1. Example-1
1. Parametric test - F test for the following data 19,22,24,27,24,18,20,19,25 26,37,40,35,30,30,40,26,30,35,45, Significance Level `alpha=0.05` and One-tailed test
Solution: Step-1: Let us take the hypothesis that the two sample have same variance Null Hypothesis `H_0 : S_1^2 = S_2^2`
Alternative Hypothesis `H_1 : S_1^2 != S_2^2`
Step-2: Calculate `S_1^2` and `S_2^2`
`bar x_1=22` and Variance `S_(1)^2=10` for `19,22,24,27,24,18,20,19,25``x` | `x - bar x = x - 22` | `(x - bar x)^2` | 19 | -3 | 9 | 22 | 0 | 0 | 24 | 2 | 4 | 27 | 5 | 25 | 24 | 2 | 4 | 18 | -4 | 16 | 20 | -2 | 4 | 19 | -3 | 9 | 25 | 3 | 9 | --- | --- | --- | `sum x=198` | `sum (x - bar x)=0` | `sum (x - bar x)^2=80` |
Mean `bar x = (sum x)/n` `=(19 + 22 + 24 + 27 + 24 + 18 + 20 + 19 + 25)/9` `=198/9` `=22`
Sample Variance `S^2 = (sum (x - bar x)^2)/(n-1)` `=80/8` `=10`
`bar x_2=34` and Variance `S_(2)^2=38` for `26,37,40,35,30,30,40,26,30,35,45``x` | `x - bar x = x - 34` | `(x - bar x)^2` | 26 | -8 | 64 | 37 | 3 | 9 | 40 | 6 | 36 | 35 | 1 | 1 | 30 | -4 | 16 | 30 | -4 | 16 | 40 | 6 | 36 | 26 | -8 | 64 | 30 | -4 | 16 | 35 | 1 | 1 | 45 | 11 | 121 | --- | --- | --- | `sum x=374` | `sum (x - bar x)=0` | `sum (x - bar x)^2=380` |
Mean `bar x = (sum x)/n` `=(26 + 37 + 40 + 35 + 30 + 30 + 40 + 26 + 30 + 35 + 45)/11` `=374/11` `=34`
Sample Variance `S^2 = (sum (x - bar x)^2)/(n-1)` `=380/10` `=38`
Step-3: `F=("Larger estimate of variance")/("Smaller estimate of variance")`
`=38/10`
`=3.8`
Step-4: `df_1=8,df_2=10,F_(0.05)=3.0717`
As calculated `F=3.8 > 3.0717`
So, `H_0` is rejected.
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
|