Question

Data Set A:

1.32, 1.01, 0, 2.21, 1.69, 1.73, 2.01, 0, 0.73, 0.91, 0, 3.03, 2.22, 1.23, 3.71, 0, 0.45, 2.18, 3.12, 1.91

Data Set B:

0, 0.63, 2.11, 1.37, 0, 1.11, 2.93, 0, 3.11, 2.61, 0, 0.38, 0.98, 1.55, 1.83, 0, 3.46, 2.31, 0, 1.49

- Interpret on comment on the nonparametric estimates of central tendency differences.

Answer 1

Solution:
Hypothesis,
H₀ : M₁= M₂ VS H₁ : M_{1 not equal to} M₂

_{(M = Median)}

#R program:

>x=c( 1.32, 1.01, 0, 2.21, 1.69, 1.73, 2.01, 0, 0.73, 0.91, 0, 3.03, 2.22, 1.23, 3.71, 0, 0.45, 2.18, 3.12, 1.91)
> x
[1]  1.32 1.01 0 2.21 1.69 1.73 2.01 0 0.73 0.91 0 3.03 [13] 2.22 1.23 3.71 0, 0.45 2.18 3.12 1.91

>y=c( 0, 0.63, 2.11, 1.37, 0, 1.11, 2.93, 0, 3.11, 2.61, 0, 0.38, 0.98, 1.55, 1.83, 0, 3.46, 2.31, 0, 1.49 )
> y
[1] 0 0.63 2.11 1.37 0 1.11 2.93 0 3.11 2.61 0 0.38 0.98 [14] 1.55 1.83 0 3.46 2.31 0 1.49

> wilcox.test(x,y)
Wilcoxon rank sum test with continuity correction
data: x and y
W = 220, p-value = 0.595
alternative hypothesis: true location shift is not equal to 0

Output:
The P-value of wilcox Mann-Whitney U-test is 0.595. This is greater than 0.05. Hence we accept the null hypothesis. Thus we conclude that the median of first population is equal to median of second population.