研究货运总量 y （万吨）与工业总产值 x1（亿元）、农业总产值 x2（亿元）,居民非商品支出 X3 （亿元）的关系。数据见表3-9。
(1）计算出 y , x1 ,x2, x3 的相关系数矩阵。
(2）求 y 关于 x1 ,x2, x3 的三元线性回归方程。
(3）对所求得的方程做拟合优度检验。
(4）对回归方程做显著性检验。
(5）对每一个回归系数做显著性检验。
(6）如果有的回归系数没通过显著性检验，将其剔除，重新建立回归方程归方程的显著性检验和回归系数的显著性检验。
(7）求出每一个回归系数的置信水平为95％的置信区间
8）求标准化回归方程。
(9）求当X01=75,X02=42,X03=3.1时的，给定置信水平为95%，用算精确置信区间，手工计算近似预测区间
（10）结合回归方程对问题做一些基本分析
表3-9
注：每一小问的运行结果我以备注的形式 放在代码段里面
#导入需要的库和数据
import numpy as np
import statsmodels.api as sm
import statsmodels.formula.api as smf
from statsmodels.stats.api import anova_lm
import matplotlib.pyplot as plt
import pandas as pd
from patsy import dmatrices
# Load data
df = pd.read_csv('C:\\Users\\joyyiyi\\Desktop\\zy3.11.csv',encoding='gbk')
#解决第(!)问
#计算相关系数
cor_matrix = df.corr(method='pearson') # 使用皮尔逊系数计算列与列的相关性
# cor_matrix = df.corr(method='kendall')
# cor_matrix = df.corr(method='spearman')
print(cor_matrix)
'''
结果：
C:\Users\joyyiyi\AppData\Local\Programs\Python\Python39\python.exe C:/Users/joyyiyi/PycharmProjects/pythonProject6/0.py
y x1 x2 x3
y 1.000000 0.555653 0.730620 0.723535
x1 0.555653 1.000000 0.112951 0.398387
x2 0.730620 0.112951 1.000000 0.547474
x3 0.723535 0.398387 0.547474 1.000000
Process finished with exit code 0
'''#解决第（2）（3）（4）（5)问
result = smf.ols('y~x1+x2+x3',data=df).fit()
#print(result.params) #输出回归系数
print(result.summary())
print("\n")
print(result.pvalues) #输出p值
#
'''
运行结果：
C:\Users\joyyiyi\AppData\Local\Programs\Python\Python39\python.exe C:/Users/joyyiyi/PycharmProjects/pythonProject6/0.py
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.806
Model: OLS Adj. R-squared: 0.708
Method: Least Squares F-statistic: 8.283
Date: Wed, 09 Nov 2022 Prob (F-statistic): 0.0149
Time: 11:15:30 Log-Likelihood: -43.180
No. Observations: 10 AIC: 94.36
Df Residuals: 6 BIC: 95.57
Df Model: 3
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Intercept -348.2802 176.459 -1.974 0.096 -780.060 83.500
x1 3.7540 1.933 1.942 0.100 -0.977 8.485
x2 7.1007 2.880 2.465 0.049 0.053 14.149
x3 12.4475 10.569 1.178 0.284 -13.415 38.310
==============================================================================
Omnibus: 0.619 Durbin-Watson: 1.935
Prob(Omnibus): 0.734 Jarque-Bera (JB): 0.562
Skew: 0.216 Prob(JB): 0.755
Kurtosis: 1.922 Cond. No. 1.93e+03
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 1.93e+03. This might indicate that there are
strong multicollinearity or other numerical problems.
Intercept 0.095855
x1 0.100197
x2 0.048769
x3 0.283510
dtype: float64
Process finished with exit code 0
'''
'''
(2)回答：线性方程：
Y=-348.2802+3.7540x1+7.1007x2+12.4475x3
(3)回答：R方=0.806，调整后R方=0.708
#或者说R=0.806>R0.05(8)=0.632，所以接受原假设，说明x与y有显著的线性关系
#或者说调整后的决定系数为0.708，说明回归方程对样本观测值的拟合程度较好。
(4）回答：做(F检验）
#原假设H0=β1=β2=β3=0
# F=8.283>F0.05(3,6)=4.76，或者说P=0.0149<α=0.05，说明拒绝原假设H0，x与y有显著的线性关系
(5)x1,x2,x3的t值分别为：
t1=1.942<t0.05(8)=1.943或者α=0.100>α=0.05，所以接受原假设，说明x1对y没有显著的影响
t2=2.465>t0.05(8)=1.943或者α=0.049<α=0.05，所以拒绝原假设，说明x1对y有显著的影响
t3=1.178<t0.05(8)=1.943或者α=0.284>α=0.05，所以接受原假设，说明x1对y没有显著的影响
'''
#在第（5）中发现除了x2外其他回归系数都未通过显著性检验，首先剔除x3看看效果
result = smf.ols('y~x1+x2',data=df).fit()
#print(result.params) #输出回归系数
print(result.summary())
print("\n")
print(result.pvalues) #输出p值
'''
运行结果：
C:\Users\joyyiyi\AppData\Local\Programs\Python\Python39\python.exe C:/Users/joyyiyi/PycharmProjects/pythonProject6/0.py
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.761
Model: OLS Adj. R-squared: 0.692
Method: Least Squares F-statistic: 11.12
Date: Wed, 09 Nov 2022 Prob (F-statistic): 0.00672
Time: 11:49:08 Log-Likelihood: -44.220
No. Observations: 10 AIC: 94.44
Df Residuals: 7 BIC: 95.35
Df Model: 2
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Intercept -459.6237 153.058 -3.003 0.020 -821.547 -97.700
x1 4.6756 1.816 2.575 0.037 0.381 8.970
x2 8.9710 2.468 3.634 0.008 3.134 14.808
==============================================================================
Omnibus: 1.265 Durbin-Watson: 1.895
Prob(Omnibus): 0.531 Jarque-Bera (JB): 0.631
Skew: -0.587 Prob(JB): 0.730
Kurtosis: 2.630 Cond. No. 1.63e+03
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 1.63e+03. This might indicate that there are
strong multicollinearity or other numerical problems.
Intercept 0.019859
x1 0.036761
x2 0.008351
dtype: float64
Process finished with exit code 0
'''
#第（6）问回答：
在剔除x3后，回归方程Y=-459.6237+4.6756x1+8.9710x2
的拟合优度R2=0.761，F值=11.12（有所提高），回归系数的P值均小于0.05 因此回归系数均通过显著性t检验
#第（7）问回答：
通过summary（）输出的回归结果最右边“[0.025 0.975]”这个位置可以看到
常数项，x1，x2的回归系数置信水平为95%的置信区间分别为：[-821.547,-97.700],[0.381,8.970],[3.134,14.808]
#标准化
dfnorm = (df-df.mean())/df.std()
new = pd.Series({'x1': 4000,'x2': 3300,'x3': 113000,'x4': 50.0,'x5': 1000.0})
newnorm = (new-df.mean())/df.std()
#标准化后构建无截距模型
resultnorm = smf.ols('y~x1+x2',data=dfnorm).fit()
print(resultnorm.summary())
'''
运行结果：
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.761
Model: OLS Adj. R-squared: 0.692
Method: Least Squares F-statistic: 11.12
Date: Fri, 11 Nov 2022 Prob (F-statistic): 0.00672
Time: 22:51:34 Log-Likelihood: -6.5156
No. Observations: 10 AIC: 19.03
Df Residuals: 7 BIC: 19.94
Df Model: 2
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Intercept -8.327e-17 0.175 -4.75e-16 1.000 -0.415 0.415
x1 0.4792 0.186 2.575 0.037 0.039 0.919
x2 0.6765 0.186 3.634 0.008 0.236 1.117
==============================================================================
Omnibus: 1.265 Durbin-Watson: 1.895
Prob(Omnibus): 0.531 Jarque-Bera (JB): 0.631
Skew: -0.587 Prob(JB): 0.730
Kurtosis: 2.630 Cond. No. 1.12
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
Process finished with exit code 0
'''
第(8)问：
标准化后的回归方程为：
y=0.4792x1+0.6765x2-8.327e-17 # Load data
df = pd.read_csv('C:\\Users\\joyyiyi\\Desktop\\zy3.11.csv',encoding='gbk')
# print(df)
result = smf.ols('y~x1+x2',data=df).fit()
#标准化
dfnorm = (df-df.mean())/df.std()
new = pd.Series({'x1': 75,'x2': 42})
newnorm = (new-df.mean())/df.std()
#标准化后构建无截距模型
resultnorm = smf.ols('y~x1+x2',data=dfnorm).fit()
#单值预测
predictnorm = resultnorm.predict(pd.DataFrame({'x1': [newnorm['x1']],'x2': [newnorm['x2']]}))
#因为单值预测是基于标准化后的模型，需要对y值还原，y值还原方法：
ypredict = predictnorm*df.std()['y'] + df.mean()['y']
print("ypredict:")
print(ypredict)
#区间
predictions = result.get_prediction(pd.DataFrame({'x1': [75],'x2': [42]}))
print('置信水平为95%，区间预测：')
print(predictions.summary_frame(alpha=0.05))
#近似预测区间：
ylow=267.83-2*np.sqrt(result.scale)
yup=267.83+2*np.sqrt(result.scale)
print(ylow,yup)
'''
运行结果：
C:\Users\joyyiyi\AppData\Local\Programs\Python\Python39\python.exe C:/Users/joyyiyi/PycharmProjects/pythonProject6/回归作业.py
ypredict:
0 267.829001
dtype: float64
置信水平为95%，区间预测：
mean mean_se ... obs_ci_lower obs_ci_upper
0 267.829001 11.782559 ... 204.435509 331.222493
[1 rows x 6 columns]
219.66776823691464 315.99223176308533
Process finished with exit code 0
'''
第（9）问：
y0的预测值为267.829；
y0预测值的置信水平为95%的精确置信区间为：[204.44,331.22]，
y0近似预测区间为：[219.67,315.99]
在做这次作业的时候因为不确定答案对不对，参考了csdn的另一位朋友的文章：
R语言之多元线性回归xt3.11_princess yang的博客-CSDN博客_为了研究货运量y与工业总产值x1文章来源：https://www.toymoban.com/news/detail-437423.html
这篇写的很好，比我更有条理哦文章来源地址https://www.toymoban.com/news/detail-437423.html
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