是否可以在svyratio - 调查包中使用汇总分母?

时间:2016-06-06 16:46:04

标签: r survey

以下是我认为解释问题的最佳方法。这不是我第一次遇到这种情况。

Lookingfor_job:是一个分类变量,用于定义失业者寻找工作的方式。类似于:课程交付,寻找代理商和致电家庭成员。它假设(1,2,...,12)区间内的值。

在这里,我想用州和lookingfor_job计算失业总人数除以按群体划分的失业总人数。最后,我需要按州的百分比,以失业者寻找工作的方式。

预期结果: x%的寻找工作的人在“打电话给朋友”选项(job_find =='2')中尝试使用状态Y.
我想的是我可以为所有类别做这件事。

svyby(~unemployed,
  ~state+lookingfor_job, # total unemployed population per state and way looking for a job
  denominator = ~svyby(~unemployed, ~state, desocup.pnad), #total unemployed population per state
  design = desocup.pnad,
  svyratio,
  vartype = 'ci')

我想某种程度上我可以计算分离然后分开。但是,我在复杂调查中的知识无法帮助我。

svyby(~unemployed,
  ~state+lookingfor_job,
  design = desocup.pnad,
  svytotal,
  vartype= 'ci') -> findjob

svyby(~unemployed,
  ~state,
  design = desocup.pnad,
  svytotal,
  vartype= 'ci') -> total

1 个答案:

答案 0 :(得分:0)

忘记状态和svyby,这是全国范围内估算的svyratio吗?

# among all unemployed nationwide, what share are looking for a job?
svyratio( ~ seeking_job , denominator = ~ unemployed , design = your_design )

如果您正在寻找,那么您可以通过此配置将状态分解为svyratio

# among all unemployed (broken out by state), what share are looking for a job?
svyby( ~ seeking_job , denominator = ~ unemployed , by = ~ state , design = your_design , svyratio )

请注意,在某些情况下,svyby + svyratio中存在错误。您可能需要将所有变量添加到分母中,如下所示:

svyby( ~ find_job , denominator = ~ find_job + unemployed , by = ~ state , design = pnad , svyratio )