed below:

For treatment combination AABB:

\(1 + 2 = 3\)

For treatment combination ABAB:

\(1 + 3 = 4\)

For treatment combination ABBA:

\(1 + 4 = 5\)

For treatment combination BABA:

\(2 + 4 = 6\)

For treatment combination BAAB:

\(2 + 3 = 5\)

For treatment combination BBAA:

\(3 + 4 = 7\)

The following table compiles the ranks as well as the sum of the ranks:

b.

The total number of combinations = 6.

The different possible sums are 3,4,5,6, and 7.

Consider the outcomes to be equally likely.

The probability of getting the sum equal to three is computed as follows:

\(\begin{array}{c}P\left( 3 \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{times}}\;{\rm{3}}\;{\rm{occurs}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \frac{1}{6}\end{array}\)

The probability of getting the sum equal to four is computed as follows:

\(\begin{array}{c}P\left( 4 \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{times}}\;4\;{\rm{occurs}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \frac{1}{6}\end{array}\)

The probability of getting the sum equal to five is computed as follows:

\(\begin{array}{c}P\left( 5 \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{times}}\;5\;{\rm{occurs}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \frac{2}{6}\end{array}\)

The probability of getting the sum equal to six is computed as follows:

\(\begin{array}{c}P\left( 6 \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{times}}\;6\;{\rm{occurs}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \frac{1}{6}\end{array}\)

The probability of getting the sum equal to seven is computed as follows:

\(\begin{array}{c}P\left( 7 \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{times}}\;7\;{\rm{occurs}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{outcomes}}}}\\ = \frac{1}{6}\end{array}\)

The table below summarizes the probability values:

c.

The null hypothesis is as follows:

There is no difference between treatments A and B.

The alternative hypothesis is as follows:

There is a difference between treatments A and B.

Criteria of rejection of null hypothesis:

The level of significance\(\left( \alpha \right)\)is given to be equal to 0.10.

If the probability value of any of the fivepossible sums is less than 0.10, the null hypothesis is rejected.

Since no probability value (for any of the fivesums) is less than 0.10, there is a failure to rejectthe null hypothesis.

Therefore, it is impossible to reject the null hypothesis at a 0.10 level of significance.