Hypothesis tests for the three comparisons (1 and 2, 1 and 3, and 2 and 3) Use a 2-tail test
Create own scenario and perform the test?
Solution:Using Minitab
a)MTB > TwoT 15 37.533 27.318 15 5.248
0.7622;
SUBC> Confidence 95.0;
SUBC> Test 0.0;
SUBC> Alternative 0.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 15 37.5 27.3 7.1
2 15 5.248 0.762 0.20
Difference = μ (1) - μ (2)
Estimate for difference: 32.29
95% CI for difference: (17.15, 47.42)
T-Test of difference = 0 (vs ≠): T-Value = 4.58 P-Value = 0.000 DF
= 14
b)MTB > TwoT 15 37.533 27.318 15 142.47
45.514;
SUBC> Confidence 95.0;
SUBC> Test 0.0;
SUBC> Alternative 0.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 15 37.5 27.3 7.1
2 15 142.5 45.5 12
Difference = μ (1) - μ (2)
Estimate for difference: -104.9
95% CI for difference: (-133.4, -76.5)
T-Test of difference = 0 (vs ≠): T-Value = -7.66 P-Value = 0.000 DF
= 22
c)MTB > TwoT 15 5.2473 0.7622 15 142.47
45.514;
SUBC> Confidence 95.0;
SUBC> Test 0.0;
SUBC> Alternative 0.
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 15 5.247 0.762 0.20
2 15 142.5 45.5 12
Difference = μ (1) - μ (2)
Estimate for difference: -137.2
95% CI for difference: (-162.4, -112.0)
T-Test of difference = 0 (vs ≠): T-Value = -11.68 P-Value = 0.000
DF = 14
Using Excel:
Conclusion:
from above all test P-value << .
We Reject all the hypothesis.
Hypothesis tests for the three comparisons (1 and 2, 1 and 3, and 2 and 3)...
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