Considering the series of experiments in which oats are present in the recipe:

Question

BlackTom AI · Accepted Answer

A post-hoc pairwise comparison following an ANOVA shows that biscuits baked for 12 minutes are significantly less stable than those baked for 16 minutes but not significantly different from those baked for 14 minutes. 1) Relevant knowledge points and concepts - Use ANOVA to test whether mean stability differs across the three cooking-time groups (12, 14, 16 min). ANOVA tests the null hypothesis that all group means are equal. - If ANOVA is significant, perform multiple comparison (post-hoc) tests (e.g., Tukey HSD) to find which specific pairs of means differ while controlling Type I error. - Interpret p-values: p < 0.05 (typical threshold) indicates a statistically significant pairwise difference. Confidence intervals for mean differences that do not include zero also indicate significance. 2) Complete reasoning process step by step - Step 1: ANOVA compares between-group variance (differences among 12, 14, 16 min group means) to within-group variance (trial-to-trial variability). Suppose the ANOVA rejects the null → proceed to pairwise tests. - Step 2: Conduct pairwise comparisons: 12 vs 14, 12 vs 16, and 14 vs 16. Each pair yields a mean difference, standard error, test statistic and p-value (or a 95% CI for the difference). - Step 3: Examine results: the 12 vs 16 comparison produces p < 0.05 (or a 95% CI excluding 0), so that difference is significant. The 12 vs 14 comparison produces p > 0.05 (CI includes 0), so it is not significant. The 14 vs 16 comparison is not required for the stated conclusion but would be checked similarly. 3) Calculations (general form) - For a pairwise test: difference = mean16 − mean12; SE = sqrt(SE16^2 + SE12^2 − 2cov) (for independent groups cov = 0). t = difference / SE. Compare t to critical value or compute p. Result: p(12 vs 16) < 0.05; p(12 vs 14) > 0.05. 4) Why the correct answer is correct - The post-hoc results directly show a significant deficit in stability for the 12-minute group compared with the 16-minute group, but the 12-minute group's mean is not sufficiently different from the 14-minute group's mean to reach significance. That pattern matches option d. 5) Why other options are incorrect - a: Claims 16 is significantly more stable than both 12 and 14; incorrect because 16 vs 14 was not significant. - b: Claims 12 is significantly less stable than both 14 and 16; incorrect because 12 vs 14 was not significant. - c: Claims no significant differences at all; incorrect because 12 vs 16 is significant. 6) Conclusion The statistical evidence supports option d: biscuits cooked for 12 minutes are significantly less stable than those cooked for 16 minutes but not 14 minutes. This matches the provided answer.

Considering the series of experiments in which oats are present in the recipe:单项选择题

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