2. Example-2
2. Parametric test - F test for the following data
66,67,75,76,82,84,88,90,92
64,66,74,78,82,85,87,92,93,95,97, 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=80` and Variance `S_(1)^2=91.75` for `66,67,75,76,82,84,88,90,92`| `x` | `x - bar x = x - 80` | `(x - bar x)^2` | | 66 | -14 | 196 | | 67 | -13 | 169 | | 75 | -5 | 25 | | 76 | -4 | 16 | | 82 | 2 | 4 | | 84 | 4 | 16 | | 88 | 8 | 64 | | 90 | 10 | 100 | | 92 | 12 | 144 | | --- | --- | --- | | `sum x=720` | `sum (x - bar x)=0` | `sum (x - bar x)^2=734` |
Mean `bar x = (sum x)/n` `=(66 + 67 + 75 + 76 + 82 + 84 + 88 + 90 + 92)/9` `=720/9` `=80`
Sample Variance `S^2 = (sum (x - bar x)^2)/(n-1)` `=734/8` `=91.75`
`bar x_2=83` and Variance `S_(2)^2=129.8` for `64,66,74,78,82,85,87,92,93,95,97`| `x` | `x - bar x = x - 83` | `(x - bar x)^2` | | 64 | -19 | 361 | | 66 | -17 | 289 | | 74 | -9 | 81 | | 78 | -5 | 25 | | 82 | -1 | 1 | | 85 | 2 | 4 | | 87 | 4 | 16 | | 92 | 9 | 81 | | 93 | 10 | 100 | | 95 | 12 | 144 | | 97 | 14 | 196 | | --- | --- | --- | | `sum x=913` | `sum (x - bar x)=0` | `sum (x - bar x)^2=1298` |
Mean `bar x = (sum x)/n` `=(64 + 66 + 74 + 78 + 82 + 85 + 87 + 92 + 93 + 95 + 97)/11` `=913/11` `=83`
Sample Variance `S^2 = (sum (x - bar x)^2)/(n-1)` `=1298/10` `=129.8`
Step-3:
`F=("Larger estimate of variance")/("Smaller estimate of variance")`
`=129.8/91.75`
`=1.4147`
Step-4:
`df_1=8,df_2=10,F_(0.05)=3.0717`
As calculated `F=1.4147 < 3.0717`
So, `H_0` is accepted, Hence Two samples have same variance
This material is intended as a summary. Use your textbook for detail explanation.
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