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# Table 3 Adding premises to NHST to conclude HT. Comparison of group means is used as an example. HT (in bold) is defined in the text

Perform NHST: if P-value ≥ α, then fail to reject H0.
If P-value < α, H0 is rejected and conclude
HA: ¬{(μ1 = μ2) [($$\overline{\boldsymbol{x}}$$ 1$$\overline{\boldsymbol{x}}$$ 2) due to chance alone]}
Further steps Aim to conclude HT Assume “there is no bias” Aim to conclude HB
Additional premises (1) ¬{(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to chance alone]}
(2) ¬{(μ1 ≠ μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to (μ1 ≠ μ2) alone nor chance alone]}
(1) ¬{(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to bias]} (1) ¬ {(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to bias alone nor chance alone]}
(2) (μ1 = μ2)
Reasoning HA: ¬{(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to chance alone]} ≡
HA: ({(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to chance alone]} {(μ1 ≠ μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to (μ1 ≠ μ2) alone nor chance alone]} {(μ1  μ2) [($$\overline{\boldsymbol{x}}$$ 1 $$\overline{\boldsymbol{x}}$$ 2) due to (μ1  μ2) alone]}).
Use 2 steps of disjunction elimination with (1) and (2)
HA: ¬{(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to chance alone]} ≡
HA: ({(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to bias nor chance alone]} {(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to bias]} {(μ1  μ2) [($$\overline{\boldsymbol{x}}$$ 1 $$\overline{\boldsymbol{x}}$$ 2) due to (μ1  μ2) alone]} {(μ1 ≠ μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to (μ1 ≠ μ2) alone]}).
Use disjunction elimination with (1)
HA: ¬{(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to chance alone]} ≡
HA: ({(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to bias alone]} {(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to bias alone nor chance alone]} (μ1 ≠ μ2)).
Use 2 steps of disjunction elimination with (1) and (2)
Conclusion Therefore {(μ1  μ2) [($$\overline{\boldsymbol{x}}$$ 1 $$\overline{\boldsymbol{x}}$$ 2) due to (μ1  μ2) alone]}, i.e., HT Therefore ({(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to bias nor chance alone]} {(μ1 ≠ μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) not due to (μ1 ≠ μ2) alone]} {(μ1  μ2) [($$\overline{\boldsymbol{x}}$$ 1 $$\overline{\boldsymbol{x}}$$ 2) due to (μ1  μ2) alone]}) Therefore {(μ1 = μ2) [($$\overline{x}$$ 1$$\overline{x}$$ 2) due to bias alone]}, i.e., HB