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
From: A logical analysis of null hypothesis significance testing using popular terminology
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 |