Simulacion
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)...
Regístrate para leer el documento completo.