问题描述
我使用以下输出运行了负二项式回归模型。
Call:
glm.nb(formula = n_Finance ~ gender + year + party.Conservative +
party.LibDem + leadership + years.in.parliament,data = Finance_UK,init.theta = 0.2537048556,link = log)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.5126 -1.2119 -0.6583 -0.0299 4.3157
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 147.295781 10.129801 14.541 < 2e-16 ***
gender -0.344608 0.052599 -6.552 5.69e-11 ***
year -0.072873 0.005060 -14.401 < 2e-16 ***
party.Conservative 0.413795 0.045723 9.050 < 2e-16 ***
party.LibDem -0.020742 0.073979 -0.280 0.779
leadership 0.950248 0.043269 21.962 < 2e-16 ***
years.in.parliament -0.001248 0.002733 -0.457 0.648
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(dispersion parameter for Negative Binomial(0.2537) family taken to be 1)
Null deviance: 11014 on 10674 degrees of freedom
Residual deviance: 10169 on 10668 degrees of freedom
AIC: 50414
Number of Fisher Scoring iterations: 1
Theta: 0.25370
Std. Err.: 0.00426
2 x log-likelihood: -50398.09200
我的解释变量如下:
gender
-女性编码为1,男性编码为0
year
-分类变量,范围为1997-2010
party.Conservative
-如果该成员是保守党成员则为1
party.LibDem
-如果该成员参加了自由民主党,则为1
leadership
-如果该成员担任领导职务,则为1;否则,则为0
years.in.parliament
-范围从0到31
我的因变量是计数变量n_Finance
,是成员在给定年份参加与金融相关的辩论的次数。
我已经解释了这些结果,但是我想在性别和年份之间进行互动,以了解男人或女人在一段时间内(1997年至2010年之间)是否越来越多地参与与金融相关的辩论。但是,当我添加gender*year
交互时,我的结果在统计上就变得微不足道了,我不确定如何阅读这些内容。我以前只在OLS中使用带有二进制和连续变量的交互,所以我不确定如何使用它。
Call:
glm.nb(formula = n_Finance ~ party.Conservative + party.LibDem +
leadership + years.in.parliament + gender * year,init.theta = 0.2537585135,link = log)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.5162 -1.2150 -0.6609 -0.0267 4.3501
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 153.835095 11.172623 13.769 <2e-16 ***
party.Conservative 0.410673 0.045722 8.982 <2e-16 ***
party.LibDem -0.023913 0.073980 -0.323 0.747
leadership 0.952143 0.043264 22.008 <2e-16 ***
years.in.parliament -0.001223 0.002733 -0.447 0.655
gender -32.394298 25.114414 -1.290 0.197
year -0.076136 0.005580 -13.644 <2e-16 ***
gender:year 0.015993 0.012533 1.276 0.202
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
(dispersion parameter for Negative Binomial(0.2538) family taken to be 1)
Null deviance: 11016 on 10674 degrees of freedom
Residual deviance: 10169 on 10667 degrees of freedom
AIC: 50414
Number of Fisher Scoring iterations: 1
Theta: 0.25376
Std. Err.: 0.00426
2 x log-likelihood: -50396.39100
解决方法
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