# 例: H₀: μ = 100, H₁: μ ≠ 100, 抽 36 个, X̄ = 104, s = 12
import numpy as np
from scipy.stats import t, norm

n, xbar, mu0, s = 36, 104, 100, 12
# 检验统计量
t_stat = (xbar - mu0) / (s / np.sqrt(n))
# p 值 (双侧)
p_value = 2 * (1 - t.cdf(abs(t_stat), df=n-1))
print(f"t = {t_stat:.3f}, p = {p_value:.3f}")
# t = 2.0, p ≈ 0.053 → 不显著 (临界)

# 决策
if p_value < 0.05:
    print("拒绝 H₀: 平均值显著不等于 100")
else:
    print("不能拒绝 H₀: 证据不足")

# 双侧
p_two = 2 * (1 - t.cdf(abs(t_stat), df=n-1))
# 单侧 (右)
p_one = 1 - t.cdf(t_stat, df=n-1)
print(f"双侧 p = {p_two:.4f}, 单侧 p (右) = {p_one:.4f}")
# 单侧 p 是双侧的一半

# 同一数据, 两种报告
xbar, s, n, mu0 = 104, 12, 36, 100

# 1. 假设检验
t_stat = (xbar - mu0) / (s / np.sqrt(n))
p = 2 * (1 - t.cdf(abs(t_stat), df=n-1))
print(f"检验: t={t_stat:.2f}, p={p:.3f}")

# 2. 95% 置信区间
t_crit = t.ppf(0.975, n-1)
margin = t_crit * s / np.sqrt(n)
ci = (xbar - margin, xbar + margin)
print(f"95% CI: [{ci[0]:.2f}, {ci[1]:.2f}]")
# CI = [99.94, 108.06] 不包含 100, 所以拒绝 H₀ (但临界)

# 例子
def cohens_d(x, y):
    nx, ny = len(x), len(y)
    pooled_std = np.sqrt(((nx-1)*x.std(ddof=1)**2 + (ny-1)*y.std(ddof=1)**2) / (nx + ny - 2))
    return (x.mean() - y.mean()) / pooled_std

treatment = np.random.normal(0.5, 1, 1000000)  # 处理组均值 0.5
control = np.random.normal(0.0, 1, 1000000)     # 对照组均值 0.0
d = cohens_d(treatment, control)
print(f"Cohen's d = {d:.4f}")  # 0.5 (中等效应, 但 n=100万可能"显著")

# d 解释: 0.2 小, 0.5 中, 0.8 大

# Bonferroni 修正
n_tests = 20
alpha_individual = 0.05 / n_tests
print(f"Bonferroni 阈值: {alpha_individual}")  # 0.0025

# BH FDR
from statsmodels.stats.multitest import multipletests
p_values = [0.001, 0.008, 0.039, 0.041, 0.042, 0.06, 0.074, 0.205]
rejected, p_adjusted, _, _ = multipletests(p_values, alpha=0.05, method="fdr_bh")
print("原始 p:", p_values)
print("BH 调整后 p:", p_adjusted.round(3))
print("是否拒绝:", rejected)

import numpy as np
from scipy.stats import ttest_ind, proportion_ztest
import matplotlib.pyplot as plt

# 模拟: 两种 UI 设计, 各 5000 用户
np.random.seed(42)
n_a, n_b = 5000, 5000
# A 设计: 5% 转化率
conv_a = np.random.binomial(1, 0.05, n_a)
# B 设计: 6% 转化率
conv_b = np.random.binomial(1, 0.06, n_b)

# === 方法 1: 比例 z 检验 ===
p_a, p_b = conv_a.mean(), conv_b.mean()
count_a, count_b = conv_a.sum(), conv_b.sum()
z_stat, p_z = proportion_ztest([count_b, count_a], [n_b, n_a])
print(f"比例 z 检验: z = {z_stat:.3f}, p = {p_z:.4f}")

# === 方法 2: 卡方检验 ===
from scipy.stats import chi2_contingency
table = np.array([[count_b, n_b - count_b], [count_a, n_a - count_a]])
chi2, p_chi, _, _ = chi2_contingency(table)
print(f"卡方检验: χ² = {chi2:.3f}, p = {p_chi:.4f}")

# === 方法 3: 95% 置信区间差值 ===
diff = p_b - p_a
se = np.sqrt(p_a*(1-p_a)/n_a + p_b*(1-p_b)/n_b)
ci = (diff - 1.96*se, diff + 1.96*se)
print(f"B - A 差异 95% CI: [{ci[0]:.4f}, {ci[1]:.4f}]")
# CI 不包含 0 → 显著

# === 效应大小: Cohen's h (比例专用) ===
def cohens_h(p1, p2):
    return 2 * (np.arcsin(np.sqrt(p1)) - np.arcsin(np.sqrt(p2)))
h = cohens_h(p_b, p_a)
print(f"Cohen's h = {h:.3f}")  # 0.043 (小效应)

比例 z 检验: z = 2.448, p = 0.0144
卡方检验: χ² = 5.991, p = 0.0144
B - A 差异 95% CI: [0.0022, 0.0203]
Cohen's h = 0.043