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6,3 |
38038,4 |
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8216 |
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Wykorzystujemy funkcję 'transponuj' Excela
Macierz X transponowana ma postać: |
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1 |
6,5 |
38183,2 |
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7290 |
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12,2 |
38309,2 |
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5125 |
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Xt |
6,3 |
6,5 |
12,2 |
14,3 |
16,4 |
16 |
14,9 |
13,2 |
10,3 |
10,4 |
13,1 |
15,1 |
17,5 |
18,7 |
18,9 |
19 |
17,6 |
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14,3 |
38418,1 |
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4784 |
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38038,4 |
38183,2 |
38309,2 |
38418,1 |
38504,7 |
38580,7 |
38609,4 |
38639,3 |
38660 |
38667 |
38653,6 |
38254 |
38242,2 |
38218,3 |
38190,6 |
38173,8 |
38157,1 |
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16,4 |
38504,7 |
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4320 |
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1 |
16 |
38580,7 |
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4550 |
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1 |
14,9 |
38609,4 |
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4890 |
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1 |
13,2 |
38639,3 |
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4998 |
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Wykorzystujemy zmienne występujące w metodzie Hellwiga
Macierz X |
1 |
10,3 |
38660 |
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Macierz Y |
4700 |
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1 |
10,4 |
38667 |
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4511 |
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1 |
13,1 |
38653,6 |
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4380 |
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1 |
15,1 |
38254 |
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4781 |
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1 |
17,5 |
38242,2 |
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4710 |
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1 |
18,7 |
38218,3 |
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4965 |
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1 |
18,9 |
38190,6 |
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5130 |
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1 |
19 |
38173,8 |
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5198 |
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1 |
17,6 |
38157,1 |
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5556 |
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Wykorzystujemy funkcję 'macierz.iloczyn' Excela
Mnożymy macierz Xt przez macierz X: |
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Wykorzystujemy funkcję 'macierz.iloczyn' Excela
Mnożymy macierz Xt przez macierz Y |
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17 |
240,4 |
652499,6 |
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88104 |
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XTX |
240,4 |
3654,26 |
9225593,46 |
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XTY |
1206235,2 |
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652499,6 |
9225593,46 |
25045216785,38 |
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3379352177,7 |
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Wykorzystujemy funkcję 'wyznacznik.macierzy' Excela
Liczymy wyznacznik macierzy XTX |
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Det XTX |
3261310869,57 |
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Wykorzystujemy funkcję 'macierz.odw' Excela
Tworzymy macierz odwrotną do macierzy XTX |
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W wyniku pomnożenia macierzy odwrotnej XtX-1 i macierzy XTY otrzymuje wektor ocen parametrów strukturalnych szacowanego modelu
Obliczam a |
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1965,5162 |
-0,3600 |
-0,0511 |
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136143,7456 |
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Macierz odwrotna (XTX)-1 |
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-0,3600 |
0,0040 |
0,0000 |
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a |
-175,6509 |
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-0,0511 |
0,0000 |
0,0000 |
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-3,3473 |
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Możemy zatem zapisać: |
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Y=136143,7456 - 175,6509X1 - 3,3473X2 |
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Yi |
Powstaje poprzez pomnożenie macierzy X przez a
Ỷi |
ei |
ei2 |
(Ỷi-Ỹ)2 |
(Yi-Ỹ)2 |
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8216,0000 |
7711,1332 |
504,8668 |
254890,5007 |
6393539,5636 |
9201586,9343 |
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7290,0000 |
7191,3137 |
98,6863 |
9738,9886 |
4034977,9361 |
4441184,3460 |
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5125,0000 |
5768,3438 |
-643,3438 |
413891,2014 |
343109,5419 |
3316,4048 |
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4784,0000 |
5034,9558 |
-250,9558 |
62978,8062 |
21795,3403 |
158872,5813 |
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4320,0000 |
4376,2126 |
-56,2126 |
3159,8608 |
650241,6021 |
744058,4637 |
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4550,0000 |
4192,0780 |
357,9220 |
128108,1370 |
981110,4679 |
400167,8754 |
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4890,0000 |
4289,2264 |
600,7736 |
360928,9162 |
798095,3654 |
85607,8754 |
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4998,0000 |
4487,7485 |
510,2515 |
260356,5683 |
482802,2238 |
34072,8166 |
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4700,0000 |
4927,8469 |
-227,8469 |
51914,1872 |
64893,1732 |
232891,4048 |
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4511,0000 |
4886,8507 |
-375,8507 |
141263,7119 |
87460,7188 |
451030,7578 |
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4380,0000 |
4457,4472 |
-77,4472 |
5998,0675 |
525829,5336 |
644147,8754 |
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4781,0000 |
5443,7274 |
-662,7274 |
439207,5522 |
68193,6419 |
161273,1107 |
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4710,0000 |
5061,6635 |
-351,6635 |
123667,1978 |
14622,7983 |
223339,6401 |
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4965,0000 |
4930,8830 |
34,1170 |
1163,9721 |
63355,5431 |
47344,6401 |
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5130,0000 |
4988,4731 |
141,5269 |
20029,8750 |
37680,7016 |
2765,5225 |
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5198,0000 |
5027,1426 |
170,8574 |
29192,2350 |
24163,3308 |
237,5225 |
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5556,0000 |
5328,9538 |
227,0462 |
51549,9831 |
21422,8746 |
139436,3460 |
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88104 |
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0,0000 |
2358040 |
14613294 |
16971334 |
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Obliczam: |
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Wariancja odchyleń losowych
Se2 |
168431,411491898 |
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Odchylenie standardowe
Se |
Przeciętna różnica między zaobserwowanymi wartościami zmiennej objaśnianej i wartościami teoretycznymi wynosi 410,404
410,403961350153 |
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Ỹ |
5182,58823529412 |
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Współczynnik zmienności losowej
Ve |
7,92% |
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Wartości te wskazują, że zmienne objaśniające modelu wyjaśniają |
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Współczynnik detreminacji
R2 |
0,861057489969317 |
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zmienność zmiennej objaśnianej w stopniu 0,861057; udział czynników |
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Współczynnik zbieżności
φ |
0,13894251003135 |
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przypadkowych w tej zmienności wynosi 0,138943 |
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*Weryfikacja - R2+φ=1 |
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Test istotności współczynnika koleracji wielorakiej: |
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Wyliczamy F*: |
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n1=k=2 |
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n2=n-k-1=14 |
Wartość na n1=2 oraz n2=14 odczytujemy z tablic Fishera
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F*=3,74 |
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Wyliczamy F: |
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R2= |
0,861057 |
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F= |
43,3803718071439 |
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Nie ma podstaw do odrzucenia hipotezy H1 - wśród zmiennych objaśniających istnieje przynajmniej jedna zmienna |
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która w istotny sposób oddziaływuje na zmienną objaśnianą. |
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Macierz wariancji i kowariancji ocen parametrów strukturalnych: |
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331054660,3 |
-60635,4 |
-8602,6 |
Macierz wariancji i kowariancji:
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D2(a) |
-60635,4 |
669,2 |
1,3 |
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-8602,6 |
1,3 |
0,2 |
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WARIANCJE: |
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STANDARDOWE BŁĘDY SZACUNKU: |
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D2(ao)= |
331054660,3 |
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D(ao)= |
18194,9075379318 |
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D2(a1)= |
669,2 |
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D(a1)= |
25,8686379068605 |
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D2(a2)= |
0,2 |
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D(a2)= |
0,472903394904892 |
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Model ma postać: |
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Y=136143,7456 - 175,6509X2 - 3,3473X3 |
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18194,9075379318 |
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25,8686379068605 |
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0,472903394904892 |
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Badanie istotności parametrów strukturalnych: |
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Obliczamy t: |
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t0= |
7,48251923326207 |
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t1= |
6,79011011837682 |
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t2= |
7,07818982917895 |
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Obliczamy t*: |
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n-k-1=17-2-1=14 |
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Odczytujemy z tablic t-studenta: |
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t*=2,145 |
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t0 > t* t1 > t* t2 > t* |
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Na poziomie istotności α = 0,05 hipoteze zerowa H0 należy odrzucic na rzecz hipotezy alternatywnej H1. |
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Oznacza to, że parametr α2 różni się w sposób istotny od zera. |
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Wszystkie zmienne objaśniające oddziaływują w sposób istotny na zmienną objaśnianą Y. |
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Badanie stałości wariancji: |
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i |
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1 |
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2 |
98,686314053557 |
96,8591625909006 |
9381,69737781051 |
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3 |
-643,34376610961 |
-645,170917572267 |
416245,512881041 |
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4 |
-250,955785413127 |
-252,782936875784 |
63899,2131755464 |
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5 |
-56,2126391321362 |
-58,0397905947927 |
3368,61729228738 |
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6 |
357,921970563126 |
356,094819100469 |
126803,520190196 |
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suma: |
10,9629087759386 |
0 |
872747,463821645 |
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średnia: |
1,82715146265643 |
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12 |
-662,727358903459 |
-589,253477375319 |
347219,660598906 |
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13 |
-351,66347235124 |
-278,189590823099 |
77389,4484423234 |
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14 |
34,1170350738685 |
107,590916602009 |
11575,8053352604 |
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15 |
141,526940740994 |
215,000822269135 |
46225,353576404 |
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16 |
170,857352698979 |
244,331234227119 |
59697,7520189473 |
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17 |
227,046213572015 |
300,520095100156 |
90312,3275590067 |
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suma: |
-440,843289168843 |
0 |
632420,347530847 |
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średnia: |
-73,4738815281404 |
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Należy zweryfikować hipotezę: |
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wobec hipotezy: |
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Sprawdzianem tej hipotezy jest statystyka: |
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Liczymy wariancje resztowe: |
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Wartości krytyczne odczytujemy z tablic Fishera: |
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F*= |
m1=n2-k-1 |
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m2=n1-k-1 |
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Wyliczenie: |
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n1=6 |
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m1=3 |
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n2=6 |
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m2=3 |
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k=2 |
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Odczytujemy z tablic Fishera
F* = 9,28 |
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290915,821273881 |
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210806,782510282 |
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F= |
1,38001167614088 |
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Jeśli F≤F*, nie ma podstaw do odrzucenia hipotezy H0, wariancja odchyleń losowych jest stała w czasie. |
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Zatem: F≤F*, nie ma podstaw do odrzucenia hipotezy H0, wariancja jest stała w czasie. |
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Badanie losowości: |
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Badanie losowego kształtowania się składnika losowego sprowadza się do weryfikowania hipotezy: |
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wobec hipotezy: |
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504,86681481413 |
a |
Oblicza się reszty modelu |
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98,686314053557 |
a |
Jeśli: |
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-643,34376610961 |
b |
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przyporządkowujemy symbol a. |
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-250,955785413127 |
b |
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przyporządkowujemy symbol b. |
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-56,2126391321362 |
b |
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357,921970563126 |
a |
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a = 9 |
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b = 8 |
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600,773598117259 |
a |
Oblicza się liczbę serii ke. Seria to podciąg złożony wyłącznie z symboli a lub b. |
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510,251475547833 |
a |
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e |
-227,846850372793 |
b |
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ke = 5 |
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-375,850651011788 |
b |
Możemy zapisać: |
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k=5 |
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-77,4471913894959 |
b |
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n1 = 9 |
Liczba reszt dodatnich |
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-662,727358903459 |
b |
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n2 = 8 |
Liczba reszt ujemnych |
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-351,66347235124 |
b |
Odczytujemy z tablic rozkładu dla testu serii: |
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34,1170350738685 |
a |
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a/2 |
1-a/2 |
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141,526940740994 |
a |
poziom istotności |
0,025 |
0,975 |
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170,857352698979 |
a |
n1-liczba el. dodatnich |
9 |
9 |
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227,046213572015 |
a |
n2-liczba el. ujemnych |
8 |
8 |
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Jeżeli k-liczba serii mieści się w przedziale <k1,k2> to rozkład jest losowy; |
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K* z tablic rozkładu serii |
wartość krytyczną z rozkładu liczby serii lewostronnego
K*1 |
wartość krytyczną z rozkładu liczby serii prawostronnego
K*2 |
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jeżeli k<klewo i k>kprawo to odrzucamy rozkład reszt i rozkład nie jest losowy. |
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5 |
14 |
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W naszym przypadku: 5 ≤ 5 ≤14,zatem nie mamy podstaw do odrzucenia hipotezy zerowej. |
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Badanie symetrii składnika losowego: |
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Werifikujemy hipotezę: |
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wobec: |
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gdzie: r - liczba reszt dodatnich n - liczba obserwacji |
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t* n-1, α - odczytujemy z tablic t-studenta |
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Wyliczamy: |
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n=17 |
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r=9 |
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α=0,05 |
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tα= |
0,235702260395516 |
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t* = 2,120 |
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Jeśli tα≤t* to nie ma podstaw do odrzucenia hipotezy H0, czyli składnik losowy jest symetryczny. Jeśli tα>t* to hipotezę H0 należy odrzucić na rzecz hipotezy alternatywnej H1 , czyli składnik losowy nie jest symetryczny. |
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W naszym przypadku tα≤t* więc nie ma podstaw do odrzucenia hipotezy H0. |
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Rozkład jest symetryczny. |
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Badanie normalności rozkładu składnika losowego - Hellwiga: |
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Weryfikujemy hipotezę: |
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H0: |
dystrybuanta rozkładu reszt jest równa Fn(Ei) |
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wobec: |
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H1: |
dystrybuanta rozkładu reszt jest różna Fn(Ei) |
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ei |
e2 |
Ui |
Ui rosnąco |
F |
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504,86681481413 |
254890,500700565 |
1,31510750026579 |
-1,72631215752836 |
0,0421 |
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98,686314053557 |
9738,98858147728 |
0,257064057246857 |
-1,67582060705022 |
0,0469 |
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-643,34376610961 |
413891,201392097 |
-1,67582060700796 |
-0,979038422900477 |
0,1638 |
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-250,955785413127 |
62978,8062323195 |
-0,653704751328129 |
-0,916034202505235 |
0,1798 |
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standaryzacja zmiennych |
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-56,2126391321362 |
3159,86079819978 |
-0,146426069536028 |
-0,653704751373009 |
0,2567 |
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357,921970563126 |
128108,137011791 |
0,932336715715354 |
-0,593509204942506 |
0,2764 |
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600,773598117259 |
360928,916194758 |
1,56493126833172 |
-0,201739110805752 |
0,4201 |
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Ui=ei/Se |
zmienna standaryzowana |
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510,251475547833 |
260356,568298741 |
1,32913378900074 |
-0,146426069582993 |
0,4418 |
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Se2=(S(ei-eśr)2)/n |
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-227,846850372793 |
51914,1872248018 |
-0,593509204893002 |
0,088870108660583 |
0,5354 |
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-375,850651011788 |
141263,711865985 |
-0,979038422850897 |
0,257064057208117 |
0,6014 |
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n=17 |
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-77,4471913894959 |
5998,0674541212 |
-0,201739110756058 |
0,368657902987945 |
0,6438 |
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-662,727358903459 |
439207,552239155 |
-1,72631215748682 |
0,445059527377263 |
0,6719 |
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ocena wariancji
Se2 |
Dla danego ciągu reszt e1,e2,...,en szacujemy odchylenia standardowe
Se |
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-351,66347235124 |
123667,197786131 |
-0,916034202463653 |
0,591423657871708 |
0,7229 |
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147377,485055411 |
383,897753386772 |
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34,1170350738685 |
1163,97208223157 |
0,088870108701824 |
0,932336715666758 |
0,8244 |
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141,526940740994 |
20029,8749555049 |
0,368657903028694 |
1,31510750023012 |
0,9058 |
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170,857352698979 |
29192,2349713031 |
0,445059527417557 |
1,32913378895128 |
0,9081 |
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227,046213572015 |
51549,9830973892 |
0,59142365791151 |
1,56493126828271 |
0,9412 |
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2358039,76088657 |
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Dł. Celi |
1/n |
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S |
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0,0588 |
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1 |
0,0000 |
0,0588 |
** |
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2 |
0,0588 |
0,1176 |
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S - liczba pustych cel |
6 |
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3 |
0,1176 |
0,1765 |
* |
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4 |
0,1765 |
0,2353 |
* |
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5 |
0,2353 |
0,2941 |
** |
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z tablic rozkładu hellwiga |
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6 |
0,2941 |
0,3529 |
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a=0,05 |
n=17 |
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7 |
0,3529 |
0,4118 |
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8 |
0,4118 |
0,4706 |
** |
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Klewo |
Kprawo |
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9 |
0,4706 |
0,5294 |
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3 |
8 |
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10 |
0,5294 |
0,5882 |
* |
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11 |
0,5882 |
0,6471 |
** |
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jeżeli S<Slewo i S>Sprawo to odrzucamy H0 i rozkład nie jest zgodny z rozkładem normalnym |
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12 |
0,6471 |
0,7059 |
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jeżeli S-liczba pustych cel mieści się w przedziale <Slewo,Sprawo> to rozkład jest normalny |
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13 |
0,7059 |
0,7647 |
* |
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14 |
0,7647 |
0,8235 |
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S mieści się w przedziale <Klewo, Kprawo> |
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15 |
0,8235 |
0,8824 |
* |
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16 |
0,8824 |
0,9412 |
* |
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17 |
0,9412 |
1,0000 |
* |
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Zatem brak podstaw do odrzucenia hipotezy Ho, rozkład jest zgodny z rozkładem normalnym. |
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Test autokorelacji Durbina Watsona: |
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Jest to najpopularniejszy test wykrywający autokorelację. |
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Weryfikujemy hipotezę: |
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H0 = q1 = 0 brak autokorelacji przeciwko |
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wobec: |
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H1 = q1= 0 występuje autokorelacja |
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jeżeli d należy do przedziału <0,4>-2 to przeprowadzamy dalszy test |
Statystyka dana jest wzorem: |
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jeżeli d = 2 to występuje brak autokorelacji |
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jeżeli d >4 i d<0 to nie ma podtsaw do odrzucenia H0 |
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jeżeli d należy d przedziału <0,2) to na tym etapie zakładamy, że istnieje autokorelacja dodatnia i stawiamy hipotezę H1: q1>0 |
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jeżeli d należy d przedziału (2,4> to na tym etapie zakładamy, że istnieje autokorelacja ujemna i stawiamy hipotezę H1: q1<0 |
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et |
et-1 |
et-et-1 |
(et-et-1)2 |
et2 |
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504,86681481413 |
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254890,500700565 |
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98,686314053557 |
504,86681481413 |
-406,180500760573 |
164982,59919811 |
9738,98858147728 |
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-643,34376610961 |
98,686314053557 |
-742,030080163167 |
550608,639866957 |
413891,201392097 |
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-250,955785413127 |
-643,34376610961 |
392,387980696483 |
153968,327395064 |
62978,8062323195 |
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-56,2126391321362 |
-250,955785413127 |
194,743146280991 |
37924,8930234194 |
3159,86079819978 |
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357,921970563126 |
-56,2126391321362 |
414,134609695262 |
171507,474947447 |
128108,137011791 |
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600,773598117259 |
357,921970563126 |
242,851627554133 |
58976,9130056915 |
360928,916194758 |
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510,251475547833 |
600,773598117259 |
-90,5221225694259 |
8194,25467447416 |
260356,568298741 |
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-227,846850372793 |
510,251475547833 |
-738,098325920626 |
544789,138726831 |
51914,1872248018 |
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-375,850651011788 |
-227,846850372793 |
-148,003800638995 |
21905,1250035874 |
141263,711865985 |
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-77,4471913894959 |
-375,850651011788 |
298,403459622292 |
89044,6247145528 |
5998,0674541212 |
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-662,727358903459 |
-77,4471913894959 |
-585,280167513964 |
342552,874485173 |
439207,552239155 |
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-351,66347235124 |
-662,727358903459 |
311,06388655222 |
96760,7415169722 |
123667,197786131 |
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34,1170350738685 |
-351,66347235124 |
385,780507425108 |
148826,599909174 |
1163,97208223157 |
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141,526940740994 |
34,1170350738685 |
107,409905667126 |
11536,8878354208 |
20029,8749555049 |
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170,857352698979 |
141,526940740994 |
29,3304119579843 |
860,27306562507 |
29192,2349713031 |
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227,046213572015 |
170,857352698979 |
56,1888608730369 |
3157,1880862095 |
51549,9830973892 |
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2405596,55545471 |
2358039,76088657 |
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zatem d = |
1,02016793582406 |
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z tablic Durbina Watsona odczytujemy wartości du i dl, |
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gdzie: |
a=0,05 |
n=17 |
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k - liczba zmiennych = 2 |
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n - liczba obserwacji = 17 |
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a=0,05 |
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jeżeli d należy do <dl,du> to brak rozstrzygnięcia testu |
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Czyli: du = 1,54 dl = 1,02 |
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jeżeli d < dl to odrzucamy H0 na rzecz H1 istnieje autokorelacja rzędu I dodatnia lub ujemna |
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jeżeli d > du to nie ma podstaw do dorzucenia H0, nie występuje autokorelacja dodatnia, lub ujemna rzędu I |
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d<dl |
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Brak rozstrzygnięcia testu. |
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Statystyka testu delta: |
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Weryfikujemy hipotezę: |
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H0 |
Net,et-t=0 |
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wobec hipotezy: |
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H1: |
Net,et-t≠0 |
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obliczamy autokorelacje od I do VII rzedu |
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et |
et-1 |
et-2 |
et-3 |
et-4 |
et-5 |
et-6 |
et-7 |
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504,86681481413 |
- |
- |
- |
- |
- |
- |
- |
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98,686314053557 |
504,86681481413 |
- |
- |
- |
- |
- |
- |
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-643,34376610961 |
98,686314053557 |
504,86681481413 |
- |
- |
- |
- |
- |
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-250,955785413127 |
-643,34376610961 |
98,686314053557 |
504,86681481413 |
- |
- |
- |
- |
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-56,2126391321362 |
-250,955785413127 |
-643,34376610961 |
98,686314053557 |
504,86681481413 |
- |
- |
- |
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357,921970563126 |
-56,2126391321362 |
-250,955785413127 |
-643,34376610961 |
98,686314053557 |
504,86681481413 |
- |
- |
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|
600,773598117259 |
357,921970563126 |
-56,2126391321362 |
-250,955785413127 |
-643,34376610961 |
98,686314053557 |
504,86681481413 |
- |
|
|
|
|
|
|
510,251475547833 |
600,773598117259 |
357,921970563126 |
-56,2126391321362 |
-250,955785413127 |
-643,34376610961 |
98,686314053557 |
504,86681481413 |
|
|
|
|
|
|
-227,846850372793 |
510,251475547833 |
600,773598117259 |
357,921970563126 |
-56,2126391321362 |
-250,955785413127 |
-643,34376610961 |
98,686314053557 |
|
|
|
|
|
|
-375,850651011788 |
-227,846850372793 |
510,251475547833 |
600,773598117259 |
357,921970563126 |
-56,2126391321362 |
-250,955785413127 |
-643,34376610961 |
|
|
|
|
|
|
-77,4471913894959 |
-375,850651011788 |
-227,846850372793 |
510,251475547833 |
600,773598117259 |
357,921970563126 |
-56,2126391321362 |
-250,955785413127 |
|
|
|
|
|
|
-662,727358903459 |
-77,4471913894959 |
-375,850651011788 |
-227,846850372793 |
510,251475547833 |
600,773598117259 |
357,921970563126 |
-56,2126391321362 |
|
|
|
|
|
|
-351,66347235124 |
-662,727358903459 |
-77,4471913894959 |
-375,850651011788 |
-227,846850372793 |
510,251475547833 |
600,773598117259 |
357,921970563126 |
|
|
|
|
|
|
34,1170350738685 |
-351,66347235124 |
-662,727358903459 |
-77,4471913894959 |
-375,850651011788 |
-227,846850372793 |
510,251475547833 |
600,773598117259 |
|
|
|
|
|
|
141,526940740994 |
34,1170350738685 |
-351,66347235124 |
-662,727358903459 |
-77,4471913894959 |
-375,850651011788 |
-227,846850372793 |
510,251475547833 |
|
|
|
|
|
|
170,857352698979 |
141,526940740994 |
34,1170350738685 |
-351,66347235124 |
-662,727358903459 |
-77,4471913894959 |
-375,850651011788 |
-227,846850372793 |
|
|
|
|
|
|
227,046213572015 |
170,857352698979 |
141,526940740994 |
34,1170350738685 |
-351,66347235124 |
-662,727358903459 |
-77,4471913894959 |
-375,850651011788 |
|
|
|
|
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|
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|
Nr rzędu (t) |
ret,et-t |
moduł ret,et-t |
(ret,et-t)^2 |
statystyka t |
t* |
autokorelacja (t>t*) |
|
|
|
|
|
|
1 |
0,453737273619722 |
0,453737273619722 |
0,205877513471859 |
1,90513049231222 |
2,145 |
nie zachodzi |
|
|
|
|
|
|
|
2 |
0,453737273619722 |
0,453737273619722 |
0,205877513471859 |
1,83582967824966 |
2,16 |
nie zachodzi |
|
|
|
|
|
|
|
3 |
0,451810876857128 |
0,451810876857128 |
0,204133068446407 |
1,7543937636346 |
2,179 |
nie zachodzi |
|
|
|
|
|
|
|
4 |
0,556521811118109 |
0,556521811118109 |
0,309716526250181 |
2,22159451455815 |
2,201 |
zachodzi |
|
|
|
|
|
|
|
5 |
0,530380093541231 |
0,530380093541231 |
0,281303043624805 |
1,97840095385842 |
2,228 |
nie zachodzi |
|
|
|
|
|
|
|
6 |
0,529185829306047 |
0,529185829306047 |
0,280037641938329 |
1,87100335115711 |
2,262 |
nie zachodzi |
|
|
|
|
|
|
|
7 |
0,565030017956327 |
0,565030017956327 |
0,319258921191727 |
1,93698177780505 |
2,306 |
nie zachodzi |
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t* |
z tablic rozkładu studenta |
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|
t* |
n = t-t-2 |
a = 0,05 |
|
Dla rzędu 4 zachodzi zjawisko autkorelacji. Odrzucamy H0 na rzecz H1. Współczynnik korelacji między et a et-t jest istotnie różny od zera. |
|
|
|
|
|
2,145 |
14 |
0,05 |
|
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|
2,16 |
13 |
0,05 |
|
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|
2,179 |
12 |
0,05 |
|
Dla rzędów: 2,3,4,5,6,7 nie zachodzi zjawisko autkorelacji. Nie ma podstaw do odrzucenia H0. Współczynnik korelacji między et a et-t jest nieistotnie różny od zera. |
|
|
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|
2,201 |
11 |
0,05 |
|
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2,228 |
10 |
0,05 |
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2,262 |
9 |
0,05 |
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2,306 |
8 |
0,05 |
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