非参中多样本的趋势秩检验的计算机实现

一般来说，年龄越大的人的β脂蛋白的含量越大。现观察三组人，他们都是男性。第一组人的年龄在20到30岁之间，第二组人的年龄在30到40岁之间，第三组人的年龄在40到50岁之间。他们的β脂蛋白的测量值如下表，问这三组人的测量值是否符合人们的经验：年龄越大的人的β脂蛋白的含量越大？

 1.首先在mathematica中算出一些必要的值，之后化为标准正态分布：

a = {};
Subscript[m, 1] = {260, 200, 240, 170, 270, 205, 190, 200, 250, 200};
Subscript[m, 2] = {310, 310, 190, 225, 170, 210, 280, 210, 280, 240};
Subscript[m, 3] = {320, 260, 360, 310, 270, 380, 240, 295, 260, 250};
t = Sort[Join[Subscript[m, 1], Subscript[m, 2], Subscript[m, 3]]];
Do[Subscript[b, i] = {}, {i, 1, 3}];
Do[Do[AppendTo[Subscript[b, i], 
    Position[t, Subscript[m, i][[j]]]], {j, 1, 
    Length[Subscript[m, i]]}], {i, 1, 3}];
Do[Subscript[t, i] = 
   Table[Apply[Plus, Subscript[b, i][[j]]]/
    Length[Subscript[b, i][[j]]], {j, 1, 
     Length[Subscript[m, i]]}], {i, 1, 3}];
m = Split[t];
a = {};
For[i = 1, i <= Length[m], i++, 
  If[Length[m[[i]]] > 1, AppendTo[a, Length[m[[i]]]]
   ]];
Print[a];
Do[Subscript[n, i] = Length[Subscript[t, i]], {i, 1, 3}];
Do[Subscript[R, i] = Apply[Plus, Subscript[t, i]][[1]], {i, 1, 3}];
Do[Subscript[w, i] = 2 \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(t = 1\), \(i\)]
\*SubscriptBox[\(n\), \(t\)]\) - Subscript[n, i], {i, 1, 3}];
Do[Print[{Subscript[n, i], Subscript[R, i], Subscript[w, i]}], {i, 1, 
  3}]
T = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(3\)]\((
\*SubscriptBox[\(w\), \(i\)]*
\*SubscriptBox[\(R\), \(i\)])\)\)
n = \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(3\)]
\*SubscriptBox[\(n\), \(i\)]\);
ET = n^2  (n + 1)/2
Do[Subscript[k, i] = (a[[i]])^3 - a[[i]], {i, 1, Length[a]}];
DT = N[(n*(n^2 - 1) - \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(Length[a]\)]
\*SubscriptBox[\(k\), \(i\)]\))*(\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(3\)]\(
\*SubscriptBox[\(n\), \(i\)]*
\*SubscriptBox[\(w\), \(i\)]*
\*SubscriptBox[\(w\), \(i\)]\)\) - n^3)/(12 (n - 1)), 9]
y = -Abs[(T - (ET))/Sqrt[(DT)]]

 

2.然后将算出的y值代入SAS程序中

data;
    p=probnorm(-1.768463653);
    put p=;
run;

 

3.最后算出p值为0.038491711。