Question

In: Statistics and Probability

2. Take data sets A and B and delete duplicated values such that each value is...

2. Take data sets A and B and delete duplicated values such that each value is unique even when pooling the two data sets. Just like with the previous problem, treat data sets A and B as hypothetical data on the weights of children whose parents smoke cigarettes, and those whose parents do not respectively.

a) Calculate the expected value of the wilcoxon Rank-Sum test statistic E(Wx) assuming the null hypothesis of equal medians being true.

b) Conduct a Wilcoxon-Rank-Sum test on the data

Note: Data set A - 12.36, 12.39, 12.44, 12.50, 12.61, 12.80,12.82, 12.87, 12.89, 12.95, 13.25

Data set B - 12.41, 12.56, 12.61, 12.64, 12.70, 12.85, 13.05, 13.08

Solutions

Expert Solution

b)

First, we put both samples together and organize it in ascending order, which is shown in the table below:

Sample Value
1 12.36
1 12.39
2 12.41
1 12.44
1 12.50
2 12.56
1 12.61
2 12.61
2 12.64
2 12.70
1 12.80
1 12.82
2 12.85
1 12.87
1 12.89
1 12.95
2 13.05
2 13.08
1 13.25

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)
1 12.36 1 1
1 12.39 2 2
2 12.41 3 3
1 12.44 4 4
1 12.50 5 5
2 12.56 6 6
1 12.61 7 7.5
2 12.61 8 7.5
2 12.64 9 9
2 12.70 10 10
1 12.80 11 11
1 12.82 12 12
2 12.85 13 13
1 12.87 14 14
1 12.89 15 15
1 12.95 16 16
2 13.05 17 17
2 13.08 18 18
1 13.25 19 19

The sum of ranks for sample 1 is:

R1 ​= 1+2+4+5+7.5+11+12+14+15+16+19 = 106.5

and the sum of ranks of sample 2 is:

R2​ = 3+6+7.5+9+10+13+17+18 = 83.5

Hence, the test statistic is R = R1​=106.5.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0​: Median (Difference) = 0

H1: Median (Difference) ≠ 0

(2) Rejection Region

The critical value for the signficance level provided and the type of tail is Rc ​= 46, and the null hypothesis is rejected if R ≤ 46.

(3) Decision about the null hypothesis

Since in this case R = 106.5 > 46, there is not enough evidence to claim that the population median of differences is different than 0, at the 0.05 significance level.


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