1. Example-1
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.
This material is intended as a summary. Use your textbook for detail explanation.
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