例：随机抽取的22个企业的纳税额（丛小到大）1.001.351.992.052.062.102.302.612.862.952.983.233.734.034.825.24 6.106.646.816.867.119.00中位数的95%置信区间(2.10,6.10)，实际置信度为98%实际置信度置信区间对称(1, 9)1-2*pbinom(0,22,0.5)=0.9999995(1.35, 7.11)1-2*pbinom(1,22,0.5)=0.999989(1.99, 6.86)1-2*pbinom(2,22,0.5)=0.9998789(2.05, 6.81)1-2*pbinom(3,22,0.5)=0.9991446(2.06, 6.64)1-2*pbinom(4,22,0.5)=0.99565651-2*pbinom(5,22,0.5)=0.9830995(2.10, 6.10)(2.30, 5.24)1-2*pbinom(6,22,0.5)=0.94752121-pbinom(6,22,0.5)-pbinom(5,22,0.5)=0.965310395%置信区间(不对称)：(2.10,5.24)或(2.30, 6.10),它们的实际置信度为96.5%
例：随机抽取的22个企业的纳税额(丛小到大) 中位数的95%置信区间(2.10,6.10), 实际置信度为98%. 实际置信度 置信区间(对称) 1-2*pbinom(0,22,0.5)=0.9999995 (1, 9) 1-2*pbinom(1,22,0.5)=0.999989 (1.35, 7.11) 1-2*pbinom(2,22,0.5)=0.9998789 (1.99, 6.86) 1-2*pbinom(3,22,0.5)=0.9991446 (2.05, 6.81) 1-2*pbinom(4,22,0.5)=0.9956565 (2.06, 6.64) 1-2*pbinom(5,22,0.5)=0.9830995 (2.10, 6.10) 1-2*pbinom(6,22,0.5)=0.9475212 (2.30, 5.24) 1-pbinom(6,22,0.5)-pbinom(5,22,0.5)=0.9653103 95%置信区间(不对称)：(2.10,5.24) 或(2.30, 6.10), 它们的实际置信度为 96.5%
例：随机抽取的22个企业的纳税额(丛小到大)1.00 1.35 1.99 2.05 2.06 2.10 2.30 2.61 2.86 2.95 2.98 3.23 3.73 4.03 4.82 5.246.10 6.64 6.816.86 7.119.00下四分位点的95%置信区间(不对称)为(1.35,3.23)，实际置信度为95.6%qbinom(0.025,22,0.25) =2qbinom(1-0.025,22,0.25)=10pbinom(1,22,0.25)=0.01486506pbinom(2,22,0.25)=0.060649431-pbinom(10,22,0.25)=0.00997443= pbinom(10,22,0.25,low=F)1-pbinom(9,22,0.25) =0.02950891= pbinom(9,22,0.25,low=F)pbinom(1,22,0.25)+1-pbinom(9,22,0.25)=0.04437397
例：随机抽取的22个企业的纳税额(丛小到大) 下四分位点的95%置信区间(不对称)为 (1.35,3.23), 实际置信度为 95.6% qbinom(0.025,22,0.25) = 2 qbinom(1-0.025,22,0.25) = 10 pbinom(1,22,0.25)= 0.01486506 pbinom(2,22,0.25)= 0.06064943 1-pbinom(10,22,0.25)= 0.00997443= pbinom(10,22,0.25, low=F) 1-pbinom(9,22,0.25) = 0.02950891= pbinom(9,22,0.25, low=F) pbinom(1,22,0.25)+1-pbinom(9,22,0.25)= 0.04437397
广义符号检验特点零分布不需要对称的假定检验某假定的比例值用二项分布
广义符号检验特点 • 零分布不需要对称的假定. • 检验某假定的比例值. • 用二项分布