1. Non parametric test - Mann whitney U test for the following data
53,38,69,57,46,39,73,48,73,74,60,78
44,40,61,52,32,44,70,41,67,72,53,72, Significance Level `alpha=0.05` and One-tailed test

Solution:
Step-1: Take the hypothesis
Null Hypothesis `H_0` : The two populations are equal

Alternative Hypothesis `H_1` : The two populations are not equal

Step-2: Ranking all group values
First we assign ranks to all observations using low to high ranking process in the combined sample.

Size in Ascending Order	Rank	Name of related sample A for sample1, B for sample2	Rank for A	Rank for B
32	1	B		1
38	2	A	2
39	3	A	3
40	4	B		4
41	5	B		5
44	6.5	B		6.5
44	6.5	B		6.5
46	8	A	8
48	9	A	9
52	10	B		10
53	11.5	B		11.5
53	11.5	A	11.5
57	13	A	13
60	14	A	14
61	15	B		15
67	16	B		16
69	17	A	17
70	18	B		18
72	19.5	B		19.5
72	19.5	B		19.5
73	21.5	A	21.5
73	21.5	A	21.5
74	23	A	23
78	24	A	24
Total			167.5	132.5

The rank total for A is `R_1=167.5`

The rank total for B is `R_2=132.5`


We have `n_1=12` and `n_2=12`

Step-3: Compute test statistic
test statistic using `R_1`

`U_1 = n_1 * n_2 + (n_1(n_1+1))/2 - R_1`

`=12 * 12 + (12(12+1))/2 - 167.5`

`=144 + 78 - 167.5`

`=54.5`

`mu_U=(n_1 n_2)/2=(12*12)/2=72`

`sigma_U=sqrt((n_1 n_2 (n_1+n_2+1))/12)=sqrt((12*12*(12+12+1))/12)=17.3205`

According the limits of acceptance region, keeping in view 10% level of significance.
As the z value for 0.45 of the area under the normal curve is 1.64, we have the following limits of acceptance region.
Upper limit = `mu_U + 1.64 sigma_U=72 + 1.64 * 17.3205=100.4056`

Lower limit = `mu_U - 1.64 sigma_U=72 - 1.64 * 17.3205=43.5944`

The value of `U_1` is 54.5 which is in the acceptance region, we accept the null hypothesis and conclude that the two samples come from identical populations at 10% level.


test statistic using `R_2`

`U_2 = n_1 * n_2 + (n_2(n_2+1))/2 - R_2`

`=12 * 12 + (12(12+1))/2 - 132.5`

`=144 + 78 - 132.5`

`=89.5`

The value of `U_2` also lies in the acceptance region and as such our conclusion remians the same.

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