2. Example-2
2. Parametric test - t-test for the following data
2.9,3.1,3.5,3.4,3.0,4.0,3.7,3.0,4.0,4.0
2.7,2.8,3.0,3.5,3.7,3.2,3.0,3.1,2.9,2.8, Significance Level `alpha=0.05` and One-tailed test
Solution:
Step-1: Take the hypothesis
Null Hypothesis `H_0` : There is no significant differentiating between samples
Alternative Hypothesis `H_1` : There is significant differentiating between samples
Step-2: Calculate `S_1^2,S_2^2`
`bar x_1=3.46` and Variance `S_(1)^2=0.2004` for `2.9,3.1,3.5,3.4,3.0,4.0,3.7,3.0,4.0,4.0`| `x` | `x^2` | | 2.9 | 8.41 | | 3.1 | 9.61 | | 3.5 | 12.25 | | 3.4 | 11.56 | | 3 | 9 | | 4 | 16 | | 3.7 | 13.69 | | 3 | 9 | | 4 | 16 | | 4 | 16 | | --- | --- | | `sum x=34.6` | `sum x^2=121.52` |
Mean `bar x = (sum x)/n` `=(2.9 + 3.1 + 3.5 + 3.4 + 3 + 4 + 3.7 + 3 + 4 + 4)/10` `=34.6/10` `=3.46`
Sample Variance `S^2 = (sum x^2 - (sum x)^2/n)/(n-1)` `=(121.52 - (34.6)^2/10)/9` `=(121.52 - 119.716)/9` `=1.804/9` `=0.2004`
`bar x_2=3.07` and Variance `S_(2)^2=0.1023` for `2.7,2.8,3.0,3.5,3.7,3.2,3.0,3.1,2.9,2.8`| `x` | `x^2` | | 2.7 | 7.29 | | 2.8 | 7.84 | | 3 | 9 | | 3.5 | 12.25 | | 3.7 | 13.69 | | 3.2 | 10.24 | | 3 | 9 | | 3.1 | 9.61 | | 2.9 | 8.41 | | 2.8 | 7.84 | | --- | --- | | `sum x=30.7` | `sum x^2=95.17` |
Mean `bar x = (sum x)/n` `=(2.7 + 2.8 + 3 + 3.5 + 3.7 + 3.2 + 3 + 3.1 + 2.9 + 2.8)/10` `=30.7/10` `=3.07`
Sample Variance `S^2 = (sum x^2 - (sum x)^2/n)/(n-1)` `=(95.17 - (30.7)^2/10)/9` `=(95.17 - 94.249)/9` `=0.921/9` `=0.1023`
Step-3: Calculate `t`
`t=|x_1-x_2|/sqrt((S_1^2)/n_1 + (S_2^2)/n_2)`
`=|3.46-3.07|/sqrt(0.2004/10 + 0.1023/10)`
`=|0.39|/sqrt(0.02 + 0.0102)`
`=|0.39|/sqrt(0.0303)`
`=|0.39|/0.174`
`=2.2416`
Step-4:
Degree of freedom `= n_1 + n_2 - 2 = 10+10-2=18`
Step-5:
`df=18,t_(0.05)=1.7341`
As calculated `t=2.2416 > 1.7341`
So, `H_0` is rejected.
This material is intended as a summary. Use your textbook for detail explanation.
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