大家好：我附上了一个示例数据集，来强调一个让我困惑的问题。假设我想要计算两个比例差值的置信区间。这应该很简单。通过“以X拟合Y”（JMP v18.0.0），我得到了以下标题下的三个区间：
两个样本比例检验。
风险差异
TOST 测试
正如您所见，间隔彼此并不一致。
我的问题是：
1. 为什么它们不一样？
2. 每个公式使用什么公式？
3. 我遗漏了什么吗？
我确实在两个样本比例报告中找到了这一点（在我看来可能更明确）。
双样本比例检验报告
感谢您的博学考虑！
仅供参考，我咨询了技术支持并收到了以下信息。
这三种检验方法略有不同，因此您在置信区间中观察到了差异。所有方法均已在 SAS 文档中进行了描述： https://go.documentation.sas.com/doc/en/statcdc/14.2/statug/statug_freq_details53.htm#statug.freq.freqrdiffcl
1. 对于比例的双样本检验，JMP 使用 Agresti-Caffo 极限：脚本如下。
x1 = 19; n1 = 20;
x2 = 18; n2 = 20;
/* proportion difference */
diff = x1/n1 - x2/n2;
/* Calculation for CI for Two Sample Test for Proportion */
p1 = (x1+1)/(n1+2);
p2 = (x2+1)/(n2+2);
pdiff= p1 - p2;
lower = pdiff - normalquantile(.95)*sqrt(p1*(1-p1)/(n1+2)+p2*(1-p2)/(n2+2));
upper = pdiff + normalquantile(.95)*sqrt(p1*(1-p1)/(n1+2)+p2*(1-p2)/(n2+2));
show(lower, upper);
2. 风险差异的置信区间采用Wald区间：脚本如下。
/* Calculation for CI for Risk Difference */
p1 = (x1)/(n1);
p2 = (x2)/(n2);
diff = x1/n1 - x2/n2;
lower = diff - normalquantile(.95)*sqrt(p1*(1-p1)/(n1)+p2*(1-p2)/(n2));
upper = diff + normalquantile(.95)*sqrt(p1*(1-p1)/(n1)+p2*(1-p2)/(n2));
show(lower, upper);
3. 等效性检验中风险差异的置信限度采用Miettinen-Nurminen（分数）置信限度：脚本如下。
/* Calculation for CI for Equivalence Test */
/* mnStdEquivDiff is a function to calculate the standard deviation of se(p1-p2) */
mnStdEquivDiff=Function({n11, n21, n1, n2, n, margin },
{default local},
p1_hat=n11/n1;
p2_hat=n21/n2;
theta=n2/n1;
d=-p1_hat*margin*(1+margin);
c=margin^2+margin*(2*p1_hat+theta+1) + p1_hat + theta*p2_hat;
b=-(1+theta+p1_hat + theta*p2_hat + margin*(theta+2));
a=1+theta;
v=b^3/(3*a)^3-b*c/(6*a^2) + d/(2*a);
if(v>0, sign=1, sign=-1);
u=sign*root(b^2/(3*a)^2 -c/(3*a));
w=(pi()+ArcCosine(v/u^3))/3;
p1_tlt=2*u*cosine(w)-b/(3*a);
p2_tlt =p1_tlt -margin;
var1=p1_tlt*(1-p1_tlt)/n1;
var2=p2_tlt*(1-p2_tlt)/n2;
correct=n/(n-1);
correct=1;
std_diff=root(correct*(var1+var2));
std_diff;
);
/* Calculation using the example data */
n11=19;
n21=18;
n1=20;
n2=20;
n=40;
alpha=0.05;
p1=n11/n1;
p2=n21/n2;
diff=p1-p2;
show(diff);
maxit=100;
conv = .00000001;
z=ChiSquare Quantile( 0.9, 1 );
pdiff = diff;
correct=1;
//calculate the lower confidence limit
inc=1+diff;
converged = 0;
for (i = 0, i < maxit, i++,
inc *= 0.5;
lcl = pdiff - inc;
var = mnStdEquivDiff(n11, n21, n1, n2, n, lcl);
var = var * var;
var *= correct;
d = diff - lcl;
if( var > 0, score=(d * d) / var, score=. );
if (score < z, pdiff = lcl);
if ((inc < conv) | abs(z - score) < conv,
converged = 1;
break();
);
);
if ((i == maxit) & converged==0, lcl = MACMISSING);
//calculate the upper confidence limit
inc=1-diff;
converged = 0;
for (i = 0, i < maxit, i++,
inc *= 0.5;
ucl = pdiff + inc;
var = mnStdEquivDiff(n11, n21, n1, n2, n, ucl);
var = var * var;
var *= correct;
d = diff - ucl;
if( var > 0, score=(d * d) / var, score=. );
if (score < z, pdiff = ucl);
if ( (inc < conv) | abs(z - score) < conv,
converged = 1;
break();
);
);
if ((i == maxit) & converged==0, ucl = MACMISSING);
show(lcl, ucl);
他们还表示将“向开发团队提出两项请求。一项是给文档团队的，要求他们为每种不同的方法添加参考文献（和统计细节）。另一项是向开发团队添加一项功能请求，允许用户选择所需的方法。”
与往常一样，技术支持非常有帮助、周到且专业。
