如何计算方差差异的P值及分组群体二项分布检验的P值？

Hey there, let's walk through your two statistical questions step by step—both are focused on calculating p-values for variance-related hypotheses, so I'll ground this in practical, actionable steps instead of just abstract theory.

Calculating a p-value for variance differences depends entirely on your data's distribution and sample setup. Here are the most common methods:

• F检验（仅适用于正态分布数据）
  This is the standard test for comparing two variances when you're confident your data follows a normal distribution:

  1. Let your two samples have variances s₁²/s₂² and sample sizes n₁/n₂. The null hypothesis is that the two population variances are equal; the alternative can be two-sided (variances differ) or one-sided (one variance is larger than the other).
  2. Compute the F-statistic: F = s₁² / s₂² (usually put the larger variance in the numerator to simplify one-sided tests).
  3. Define degrees of freedom: numerator df₁ = n₁ - 1, denominator df₂ = n₂ - 1.
  4. Calculate the p-value: For a two-sided test, find the probability that an F-distribution with df₁/df₂ degrees of freedom is greater than your F-statistic, then multiply by 2. For a one-sided test, just take the single-tail probability. You can use tools like R's pf() function or Python's scipy.stats.f.cdf() to compute this.

• Levene检验（适用于非正态或不确定正态性的数据）
  A more robust alternative that doesn't rely on normality:

  1. For each data point in a group, calculate the absolute deviation from the group mean: |xᵢⱼ - x̄ᵢ| (where xᵢⱼ is the j-th value in group i, x̄ᵢ is group i's mean).
  2. Run a one-way ANOVA on these absolute deviations. The F-statistic from the ANOVA tests whether the mean absolute deviations (and thus the variances) differ across groups.
  3. The p-value from the ANOVA output is your answer—small p-values indicate significant variance differences.

• Brown-Forsythe检验（Levene的更稳健变体）
  Works almost exactly like Levene's test, but replaces the group mean with the group median when calculating absolute deviations: |xᵢⱼ - medianᵢ|. This makes it even more resistant to outliers and non-normality. Follow the same ANOVA step to get the p-value.

First, a quick clarification: You mentioned the group means follow a distribution with variance kp(1-p) under the null hypothesis—this is likely a typo. For a group of size k where each individual is a Bernoulli(p) variable, the group mean (proportion of 1s) has a variance of p(1-p)/k (since the variance of a sample mean is the population variance divided by sample size). I'll proceed with that corrected value, but the logic holds if you adjust for your original wording.

The null hypothesis states that group means are just random samples from a binomial population; the alternative is that group differences are larger than expected by chance. Here's how to calculate the p-value:

• Method 1: Chi-Square Goodness-of-Fit Test

  1. Calculate each group's observed proportion pᵢ = (number of 1s in group i) / k. The expected proportion under the null is p (use the overall population proportion if p is unknown: p̂ = total number of 1s / (total number of individuals)).
  2. Compute the chi-square statistic: χ² = Σ [k*(pᵢ - p)² / (p(1-p))]. Each term here is (observed count - expected count)² / expected count (since k*pᵢ is the observed number of 1s, k*p is the expected number).
  3. Determine degrees of freedom: If p is known, df = number of groups m - 1. If p was estimated from the data, df = m - 2.
  4. Find the p-value: This is the probability that a chi-square distribution with the calculated df is greater than your χ² statistic. A small p-value means the observed group differences are unlikely under the binomial null hypothesis.

• Method 2: Variance Ratio Test

  1. Calculate the sample variance of the group means: s² = Σ (pᵢ - p̄)² / (m - 1) (use p instead of p̄ if the population proportion is known).
  2. The null hypothesis expects the group mean variance to be σ₀² = p(1-p)/k.
  3. Compute the chi-square statistic: χ² = (m - 1)*s² / σ₀². This follows a chi-square distribution with m - 1 df (or m - 2 if p was estimated).
  4. The p-value is the right-tail probability of this chi-square distribution above your calculated statistic—again, small values suggest the group variance is larger than binomial expectations.

• Key Notes

  • If you estimate p from the data, remember to adjust the degrees of freedom to account for that extra parameter you've estimated.
  • For large numbers of groups, you could use a normal approximation for the test statistic, but the chi-square test is the more standard and reliable choice here.

内容的提问来源于stack exchange，提问作者quarague