怎樣在Excel或者Matlab中進行dw檢驗?非常急啊!急

2021-03-24 01:47:56 字數 6030 閱讀 9668

1樓:匿名使用者

dw檢驗用於檢驗隨機誤差項具有一階自迴歸形式的序列相關問題,也是就自相關檢驗

d-w= ∑(et-et-1)^2/∑et^2,et是第t期的殘差,et-1是第t-1期的殘差,∑是對t從第2期到第t期求和,^2表示平方.

在d-w小於等於2時,d-w檢驗法則規定:

如d-w>d u,認為ei無自相關;

有自相關.

如4-d-w<dl,認為ei存在負自相關;

如dl<4-d-w<du,不能確定是否

有自相關.例如:資料x,y匯入matlab可以求出殘差,時間序列為【1:

1:20】,再利用excel就可以求出dw,結合上面的dw檢驗判斷ei是否具有自相關性。如果沒有就是普通的線性迴歸,如果有,重新建立新的模型。

x =1.0e+02 *

1.273000000000000

1.300000000000000

1.327000000000000

1.294000000000000

1.350000000000000

1.371000000000000

1.412000000000000

1.428000000000000

1.455000000000000

1.453000000000000

1.483000000000000

1.466000000000000

1.502000000000000

1.531000000000000

1.573000000000000

1.607000000000000

1.642000000000000

1.656000000000000

1.687000000000000

1.717000000000000

>> y

y =20.960000000000001

21.399999999999999

21.960000000000001

21.520000000000000

22.390000000000001

22.760000000000002

23.480000000000000

23.660000000000000

24.100000000000001

24.010000000000002

24.539999999999999

24.300000000000001

25.000000000000000

25.640000000000001

26.359999999999999

26.980000000000000

27.520000000000000

27.780000000000001

28.239999999999998

28.780000000000001

>> k=polyfit(x,y,1);

>> a=k(1);

>> b=k(2);

>> scatter(x,y,'-')

>> hold on

>> a

a =0.176290106553353

>> b

b =-1.457589881004317

>> y1=a*x+b

y120.984140683237555

21.460123970931608

21.936107258625661

21.354349906999598

22.341574503698375

22.711783727460418

23.434573164329166

23.716637334814532

24.192620622508585

24.157362601197917

24.686232920857975

24.386539739717271

25.021184123309343

25.532425432314071

26.272843879838156

26.872230242119553

27.489245615056291

27.736051764230986

28.282551094546381

28.811421414206439

>> x

x =1.0e+02 *

1.273000000000000

1.300000000000000

1.327000000000000

1.294000000000000

1.350000000000000

1.371000000000000

1.412000000000000

1.428000000000000

1.455000000000000

1.453000000000000

1.483000000000000

1.466000000000000

1.502000000000000

1.531000000000000

1.573000000000000

1.607000000000000

1.642000000000000

1.656000000000000

1.687000000000000

1.717000000000000

>> y

y =20.960000000000001

21.399999999999999

21.960000000000001

21.520000000000000

22.390000000000001

22.760000000000002

23.480000000000000

23.660000000000000

24.100000000000001

24.010000000000002

24.539999999999999

24.300000000000001

25.000000000000000

25.640000000000001

26.359999999999999

26.980000000000000

27.520000000000000

27.780000000000001

28.239999999999998

28.780000000000001

>> x=[ones(20,1),x]

x =1.0e+02 *

0.010000000000000 1.273000000000000

0.010000000000000 1.300000000000000

0.010000000000000 1.327000000000000

0.010000000000000 1.294000000000000

0.010000000000000 1.350000000000000

0.010000000000000 1.371000000000000

0.010000000000000 1.412000000000000

0.010000000000000 1.428000000000000

0.010000000000000 1.455000000000000

0.010000000000000 1.453000000000000

0.010000000000000 1.483000000000000

0.010000000000000 1.466000000000000

0.010000000000000 1.502000000000000

0.010000000000000 1.531000000000000

0.010000000000000 1.573000000000000

0.010000000000000 1.607000000000000

0.010000000000000 1.642000000000000

0.010000000000000 1.656000000000000

0.010000000000000 1.687000000000000

0.010000000000000 1.717000000000000

>> [b,bint,r,rint,stats]=regress(y,x,0.05)

b =-1.457589881004317

0.176290106553353

bint =

-1.915783961023543 -0.999395800985091

0.173199098501920 0.179381114604787

r =-0.024140683237555

-0.060123970931610

0.023892741374340

0.165650093000401

0.048425496301626

0.048216272539584

0.045426835670835

-0.056637334814532

-0.092620622508583

-0.147362601197916

-0.146232920857976

-0.086539739717271

-0.021184123309343

0.107574567685930

0.087156120161843

0.107769757880448

0.030754384943709

0.043948235769015

-0.042551094546383

-0.031421414206438

rint =

-0.197380214591832 0.149098848116723

-0.234089240235379 0.113841298372159

-0.154875657364706 0.202661140113385

0.011358385565306 0.319941800435497

-0.130852274009141 0.227703266612392

-0.132455310112963 0.228887855192131

-0.137385784544954 0.228239455886624

-0.239133423038235 0.125858753409171

-0.271758881008546 0.086517635991379

-0.316600686646691 0.021875484250859

-0.315878102151843 0.023412260435891

-0.266571280710952 0.093491801276411

-0.206112831653057 0.163744585034371

-0.068631104926293 0.283780240298153

-0.090160666136194 0.264472906459881

-0.064278724936611 0.279818240697506

-0.146265745654130 0.207774515541547

-0.130937032188002 0.218833503726033

-0.213963196628595 0.128861007535830

-0.199398817443354 0.136555989030478

stats =

1.0e+04 *

columns 1 through 2

0.000099874786001 1.435738949226265

columns 3 through 4

0.000000000000000 0.000000767910880

怎樣在Excel中插入,在excel中,怎樣樣在表格中的 中插入

輸入p 大寫 文字欄選擇wingdings 2字型 也可以右擊語言欄上的軟鍵盤 選擇一類符號 放在字下邊是不太可能了,除非用藝術字的方法移動到想要的位置。你為了表示記號,你可以用其他方法標示出來啊。比如顏色。另外,按住alt鍵加回車可以在單元格內換行,那麼就可以在 租 字下面加 了 alt 4142...

excel怎麼做矩陣,怎樣在Excel中計算矩陣?

0,1 x 1 dx和 0,1 x 1 dx都是發散的 當然不能分開寫。在計算一般的無窮限反常積分,在分部積分一定要注意積分收斂性,主要的判斷方法有 1 非負函式cauchy判別法 f x g x 是比1 x高階的無窮小,積分 0,f x g x dx收斂,若是同階 等價無窮小 或低階的無窮小,積分...

excel計算日期函式,在excel怎樣用excel函式公式計算時間差

datedif a1,today y 從a1的日期值到今天相隔多少年 datedif 1973 4 1 today m 從1973 4 1至今天相隔多少月 datedif a1,b1,m a1至b1相隔多少天 公式中,前的日期放前,後的日期放後 你說的我試了,沒錯是等於1096 你為何會說不對,你試...