Summary on Goodness of Fit

Statistics

Table of Contents
Exercise-1.............................................................................................................................................1
Excerise-2.............................................................................................................................................3
Excerise-3.............................................................................................................................................5
Exercise-4.............................................................................................................................................8
Exercise-5...........................................................................................................................................11
References..........................................................................................................................................13

Exercise-1
ANSWER:
The population of the pth quintile value is computed as follows,
Let us consider the QD(p| D0 >Dn ) as the pthequation,
F (d, λ) =λ e− λd for d=0
QD(p| D0 >Dn)= d pD+=λ pX
Assume the pthquintile of Z with 0<p<1 are the individual values which consider the pthroot
Q (p), whose equation is F (QD (p)) =p.
ANSWER:
The MLE population is,
λ (QD (p))= log (Dn)
1

=[∑
1
n
log ( D 1 )−2nlog ( D2 ) − 1
λ ∑
i=0
1
exp (f ( λ e− λd ))
ANSWER:
The approximate confidence interval of QD (p) based on its MLE is equated below.
The approximate confidence interval and speculation tests carries the anticipated values as
the example of populace, and the size of the developed sample’s confidence interval is,
100(1- λ)% confidence interval which can be considered as QD ( p ) .
(QD (p) ± QD( p)λ)=[ D± λ(QD p
2 ) √ D
n ]
ANSWER:
We can calculate the pivot point estimator with the considered value QD (5) Dn, which is the
mean of the distribution.
Dn= D1+ D 2+... .. Dn
n
P(−0.5≤ √ D λ e−λd ≤ .0 .6 )=0.5
Ranging the terms equivalent to,
P(QD (5 )− 0.5
√n ≤ Q≤ QD ( 5 )+ 0.6
√ n )=0.5
Interval,
(QD (5 )− 0.5
√n ≤ Dn+ 0.6
√ n )
As our 95% confidence interval of QD ( 5 ) .
2