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- 2020-11-23
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自己后面找资料弄了下,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)]]
复制代码然后将算出的y值代入SAS程序中(mathematica中不懂怎么算p值,话说回来,求p值和求概率一样吗?(有点迷糊,基础不好啊)): - data;
- p=probnorm(-1.768463653);
- put p=;
- run;
复制代码最后算出p值为0.038491711
(PS:做完这个后发现教科书上解答有错 )
趋势的秩检验不是很主流,多样本一般都是用W-H检验,所以也没简单的菜单操作,只能通过编程实现了
希望后面有人重新改进下代码,弄成用SAS或mathematica一次性就能解决并且输出的。
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