1. Example-1
1. Non parametric test - Median test for the following data
79,86,40,50,75,38,70,73,50,40,20,80,55,61,50,80,60,30,70,50
85,80,50,55,65,50,63,75,55,45,30,85,65,80,55,75,65,50,75,62, Significance Level `alpha=0.05` and One-tailed test
Solution:
Step-1:State the hypothesis
`H_0`: There is no difference between Sample A and Sample B.
`H_1`: There is difference between Sample A and Sample B.
Step-2: Ranking all Sample values
First we assign ranks to all observations using high to low ranking process in the combined sample.
| Size in Descending Order | Rank | Name of related sample
A for sample1, B for sample2 | Rank for A | Rank for B | | 86 | 1 | A | 1 | | | 85 | 2.5 | B | | 2.5 | | 85 | 2.5 | B | | 2.5 | | 80 | 5.5 | A | 5.5 | | | 80 | 5.5 | A | 5.5 | | | 80 | 5.5 | B | | 5.5 | | 80 | 5.5 | B | | 5.5 | | 79 | 8 | A | 8 | | | 75 | 10.5 | A | 10.5 | | | 75 | 10.5 | B | | 10.5 | | 75 | 10.5 | B | | 10.5 | | 75 | 10.5 | B | | 10.5 | | 73 | 13 | A | 13 | | | 70 | 14.5 | A | 14.5 | | | 70 | 14.5 | A | 14.5 | | | 65 | 17 | B | | 17 | | 65 | 17 | B | | 17 | | 65 | 17 | B | | 17 | | 63 | 19 | B | | 19 | | 62 | 20 | B | | 20 | | 61 | 21 | A | 21 | | | 60 | 22 | A | 22 | | | 55 | 24.5 | A | 24.5 | | | 55 | 24.5 | B | | 24.5 | | 55 | 24.5 | B | | 24.5 | | 55 | 24.5 | B | | 24.5 | | 50 | 30 | A | 30 | | | 50 | 30 | A | 30 | | | 50 | 30 | A | 30 | | | 50 | 30 | A | 30 | | | 50 | 30 | B | | 30 | | 50 | 30 | B | | 30 | | 50 | 30 | B | | 30 | | 45 | 34 | B | | 34 | | 40 | 35.5 | A | 35.5 | | | 40 | 35.5 | A | 35.5 | | | 38 | 37 | A | 37 | | | 30 | 38.5 | A | 38.5 | | | 30 | 38.5 | B | | 38.5 | | 20 | 40 | A | 40 | | | Total | | | 446.5 | 373.5 |
Step-3: Find Grand Median
Median `= (20^(th) "term" + 21^(st) "term" )/2=(62 + 61)/2= 61.5`
| Sample A | Sample B | Total | | `>` Median | 8(a) | 12(b) | 20(a+b) | | `<=` Median | 12(c) | 8(d) | 20(c+d) | | Total | 20(a+c) | 20(b+d) | 40(a+b+c+d)=n |
Step-4: Computation of test statistic
`chi^2=(n(|ad-bc|-n/2)^2)/((a+b)(c+d)(a+c)(b+d))`
`chi^2=(40(|64-144|-40/2)^2)/(20xx20xx2020)`
`=(40(80-20)^2)/(20xx20xx20xx20)`
`=0.9`
Step-5: Compute the degrees of freedom (df).
`df=2-1=1`
Step-6:
The Critical value of chi-square is `chi^2(0.05,1)=3.8415`
Since the computed `chi^2`(0.9) < critical `chi^2`(3.8415)
So we accept the null hypothesis (`H_0`) and conclude that the two samples are identical.
This material is intended as a summary. Use your textbook for detail explanation.
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