here take null hypothesis as median difference = 0
and alternative hypothesis as median difference not equal to zero
First, we put both samples together and organize it in ascending order, which is shown in the table below:
| Sample | Value |
| 2 | 8 |
| 2 | 8 |
| 2 | 10 |
| 2 | 11 |
| 1 | 12 |
| 1 | 12 |
| 1 | 14 |
| 2 | 14 |
| 1 | 16 |
| 1 | 17 |
| 2 | 17 |
| 1 | 18 |
| 2 | 19 |
| 1 | 20 |
| 1 | 20 |
| 2 | 22 |
| 1 | 25 |
| 1 | 25 |
Now, that the values that are in ascending order are assigned ranks to them, taking care of assigning the average rank to values with rank ties
| Sample | Value | Rank | Rank (Adjusted for ties) |
| 2 | 8 | 1 | 1.5 |
| 2 | 8 | 2 | 1.5 |
| 2 | 10 | 3 | 3 |
| 2 | 11 | 4 | 4 |
| 1 | 12 | 5 | 5.5 |
| 1 | 12 | 6 | 5.5 |
| 1 | 14 | 7 | 7.5 |
| 2 | 14 | 8 | 7.5 |
| 1 | 16 | 9 | 9 |
| 1 | 17 | 10 | 10.5 |
| 2 | 17 | 11 | 10.5 |
| 1 | 18 | 12 | 12 |
| 2 | 19 | 13 | 13 |
| 1 | 20 | 14 | 14.5 |
| 1 | 20 | 15 | 14.5 |
| 2 | 22 | 16 | 16 |
| 1 | 25 | 17 | 17.5 |
| 1 | 25 | 18 | 17.5 |
The sum of ranks for sample 1 is:
R1=5.5+5.5+7.5+9+10.5+12+14.5+14.5+17.5+17.5=114
and the sum of ranks of sample 2 is:
R2=1.5+1.5+3+4+7.5+10.5+13+16=57
b)Hence, the test statistic is R = max ( R1, R2 )= R_1 = 114
a)(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H_0: Median (Difference) = 0
H_a: Median (Difference) not equal to 0
(2) Rejection Region
The critical value for the signficance level provided and the type of tail is R_c = 40, and the null hypothesis is rejected if R≤40.
(3) Decision about the null hypothesis
Since in this case R = 114 >40, there is not enough evidence to claim that the population median of differences is different than 0, at the 0.05 significance level.
a)
The following null and alternative hypotheses need to be tested:
H_0: Median (Difference) = 0
H_a: Median (Difference) not equal to 0
b)Hence, the test statistic is R = max ( R1, R2 )= R_1 = 114
c) wilcoxon rank-sum test will compute the p-value for sample sizes that are sufficiently large to use normal approximation. Otherwise, critical values will be used instead.
here the sample sizes are not sufficiently large so we use critical values.
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