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Table 2 APFD value for the test cases example Test suite id Test cases ordering APFD T1 A,B,C,D,E 0.5 T2 E,D,C,B,A 0.64 T3 C,E,B,A,D 0.84.
Moreover, when the W/B was 0.45 and 0.55, the MAPE value for the test specimens was 9.38% and 5.29%, respectively.
Moreover, when D was 0%, 10%and20%0%, the MAPE value for the test sample cured at high temperature was 3.05%4.47%%, respectively, while that for the test sample cured at room temperature was 7.83% 9.17%.
In all cases the value for the test and the P value were reported.
For Equation 1, the p value for the test was much less than 0.001, i.e., the evidence indicates heteroskedasticity.
In the case without reoptimizing the atomic parameters, the R 2 value for the test set falls to 0.58.
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The p-value for the test is then computed as p=frac{1+ {t_{m}>t}}{1000}, where t is the observed likelihood ratio.
Note, though, that the p-value for that test is much smaller than the p-value for the test of the null hypothesis that the true impact is zero or negative (0.67percentt).
In both cases, 1000 samples (of three times 50 iris) were generated and for each, 1000 bootstrap samples were used in the resampling method to determine a p-value for the test of homogeneity.
When the 95% interval excludes −5% but also lies above zero, a 2-sided P-value for the test of superiority was calculated.
The p-value for the test is determined by comparing the observed T statistic to a t-distribution with n-2 degrees-of-freedom.
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