Estadística Inferencial

PROBLEMA 3
Se realizo un estudio sobre la cantidad de azúcar transformada en cierto proceso a varias temperaturas. Los fatos se recolectan y se registran como sigue;
Temperatura x | Azúcar trans y | Xi yi | Xi² | Yi² |
1.0 | 8.1 | 8.1 | 1 | 65.61 |
1.1 | 7.8 | 8.58 | 1.21 | 60.84 |
1.2 | 8.5 | 10.2 | 1.44 | 72.25 |
1.3 | 9.8 | 12.74 | 1.69 | 96.04 |
1.4 | 9.5 | 13.3 | 1.96 | 90.25 |1.5 | 8.9 | 13.35 | 2.25 | 79.21 |
1.6 | 8.6 | 13.76 | 2.56 | 73.96 |
1.7 | 10.2 | 13.34 | 2.89 | 104.04 |
1.8 | 9.3 | 16.74 | 3.29 | 86.49 |
1.9 | 9.2 | 17.48 | 3.61 | 84.64 |
2.0 | 10.5 | 21 | 4 | 110.25 |
16.5 | 100.4 | 152.69 | 25.85 | 923.58 |

a) Estime la línea de regresión lineal
b) Estime la cantidad de medida de azúcar transformada que se produce cuando latemperatura codificada es 1.75

FORMULAS

x=x1+x2+x3+….+xnn y= y1+y2+y3+….+ynn

b= xi yi-n xyxi²-nx² a=y+bx y=a+bx
SUSTITUCION
x=16.511=1.5 y=100.411=9.127 b= 152.59-111.5(9.127)25.85-11(1.5)²=1.81
Y=6.412+ 1.81x

(y-Y)² |
.0148 |
.3636 |
.0070 |
1.0712 |
.3069 |
.0515 |
.5012 |
.5055 |
.1369 |
.4238 |
.2190|
=3.60 |
Error estándar de la mejor estimación de la recta
sxy=y-Y2n-2

sxy=3.6011-2= 0.6325
VARIANZAS
s²=syy-bsxyn-2 s²=7.2-1.81(1.99)11-2= .3997

Coeficiente de correlacion de person ®
r=bsxxsyy r=sxysxxsyy
Calculando sxx
sxx=xi²-xi²n=25.88-(16.5)²11=1.11
Calculando syy
syy=yi²-(yi)²n=923.97-(110.4)²11=.70
Calculando sxysxy=xiyi-xiyin=152.59-16.5(100.4)11=1.99
r=(1.81)1.17.2=.70 r=1.991.11(7.2)= 1.992.81=70

Coeficiente de correlacion de spearman
xi | yi | Ryi | Ryi | dRI | dR2 |
1 | 8.1 | 1 | 2 | -1 | 1 |
1.1 | 7.8 | 2 | 1 | 1 | 1 |
1.2 | 8.5 | 3 | 3 | 0 | 0 |
1.3 | 9.8 | 4 | 9 | -5 | 25 |
1.4 | 9.5 | 5 | 8 | -3 | 9 |
1.5 | 8.9 | 6 | 5 | 1 | 1 |
1.6 | 8.6 | 7 | 4 | 3 | 9 |
1.7 |10.2 | 8 | 10 | -2 | 4 |
1.8 | 9.3 | 9 | 7 | 2 | 4 |
1.9 | 9.2 | 10 | 6 | 4 | 16 |
2.0 | 10.5 | 11 | 11 | 0 | 0 |

FORMULA SUSTITUCION
r=1-6di²n(n²-1) r=1-420720=0.4166

Intervalos de confianza para β
b-Tα2 Ssxx <β<b+Tα2 Ssxx s=syy-bsxyn-2=7.2-1.81(1.99)11-2=0.3997=.63221.81-2.262(0.6322)1.1=0.4465
1.81+2.262(0.6322)1.1=3.1734
0.4465<β<3.1734

Intervalos de confianza para α
α-T∝2SI=1Nxi²nSxx<∝<α+T∝2SI=1Nxi²nSxx
6.412-2.262(25.85)11(1.1) = 4.3218
6.412+2.262(25.85)11(1.1) =8.5021
4.3218<∝<8.5021
Intervalo de confianza para My/Xo
Yo-T∝2S1n+(xo-x)²Sxx<MyXo<Yo+T∝2S1n+(xo-x)²Sxx
Yo= a + bXo
Yo= 6.412 + 1.81 Yo=8.222
8.222+2.2620.6322111+1-1521.1=9.0285
8.222-2.2620.6322111+1-1521.1= 7.4154
7.4154<My/Xo<9.0285

Intervalo de confianza para Yo
Yo-T∝2S1+1n+xo-x2Sxx<Yo<Yo-T∝2S1+1n+xo-x2Sxx
8.22+2.262(0.6322)1+111+(1-1.5)²1.1 =9.8638
8.22-2.262(0.6322)1+111+(1-1.5)²1.1= 6.5801
6.5801<Yo<9.8638
Pruba de hipótesis sobre la pendiente β
PLANTEAMIENTO FORMULA
Ho : β=1 t=t-βssxxH1 : β<1 SUSTITUCION
t=1.81-10.6221.1= 1.3449
∝ = .05

GRAFICA TABLA v= n-2= 11-2 =9
∝V | ∝=.05 |
9 | 1.833 |

Prueba de hipótesis sobre la pendiente α
Planteamiento formulasustitucion
t=a-∝si=1nXi²nSxx t=6.412-10.633225.8511(1.1)=20.37
Ho: α=1 tabla
∝v | ∝/2 =.05/2 =.025 |
9 | 3.690 |

Hi:α<= 1
Α = .05
Grafica

Regresion lineal multiple
Xi²Yi | Xi³ | Xi^4 |
8.1 | 1 | 1 |
9.438 | 1.331 | 1.4641 |
17.64 | 1.72 | 2.0736 |
16.56 | 2.197 | 2.85 |
18.62 |...