Badano zalezność czasu wytwarzania pewnego wyrobu (w godz) od czasu szkolenia (w tygodniach) i stażu pracy (w latach). Na podstawie wykresów przyjęto hipotezę, że zależność jest liniowa. Oszacuj i zinterpretuj parametry modelu oraz dokonaj jego statystycznej weryfikacji. |
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Analiza regresji |
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PODSUMOWANIE - WYJŚCIE |
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Y |
X1 |
X2 |
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Czas wytwarzania |
Czas szkolenia |
Staż pracy |
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0 |
0 |
1 |
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Statystyki regresji |
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13 |
0 |
0 |
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X= |
0 |
1 |
1 |
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Wielokrotność R |
0,991137195940335 |
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11 |
0 |
1 |
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1 |
0 |
1 |
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R kwadrat |
0,982352941176471 |
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10 |
1 |
0 |
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1 |
2 |
1 |
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Dopasowany R kwadrat |
0,970588235294118 |
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8 |
1 |
2 |
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2 |
1 |
1 |
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Błąd standardowy |
0,387298334620742 |
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7 |
2 |
1 |
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2 |
0 |
1 |
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Obserwacje |
6 |
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8 |
2 |
0 |
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13 |
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ANALIZA WARIANCJI |
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Y= |
11 |
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df |
SS |
MS |
F |
Istotność F |
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b=(X'X)^-1X'Y |
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10 |
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Regresja |
2 |
25,05 |
12,525 |
83,4999999999999 |
0,002344274697024 |
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8 |
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Resztkowy |
3 |
0,450000000000001 |
0,15 |
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7 |
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Razem |
5 |
25,5 |
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8 |
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MACIERZ CROSS |
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Współczynniki |
Błąd standardowy |
t Stat |
Wartość-p |
Dolne 95% |
Górne 95% |
Dolne 99,0% |
Górne 99,0% |
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y |
X1 |
X2 |
1 |
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Przecięcie |
12,55 |
0,287228132326902 |
43,6934916448801 |
2,63877374954813E-05 |
11,6359110339894 |
13,4640889660106 |
10,8723442140534 |
14,2276557859466 |
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Y |
SUMA y^2 |
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Czas szkolenia |
-2,25 |
0,193649167310371 |
-11,6189500386222 |
0,001369331236764 |
-2,86627865516349 |
-1,63372134483651 |
-3,38107530014585 |
-1,11892469985415 |
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x1 |
suma (y*x1) |
suma(x1*x1) |
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Staż pracy |
-1,2 |
0,212132034355964 |
-5,65685424949238 |
0,010937657009751 |
-1,87509944228398 |
-0,524900557716019 |
-2,43903091225361 |
0,039030912253613 |
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x2 |
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1 |
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SKŁADNIKI RESZTOWE - WYJŚCIE |
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13 |
0 |
0 |
1 |
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13 |
11 |
10 |
8 |
7 |
8 |
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11 |
0 |
1 |
1 |
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0 |
0 |
1 |
1 |
2 |
2 |
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Obserwacja |
Przewidywane Czas wytwarzania |
Składniki resztowe |
10 |
1 |
0 |
1 |
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0 |
1 |
0 |
2 |
1 |
0 |
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1 |
12,55 |
0,450000000000001 |
8 |
1 |
2 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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2 |
11,35 |
-0,35 |
7 |
2 |
1 |
1 |
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3 |
10,3 |
-0,299999999999999 |
8 |
2 |
0 |
1 |
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4 |
7,9 |
0,100000000000001 |
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5 |
6,85 |
0,150000000000001 |
macierz cross |
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6 |
8,05 |
-0,049999999999999 |
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y |
X1 |
X2 |
1 |
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Y |
567 |
48 |
34 |
57 |
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X'X= |
10 |
4 |
6 |
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x1 |
48 |
10 |
4 |
6 |
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4 |
6 |
4 |
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80 |
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x2 |
34 |
4 |
6 |
4 |
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6 |
4 |
6 |
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1 |
57 |
6 |
4 |
6 |
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Szacowanie z wykorzystaniem funkcji reglinp |
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(x'x)-1= |
0,25 |
0,00 |
-0,25 |
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parametry |
-1,2 |
-2,25 |
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0,00 |
0,30 |
-0,20 |
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błędy |
0,212132034355964 |
0,193649167310371 |
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-0,25 |
-0,20 |
0,55 |
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Rkwadr/ s |
0,982352941176471 |
0,387298334620742 |
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F/v |
83,5 |
3 |
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20 |
0 |
-20 |
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RSK/SKR |
25,05 |
0,45 |
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(x'x)-1= |
1/80 |
0 |
24 |
-16 |
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-20 |
-16 |
44 |
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b= |
-2,25 |
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-1,2 |
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12,55 |
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model hipotetyczny |
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Y=B1*X1 + B2*X2 + B0 |
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model ekonometryczny |
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Y=b1*X1 + b2*X2 +b0 |
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m. ekonometryczny po oszacowaniu |
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Y=-2,25*X1 -1,2*X2 +12,55 |
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INTERPRETACJE: |
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Jeżeli czas szkolenia wzrośnie o 1 tydzień to ceteris paribus czas wytwarania wyrobu spadnie średnio o 2,25 godziny |
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Jeżeli staż pracy wzrośnie o 1 rok to ceteris paribus czas wytwarania wyrobu spadnie średnio o 1,2 godziny |
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PROGNOZY I BŁĘDY PROGNOZ |
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Posługując się odpowiednim wzorem i formułami zapisz macierz wariancji i kowarianacji ocen parametrów |
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odchylenie standardowe składnika losowego wynosi: |
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s |
0,387298334620742 |
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(godziny) |
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D^2(b)=s^2(X'X)^-1 |
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0,04 |
0,00 |
-0,04 |
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0,00 |
0,05 |
-0,03 |
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-0,04 |
-0,03 |
0,08 |
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0,25 |
0,00 |
-0,25 |
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(X'X)^-1= |
0,00 |
0,30 |
-0,20 |
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-0,25 |
-0,20 |
0,55 |
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Jaki jest przewidywany czas wytwarzania wyrobu przez pracownika o 1.5 rocznym stażu, który uczestniczył w 2 tygodniowym kursie? |
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2 |
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X*= |
1,5 |
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X*'= |
2 |
1,5 |
1 |
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Y*= |
6,25 |
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1 |
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Błąd prognozy wyznaczany w sposób macierzowy. |
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X*'(X'X)-1= |
0,25 |
0,25 |
-0,25 |
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X*'(X'X)-1X*= |
0,625 |
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m*= |
0,493710441453288 |
błąd |
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Przedział ufności dla prognozy (alfa=0.05) |
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rozkład reszt normalny |
|
t kryt |
3,18244630528371 |
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4,678793029717 |
7,82120697028301 |
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rozkłąd reszt nieznany |
|
w |
4,47213595499958 |
|
4,04205978341804 |
8,45794021658197 |
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