在oracle中计算中位数的置信区间

Question

我正在尝试生成一个 oracle sql 查询，它不仅可以计算中位年龄，还可以计算 95% 的置信度。在这种情况下，要增加复杂性，需要跨组进行性别我有 table 的人，他们的年龄和性别。我想确定每个组的中位年龄及其 95% 的置信区间。我目前失败的尝试如下。

select gender,
       median(age),
       count(*),
       percentile_cont(ROUND((COUNT(*)/2)-1.96*sqrt(COUNT(*))/2)/COUNT(*)) 
         within GROUP (ORDER BY age) lowmedianage,
       percentile_cont(ROUND((COUNT(*)/2)+1.96*sqrt(COUNT(*))/2)/COUNT(*)) 
         within GROUP (ORDER BY age) highmedianage
  from persontable
  group by gender

我得到的错误不是 GROUP BY 表达式。

Answer 1

这里的问题是您将函数 count 作为 percentile_cont 的参数，它需要一个常量，而这个常量必须是 group by 子句的一部分。您可能可以在此处使用子查询。与此类似的内容：

select gender, median(age), count(*),
       percentile_cont(low) within GROUP (ORDER BY age) lowmedianage,
       percentile_cont(high) within GROUP (ORDER BY age) highmedianage
  from (select age, gender,
               ROUND((COUNT(*)/2)-1.96*sqrt(COUNT(*))/2)/COUNT(*) low,
               ROUND((COUNT(*)/2)+1.96*sqrt(COUNT(*))/2)/COUNT(*) high
          from persontable
         group by age, gender)
 group by gender, low, high

Answer 2

使用 this book 中的公式，我将以以下查询结束（我不确定您是否能很好地处理低范围和高范围；我的解释是您计算了序列号的范围并且您必须从这些位置查找值）。

with tab as 
-- add sequence per group
(
select gender, age,
row_number() over (PARTITION  BY gender order by  gender, age) as seq
from persontable
),
-- get count
N as (select gender, count(*) cnt from persontable group by gender),
-- calculate sequence numbers of the CI
ci_seq as (
select gender,
round(cnt/2 - (1.96 * sqrt(cnt)/2)) r,
round(1 + cnt/2 + (1.96 * sqrt(cnt)/2)) s
from n),
-- calculate median
med as (
select 
  gender,
  median(age) median_age
from  persontable
group by gender),
med2 as (
select med.gender, median_age, r, s
from med 
join ci_seq on med.gender = ci_seq.gender 
)
select gender, median_age,
(select age from tab where seq = r and gender = med2.gender) ci_from,
(select age from tab where seq = s and gender = med2.gender) ci_to
from med2
;

另请注意，该公式仅近似于 CI。您还可以检查 this thread 以了解替代计算。

在oracle中计算中位数的置信区间

calculating confidence intervals for a median in oracle

oracle

median