1. Example-1
1. Non parametric test - Sign test for the following data
7,5,2,3,8,2,4,4,3,7,6,2,10
2,1,0,1,3,2,3,5,1,4,4,3,4, Significance Level `alpha=0.05` and One-tailed test
Solution:
Step-1: The sign test with the null hypothesis:
`H_0: p=0.50`.
`H_1: p!=0.50`.
Step-2: In 4th column, we put + if difference is positive, - if difference is negative and n/a if difference is zero.
| # | A | B | Sign | | 1 | 7 | 2 | + | | 2 | 5 | 1 | + | | 3 | 2 | 0 | + | | 4 | 3 | 1 | + | | 5 | 8 | 3 | + | | 6 | 2 | 2 | n/a | | 7 | 4 | 3 | + | | 8 | 4 | 5 | - | | 9 | 3 | 1 | + | | 10 | 7 | 4 | + | | 11 | 6 | 4 | + | | 12 | 2 | 3 | - | | 13 | 10 | 4 | + |
To perform the test, we count number of + sign and number of - sign.
Number of plus sign = 10
Number of minus sign = 2
n = 10 + 2 = 12
`mu=np=12xx0.5=6`
`sigma=sqrt(npq)=sqrt(12xx0.5xx0.5)=sqrt(3)=1.7321`
Applying the z-test statistic, we get
`z=(bar x-mu)/(sigma)=(10-6)/(1.7321)=2.3094`
Step-3:
At `alpha=0.05`, the critical value of `Z_(0.05)=1.6449`
As calculated `z=2.3094 > 1.6449`
So, `H_0` is rejected.
This material is intended as a summary. Use your textbook for detail explanation.
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