1. Example-1
1. Non parametric test - Chi square test for the following data
18,36,21,9,6
12,36,45,36,21
6,9,9,3,3
3,9,9,6,3, Significance Level `alpha=0.05` and One-tailed test
Solution:
Step-1:State the hypothesis
`H_0`: two categories variables are independent.
`H_1`: two categories variables are not independent.
Step-2:Observed Frequencies
| `B_1` | `B_2` | `B_3` | `B_4` | `B_5` | Total | | `A_1` | 18 | 36 | 21 | 9 | 6 | 90 | | `A_2` | 12 | 36 | 45 | 36 | 21 | 150 | | `A_3` | 6 | 9 | 9 | 3 | 3 | 30 | | `A_4` | 3 | 9 | 9 | 6 | 3 | 30 | | Total | 39 | 90 | 84 | 54 | 33 | 300 |
Step-3:Expected Frequencies Steps
`E_(i,j)=("RowTot"_i xx "ColTot"_j)/("GrandTot"`
`E_(1,1)=(90xx39)/(300)=11.7`
`E_(1,2)=(90xx90)/(300)=27`
`E_(1,3)=(90xx84)/(300)=25.2`
`E_(1,4)=(90xx54)/(300)=16.2`
`E_(1,5)=(90xx33)/(300)=9.9`
`E_(2,1)=(150xx39)/(300)=19.5`
`E_(2,2)=(150xx90)/(300)=45`
`E_(2,3)=(150xx84)/(300)=42`
`E_(2,4)=(150xx54)/(300)=27`
`E_(2,5)=(150xx33)/(300)=16.5`
`E_(3,1)=(30xx39)/(300)=3.9`
`E_(3,2)=(30xx90)/(300)=9`
`E_(3,3)=(30xx84)/(300)=8.4`
`E_(3,4)=(30xx54)/(300)=5.4`
`E_(3,5)=(30xx33)/(300)=3.3`
`E_(4,1)=(30xx39)/(300)=3.9`
`E_(4,2)=(30xx90)/(300)=9`
`E_(4,3)=(30xx84)/(300)=8.4`
`E_(4,4)=(30xx54)/(300)=5.4`
`E_(4,5)=(30xx33)/(300)=3.3`
Step-3:Expected Frequencies
| `B_1` | `B_2` | `B_3` | `B_4` | `B_5` | Total | | `A_1` | 11.7 | 27 | 25.2 | 16.2 | 9.9 | 90 | | `A_2` | 19.5 | 45 | 42 | 27 | 16.5 | 150 | | `A_3` | 3.9 | 9 | 8.4 | 5.4 | 3.3 | 30 | | `A_4` | 3.9 | 9 | 8.4 | 5.4 | 3.3 | 30 | | Total | 39 | 90 | 84 | 54 | 33 | 300 |
Step-4: Compute Chi-square
`chi^2=sum (O_(ij)-E_(ij))^2/(E_(ij))`
`=(18-11.7)^2/11.7+(36-27)^2/27+(21-25.2)^2/25.2+(9-16.2)^2/16.2+(6-9.9)^2/9.9+(12-19.5)^2/19.5+(36-45)^2/45+(45-42)^2/42+(36-27)^2/27+(21-16.5)^2/16.5+(6-3.9)^2/3.9+(9-9)^2/9+(9-8.4)^2/8.4+(3-5.4)^2/5.4+(3-3.3)^2/3.3+(3-3.9)^2/3.9+(9-9)^2/9+(9-8.4)^2/8.4+(6-5.4)^2/5.4+(3-3.3)^2/3.3`
`=39.69/11.7+81/27+17.64/25.2+51.84/16.2+15.21/9.9+56.25/19.5+81/45+9/42+81/27+20.25/16.5+4.41/3.9+0/9+0.36/8.4+5.76/5.4+0.09/3.3+0.81/3.9+0/9+0.36/8.4+0.36/5.4+0.09/3.3`
`=3.3923+3+0.7+3.2+1.5364+2.8846+1.8+0.2143+3+1.2273+1.1308+0+0.0429+1.0667+0.0273+0.2077+0+0.0429+0.0667+0.0273`
`=23.5669`
Step-5: Compute the degrees of freedom (df).
`df=(4-1)*(5-1)=12`
Step-6:for 12 df, `p(chi^2>=23.5669)=0.0233`
Since the p-value(0.0233) < `alpha(0.05)` (one-tailed test), we reject the null hypothesis `H_0`.
This material is intended as a summary. Use your textbook for detail explanation.
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