B. Determine the characteristics of the comparison distribution.
C. Determine the cut-off value for a 5% significance level.
From the given data, the following Table is calculated:
Therapy A | Therapy B | Therapy C | Total | |
N | 10 | 10 | 10 | 30 |
28 | 26 | 52 | 106 | |
Mean | 2.8 | 2.6 | 5.2 | 3.533 |
100 | 82 | 286 | 468 | |
Std. Dev. | 1.5492 | 1.2649 | 1.3166 | 1.7953 |
From the above Table, ANOVA Table is Calculated as follows:
Source of variation | Sum of Squares | Degrees of freedom | Mean Square | F |
Between Treatments | 41.8667 | 2 | 20.9333 | 10.9535 |
Within treatments | 51.6 | 27 | 1.9111 | |
Total | 93.4667 | 29 |
The F - Ratio = 10.9535. The p - value =0.0003. The result is significant at p <0.05
Question (a)
State your hypotheses:
H0 : Null Hypothesis:
HA: Alternative Hypothesis: (At least one mean is different from other 2 means)
Question (b):
Determine the characteristic of the comparison distribution:
F Distribution with Degrees of Freedom for numerator = 2 and Degrees of Freedom for denominator = 27
Question C:
Determine the cut off value for a 5% significance level:
From Table, critical value of F is given by:
3.354
B. Determine the characteristics of the comparison distribution. C. Determine the cut-off value f...
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