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.
Any bug, improvement, feedback then
|
