【DA】Python 实现区间估计

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一个总体均值的置信区间

def mean_interval(mean=None, std=None, sig=None, n=None, confidence=0.95):
    """
    mean:样本均值
    std：样本标准差
    sig: 总体方差
    n:   样本量
    confidence：置信水平
    功能：构建总体均值的置信区间
    """
    alpha = 1 - confidence
    z_score = scipy.stats.norm.isf(alpha / 2)  # z分布临界值
    t_score = scipy.stats.t.isf(alpha / 2, df = (n-1) )  # t分布临界值
   
    if n >= 30 and sig != None:
        me = z_score*sig / np.sqrt(n)  # 误差
        lower_limit = mean - me
        upper_limit = mean + me
        
    if n >= 30 and sig == None:
        me = z_score*std / np.sqrt(n)
        lower_limit = mean - me
        upper_limit = mean + me
        
    if n < 30 and sig == None:
        me = t_score*std / np.sqrt(n)
        lower_limit = mean - me
        upper_limit = mean + me
    
    return (round(lower_limit, 3), round(upper_limit, 3))
 
mean_interval(mean=8900, std=None, sig=500, n=35, confidence=0.95)
mean_interval(mean=8900, std=500, sig=None, n=35, confidence=0.90)
mean_interval(mean=8900, std=500, sig=None, n=35, confidence=0.99)

一个总体方差的置信区间

(1) 样本均值为21，  样本标准差为2，    样本量为50；                    
(2) 样本均值为1.3， 样本标准差为0.02， 样本量为15；                        
(3) 样本均值为167， 样本标准差为31，   样本量为22；                        
Question1: 根据以上样本结果，计算总体方差的90%的置信区间？  
Question2: 根据以上样本结果，计算总体标准差的90%的置信区间？        
 
def std_interval(mean=None, std=None, n=None, confidence=0.95, para="总体标准差"):
    """
    mean:样本均值
    std：样本标准差
    n:   样本量
    confidence：置信水平
    para：总体估计参数
    功能：构建总体方差&总体标准差的置信区间
    """
    variance = np.power(std,2)
    alpha = 1 - confidence
    
    chi_score0 = scipy.stats.chi2.isf(alpha / 2, df = (n-1))
    chi_score1 = scipy.stats.chi2.isf(1 - alpha / 2, df = (n-1))
   
    if para == "总体标准差"：
        lower_limit = np.sqrt((n-1)*variance / chi_score0)
        upper_limit = np.sqrt((n-1)*variance / chi_score1)
    if para == "总体方差":
        lower_limit = (n-1)*variance / chi_score0
        upper_limit = (n-1)*variance / chi_score1
        
    return (round(lower_limit, 2), round(upper_limit, 2))
 
std_interval(mean=21, std=2, n=50, confidence=0.90)   
std_interval(mean=1.3, std=0.02, n=15, confidence=0.90)  
std_interval(mean=167, std=31, n=22, confidence=0.90)  

两个总体方差比的置信区间

data1 = [3.45, 3.22, 3.90, 3.20, 2.98, 3.70, 3.22, 3.75, 3.28, 3.50, 3.38, 3.35, 2.95, 3.45, 3.20, 3.16, 3.48, 3.12, 3.20, 3.18, 3.25]
data2 = [3.22, 3.28, 3.35, 3.38, 3.19, 3.30, 3.30, 3.20, 3.05, 3.30, 3.29, 3.33, 3.34, 3.35, 3.27, 3.28, 3.16, 3.28, 3.30, 3.34, 3.25]
 
def two_std_interval(d1, d2, confidence=0.95, para="两个总体方差比"):
    """
    d1: 数据1
    d2: 数据2
    confidence：置信水平
    para：总体估计参数
    功能：构建两个总体方差比&总体标准差比的置信区间
    """
    n1 = len(d1)
    n2 = len(d2)
    var1 = np.var(d1, ddof=1)  # ddof=1  样本方差
    var2 = np.var(d2, ddof=1)  # ddof=1  样本方差
    alpha = 1 - confidence
    
    f_score0 = scipy.stats.f.isf(alpha / 2, dfn=n1-1, dfd=n2-1)    # F分布临界值
    f_score1 = scipy.stats.f.isf(1-alpha / 2, dfn=n1-1, dfd=n2-1)  # F分布临界值
   
    if para == "两个总体标准差比":
        lower_limit = np.sqrt((var1 / var2) / f_score0)
        upper_limit = np.sqrt((var1 / var2) / f_score01)
    if para == "两个总体方差比":
        lower_limit = (var1 / var2) / f_score0
        upper_limit = (var1 / var2) / f_score1
        
    return (round(lower_limit, 2), round(upper_limit, 2))
 
two_std_interval(data1, data2, confidence=0.95, para="两个总体方差比")

全文转自：python 实现参数估计–置信区间