请问如何求两组率的差值的可信区间？

shiyiming

假设A组有403人 有效率为百分之33%  B组有401人 有效率为38%  求两组率的差值5%的可信区间~ 没找到现成的公式 求算法 谢谢各位了

shiyiming

确实没辙耶！
哪位，嘿，大哥能拿个可行性方案出来？

shiyiming

很简单的问题,教科书上有公式.两大样本率的Z检验的合并标准误算出+-1.96倍就行了.
也可以用 (p1-p2)+-1.96*sqrt(p1(1-p1)/n1+p2(1-p2)/n2)

shiyiming

楼上正确。
[code:1ein6bnj]data _null_;
   input pa na pb nb;
        dif_p = pb-pa;
        sigma = sqrt(pa*(1-pa)/na+pb*(1-pb)/nb);
        z = dif_p/sigma;
        p = 1-probnorm(z);
        _95LCI = dif_p-1.96*sigma;
        _95UCI =dif_p+1.96*sigma;
        format pa pb dif_p percent6. p pvalue6.4 _95LCI _95UCI percentn7.1;
        putlog @2 'Group A:' @20 'Proportion = ' pa @40 'sample = ' na;
        putlog @2 'Group B:' @20 'Proportion = ' pb @40 'sample = ' nb;
        putlog @2 'Between A and B: ' @20'Difference[95% CI] = '  dif_p '[' _95LCI',' _95UCI']' '   p value = ' p;
        if p < 0.05 then put @2 ' Conclusion: the difference in proportion between A and B is not significant at 5% level';
                else                                       put @2 'Conclusion: the difference in proportion between A and B is significant at 5% level';
   cards;
   0.33 403  0.38 401
   ; [/code:1ein6bnj]

shiyiming

I am wondering how come SAS did not provide two-independent sample proportion test or just I don't know it. some thoughts here:
[code:2iuauos8]data a;
        input trt $ y p n;
        ny = round(p*n);
datalines;
a 1 0.33 403
a 0 0.67 403
b 1 0.38 401
b 0 0.62 401
;
proc ttest data = a;
        class trt;
        var y;
        freq ny;
run;[/code:2iuauos8]
Here a little trouble is the test p value for the difference using T test rather than normal test we want while the confidence interval is Okay.
*****JingJu*****

shiyiming

谢谢各位 我在一本别的书上找到了公式，用SAS简单的编了下 是没有各位编的好了 呵呵！

shiyiming

SAS里我也不知道有直接对两组率比较的程序 不过U检验的公式不算复杂 自己编个~~~嘿嘿