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  • Publicado : 2 de junio de 2011
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k=3 |
m=8 |
c=7 |
g=3 |
a=13 |

X0=4 m=2g3
a= 1+4(k)
a=1+4(3)
a=13

x1=(13*4+7) mod 8= 3 r1=3/7= 0.428
x2=(13*3+7) mod 8= 6r2=6/7= 0.857
x3=(13*6+7) mod 8= 5 r3=5/7= 0.714
x4=(13*5+7) mod 8= 0 r4=0/7= 0x5=(13*0+7) mod 8= 7 r5=7/7= 1
x6=(13*7+7) mod 8= 2 r6=2/7= 0.285
x7=(13*2+7) mod 8= 1r7=1/7= 0.142
x8=(13*1+7) mod 8= 4 r8=4/7= 0.571

Prueba de Medias
H0: μri=o.5
H1: μri=o.5
n=8
r= 18[0.428+0.857+0.714+0+1+0.285+0.142+0.571]
ŕ= 0.499

LIŕ=12 –Zα/2(1/√12n)
LIŕ=12 –Z0.05/2(1/√12(8)
LIŕ=12 –1.96(1/√96)
LIŕ=0.29958331

ISŕ=12 –Zα/2(1/√12n)= ½ +Z0.05/2 (1/√12(8)
ISŕ=12 +(1.96)(1/√96)=0.700041662
Prueba deVarianza
V(r) i=1ri-r2n-1


DESCRIPTIVE STATISTICS
data points 100
minimum 2.69
maximum 28.501
mean18.7356
median 18.5885
mode 18.1715
standard deviation 4.09382
variance 16.7594coefficient of variation 21.8505
skewness 0.
kurtosis 0.

GOODNESS OF FIT

data points 100
estimatesmaximum likelihood estimates
accuracy of fit 3.e-004
level of significance 5.e-002

SUMMARY

KOLMOGOROV ANDERSON
DISTRIBUTIONChi Squared Smirnov Darling

Lognormal 0.44 (5) 4.4e-002 0.205
Uniform 81.9 (5)...
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