目录

参数估计

方差比的置信区间

均值差的置信区间

一个正态总体方差的点估计和置信区间

一个正态总体均值的点估计和置信区间

单样本t检验的SciPy实现方式

单样本t检验的statsmodels实现方式

两样本t检验SciPy的实现方式

两样本t检验statsmodels的实现方式

配对t检验的SciPy实现

方差分析

单因素方差分析的SciPy实现

事后检验

非参数方法

SciPy实现有符号秩和检验

SciPy实现秩和检验

一元线性回归

参数估计

方差比的置信区间

import numpy as npfrom scipy.stats import f# 定义一个实现方差比置信区间的函数def var_ratio_ci_est(data1,data2,alpha):n1 = len(data1) # 样本1的样本容量n2 = len(data2) # 样本2的样本容量f_lower_value = f.ppf(alpha/2,n1-1,n2-1) # 左侧临界值f_upper_value = f.ppf(1-alpha/2,n1-1,n2-1) # 右侧临界值var_ratio = np.var(data1) / np.var(data2)return var_ratio / f_upper_value,var_ratio / f_lower_valuesalary_18 = [1484, 785, 1598, 1366, 1716, 1020, 1716, 785, 3113, 1601]salary_35 = [902, 4508, 3809, 3923, 4276, 2065, 1601, 553, 3345, 2182]print(var_ratio_ci_est(salary_18,salary_35,0.05))

(0.05314530971751432, 0.8614126063098585)

均值差的置信区间

import numpy as npfrom scipy.stats import t# 两个正态总体均值差的置信区间def mean_diff_ci_t_est(data1,data2,alpha,equal=True):n1 = len(data1) # 样本1的样本容量n2 = len(data2) # 样本2的样本容量mean_diff = np.mean(data1) - np.mean(data2) # 两个样本的均值差（两个总体均值差的无偏点估计）sample1_var = np.var(data1) # 样本1的方差sample2_var = np.var(data2) # 样本2的方差if equal: # 方差未知且相等sw = np.sqrt(((n1-1)*sample1_var + (n2-1)*sample2_var) / (n1+n2-2))t_value = np.abs(t.ppf(alpha/2,n1+n2-2)) # t值return mean_diff - sw*np.sqrt(1/n1+1/n2)*t_value, \mean_diff + sw*np.sqrt(1/n1+1/n2)*t_valueelse: # 方差未知且不等df_numerator = (sample1_var/n1+sample2_var/n2)**2 # 自由度分子# 自由度分母df_denominator = (sample1_var/n1)**2/(n1-1) + (sample2_var/n2)**2/(n2-1)df = df_numerator / df_denominator # 自由度t_value = np.abs(t.ppf(alpha/2,df))return mean_diff - np.sqrt(sample1_var/n1 + sample2_var/n2)*t_value, \mean_diff + np.sqrt(sample1_var/n1 + sample2_var/n2)*t_valuesalary_18 = [1484, 785, 1598, 1366, 1716, 1020, 1716, 785, 3113, 1601]salary_35 = [902, 4508, 3809, 3923, 4276, 2065, 1601, 553, 3345, 2182]print(mean_diff_ci_t_est(salary_18,salary_35,0.05,equal=True)) # 方差相等print(mean_diff_ci_t_est(salary_18,salary_35,0.05,equal=False)) # 方差不等

(-2196.676574582891, -199.32342541710875)(-2227.5521493017823, -168.4478506982175)

一个正态总体方差的点估计和置信区间

import numpy as npfrom scipy.stats import chi2# 定义计算方差的置信区间的函数def var_ci_est(data,alpha):n = len(data) # 样本容量s2 = np.var(data) # 样本方差# chi2.ppf(alpha/2,n-1)的意思是返回左侧面积为alpha/2的卡方值chi2_lower_value = chi2.ppf(alpha/2,n-1)# # chi2.ppf(1-alpha/2,n-1)的意思是返回左侧面积为1-alpha/2的卡方值chi2_upper_value = chi2.ppf(1-alpha/2,n-1)return (n-1) * s2 / chi2_upper_value,(n-1) * s2 / chi2_lower_valuesalary_18 = [1484, 785, 1598, 1366, 1716, 1020, 1716, 785, 3113, 1601]salary_35 = [902, 4508, 3809, 3923, 4276, 2065, 1601, 553, 3345, 2182]# 样本方差是总体方差的无偏点估计,再调用函数计算方差的置信区间print(np.std(salary_18),np.var(salary_18),var_ci_est(salary_18,0.05))print(np.std(salary_35),np.var(salary_35),var_ci_est(salary_35,0.05))

631.0754629994736 398256.24 (188421.9056837747, 1327329.3204650925)1364.3074580167038 1861334.8399999999 (880629.6611156773, 6203554.546528138)

一个正态总体均值的点估计和置信区间

import numpy as npfrom scipy.stats import norm, t# 计算一个正态总体均值的置信区间的函数def mean_ci_est(data,alpha,sigma=None):n = len(data) # 样本容量sample_mean = np.mean(data) # 样本均值if sigma is None: # 方差未知的情况s = np.std(data)se = s / np.sqrt(n)# t.ppf(alpha / 2,n-1)返回的是左侧面积为alpha / 2对应的t值t_value = np.abs(t.ppf(alpha / 2,n-1))# 根据公式返回置信区间return sample_mean - se * t_value,sample_mean + se * t_valueelse: # 方差已知的情况se = sigma / np.sqrt(n) # 标准误# norm.ppf(alpha / 2)返回的是左侧面积为alpha / 2对应的z值z_value = np.abs(norm.ppf(alpha / 2))# 根据公式返回置信区间return sample_mean - se * z_value,sample_mean + se * z_valuesalary_18 = [1484, 785, 1598, 1366, 1716, 1020, 1716, 785, 3113, 1601]salary_35 = [902, 4508, 3809, 3923, 4276, 2065, 1601, 553, 3345, 2182]# 样本均值是总体均值的无偏点估计,并调用函数计算均值的置信区间print(np.mean(salary_18),mean_ci_est(salary_18,0.05))print(np.mean(salary_35),mean_ci_est(salary_35,0.05))

1518.4 (1066.9558093661096, 1969.8441906338905)2716.4 (1740.433238065152, 3692.366761934848)

假设检验

初始化设置

import pandas as pdimport scipy.stats as ssimport matplotlib# 解决绘图的兼容问题%matplotlib inlinematplotlib.rcParams['font.sans-serif'] = ['SimHei']

单样本t检验的SciPy实现方式

ccss = pd.read_excel("CCSS_sample.xlsx",sheet_name="CCSS") # 读取Excel文件中名称为CCSS的表单ccss.head()# 广州基期的消费信心指数的描述统计ccss.query("s0 == '广州' & time == 203004" ).index1.describe()from scipy import stats as ss# 单样本t检验ss.ttest_1samp(ccss.query("s0 == '广州' & time == 203004" ).index1,100)

Ttest_1sampResult(statistic=-1.3625667518512996, pvalue=0.17611075148299993)

我们可以看出P值大于0.05，接受H0,没有显著性差异

单样本t检验的statsmodels实现方式

from statsmodels.stats import weightstats as wsdes = ws.DescrStatsW(ccss.query("s0 == '广州' & time == 203004").index1)des.mean

97.16472701710536

# 置信水平为95%的置信区间（可信区间）des.tconfint_mean()

(93.03590418536487, 101.29354984884586)

# 单样本t检验des.ttest_mean(100)

(-1.3625667518512996, 0.17611075148299993, 99.0)

我们可以看出P值大于0.05，接受H0,没有显著性差异

两样本t检验SciPy的实现方式

from scipy import stats as ss# 消费的信心指数分布的对称性考察ccss.index1.plot.hist()

# 不同婚姻状况的消费信心指数分组描述ccss.groupby('s7').index1.describe()

# 方差齐性检验ss.levene(ccss.index1[ccss.s7 == '未婚'],ccss.index1[ccss.s7 == '已婚'])

LeveneResult(statistic=0.6178738290825966, pvalue=0.431337363959)

# 两样本t检验(方差齐性)ss.ttest_ind(ccss.index1[ccss.s7 == '未婚'],ccss.index1[ccss.s7 == '已婚'])

Ttest_indResult(statistic=2.4052614244262576, pvalue=0.01632071963816213)

# 两样本t检验(方差不齐性)ss.ttest_ind(ccss.index1[ccss.s7 == '未婚'],ccss.index1[ccss.s7 == '已婚'],equal_var=False)

Ttest_indResult(statistic=2.466907208318663, pvalue=0.013870358702526698)

两样本t检验statsmodels的实现方式

from statsmodels.stats import weightstats as wsd1 = ws.DescrStatsW(ccss.index1[ccss.s7 == '未婚']) # 未婚的消费信心指数数据d2 = ws.DescrStatsW(ccss.index1[ccss.s7 == '已婚']) # 已婚的消费信心指数数据comp = pareMeans(d1,d2) # 创建CompareMeans对象comp.ttest_ind() # 两组独立样本t检验(方差齐性)

(2.4052614244262576, 0.01632071963816213, 1131.0)

comp.ttest_ind(usevar='unequal') # 两组独立样本t检验(方差不齐性)

(2.4669072083186636, 0.013870358702526675, 690.0875773844764)

配对t检验的SciPy实现

import scipy.stats as ssimport pandas as pdccss_p.loc[:,['index1','index1n']].describe() # 取出4月和12月信心指数，并描述

# 用相关分析确认配对信息是否的确存在ss.pearsonr(ccss_p.index1,ccss_p.index1n)

(0.2638011798615908, 0.01301162367951006)

# 配对t检验ss.ttest_rel(ccss_p.index1,ccss_p.index1n)

Ttest_relResult(statistic=1.1616334792419984, pvalue=0.24856144386191056)

方差分析

单因素方差分析的SciPy实现

导包

import pandas as pdimport scipy.stats as ssimport matplotlib# 解决绘图的兼容问题%matplotlib inlinematplotlib.rcParams['font.sans-serif'] = ['SimHei']

ccss = pd.read_excel("CCSS_sample.xlsx",sheet_name="CCSS") # 读取Excel文件中名称为CCSS的表单# 分别提取出北京四个时间段的消费信心数据a = ccss.query("s0 == '北京' & time == 203004 ").index1b = ccss.query("s0 == '北京' & time == 203012 ").index1c = ccss.query("s0 == '北京' & time == 203112 ").index1d = ccss.query("s0 == '北京' & time == 203212 ").index1ss.levene(a,b,c,d) # 方差齐性检验

LeveneResult(statistic=0.44332330387152036, pvalue=0.7221678627997157)

ss.f_oneway(a,b,c,d) # 单因素方差分析

F_onewayResult(statistic=5.630155391280303, pvalue=0.0008777240313291846)

事后检验

# 需要先在环境中安装scikit_posthocs包：pip install scikit_posthocsimport scikit_posthocs as sp# 创建对象，该对象接收事后检验的数据，并且设置p值校正的方法（控制两两比较的α值）为'bonferroni'pc = sp.posthoc_conover(ccss,val_col='index1',group_col='time',p_adjust='bonferroni')

# 使用热力图显示比较结果heatmap_args = {'linewidths': 0.25, 'linecolor': '0.5', 'clip_on': False, 'square': True, 'cbar_ax_bbox': [0.80, 0.35, 0.04, 0.3]}sp.sign_plot(pc,**heatmap_args) # 绘制热力图

非参数方法

SciPy实现有符号秩和检验

import numpy as npfrom scipy import statsdata = [36, 32, 31, 25, 28, 36, 40, 32, 41, 26, 35, 35, 32, 87, 33, 35]data = np.array(data) # 转换为numpy数组，方便运算# 检验总体均值是否为37k = min(len(data[data>37]),len(data[data<37]))pvalue = 2*stats.binom.cdf(k,len(data),0.5)print(pvalue)

0.021270751953125

SciPy实现秩和检验

import scipy.stats as statsweight_high=[134,146,104,119,124,161,107,83,113,129] # 高蛋白人群的体重的样本weight_low=[70,118,101,85,112,132,94,100,68,86] # 低蛋白人群的体重的样本# mannwhitneyu方法针对样本容量相等或不等都是适用的stats.mannwhitneyu(weight_high,weight_low,alternative='two-sided')

MannwhitneyuResult(statistic=81.0, pvalue=0.0211339281291611)

一元线性回归

import pandas as pdimport scipy.stats as ssccss = pd.read_excel("CCSS_sample.xlsx",sheet_name="CCSS") # 读取Excel文件中名称为CCSS的表单# 建立年龄和总消费信心指数的回归方程ss.linregress(ccss.s3,ccss.index1)

LinregressResult(slope=-0.3576848241677336, intercept=108.89832302796889, rvalue=-0.2190793138600294, pvalue=6.243013355018415e-14, stderr=0.047077765734542185, intercept_stderr=1.815504369310354)

# 以k*2形式的二维数组提供数据ss.linregress(ccss.loc[:,['s3','index1']])

LinregressResult(slope=-0.3576848241677336, intercept=108.89832302796889, rvalue=-0.2190793138600294, pvalue=6.243013355018415e-14, stderr=0.047077765734542185, intercept_stderr=1.815504369310354)