Download The final exam solutions

The final exam solutions
Part I, #1, Central limit
theorem
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Let X1,X2, …, Xn be a sequence of i.i.d. random
variables each having mean μ and variance σ2
The sum of a large number of independent random
variables has a distribution that is approximately
normal
X 1  X 2  ... X n is approximat ely normal with mean n and variance n 2
X 1  X 2  ... X n  n
is approximat ely a standard normal random variable
 n
Part II #2
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If the successful probability of trial is very small, then the
accumulatively many successful trials distribute as a Poisson.
The Poisson parameter λ is the mean frequency of interest
within a large bundle of experiment
The exponential distribution is often used to describe the
distribution of the amount of time until the specific event first
occurs.
The exponential distribution seems to partition the Poisson
distribution with a parameter λ by every two occurrences into
several small time intervals and focuses on a specific time
interval.
The same distributional parameter λ,

The exponential distribution emphasizes the cycle time, 1/λ, while the
Poisson on the frequency λ.
Part I, #3
If Z and Xn2 are independent random variables,
with Z having a std. normal dist. And Xn2 having a
chi-square dist. with n degrees of freedom,
Tn 
Z12  Z 22  ...Z n2
X 2n
,

2
n
Xn / n n
Z
•For large n, the t distribution approximates to the standard normal
distribution
Part I, #4

When these two population variances are
unknown but equal, we calculate the
pooled variance estimator, Sp2, by means
of weighted average of individual sample
2
2
variance, S1 and S2
Part I, #5
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It is not correct.
He had better say that the statistic or random variables
used to obtain this confidence interval is such that 95%
of the time that it is employed it will result in an interval
covered the μ.
He can assert with probability 0.95 the interval will cover
the μ only before the sample data are observed.
Whereas after the sample data are observed and
computed the interval, he can only assert that the
resultant interval indeed contains μ with confidence 0.95.
Part I, #6
Suppose the i.i.d. random variables X1 , X 2 ,  X n , whose joint distributi on
is assumed given except for an unknown parameter  , are to be observed
and constitute d a random sample.
The value of likelihood function f (x 1 , x 2 , , x n / ) will be determined
by the observed sample (x 1 , x 2 , , x n ) if the true value of  could also
be found.
^
The maximum likelihood estimator of  , denoted by  , would maximize
the probabilit y of likelihood function of observed values.
Max f (x 1 , x 2 , , x n / ) means

df
d log f
 0, (or
 0)
dθ
dθ
Part I, #7

(a) if d(X) is the estimator of θ, and
E[d(X)]=θ, then the d(x) is unbiased.
 (x  X )
(b) the MLE estimator  
is not
n
unbiased.
 (x  X )
(c) the sample variance S  n  1 is an
2
unbiased estimator of σ

n

^
i
i 1
2

n

2
2
i 1
i
2
Part II #1
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Mean=np=90,
Var=np(1-p)=150(.6)(.4)=36,
∴standard deviation=6
P{X≦80}=p{X<80.5}=P{(X90)/6<(80.5-90)/6}=P{Z<-1.583}
=1-P{Z≦1.583} (=1-0.943=0.057)
Part II #2,3
_
if σ is known, 
( X  )
~ Z (0,1),
/ n
_
_
( X  )
 P{ z  
}  1   , P{ X  z   / n  },
/ n
   (90450  1.645(9400 / 4), )  (86584.25, ) with 95% confidence
_
if σ is unknown, 
( X  )
~ t n 1 ,
S/ n
_
_
( X  )
 P{t  ,n 1
}  1   , P{ X t  ,n 1S / n  },
S/ n
 we say   (90450  1.735(9400 / 4),  )  (86330.45, ) with 95% confidence
The lower bound of 95% confidence upper interval is smaller wh en the  is unknown
Part II #4
Suppose the deffective rate is p. So the 100 product items seem to obey
a binomial distributi on with mean 17 and variance 100(.17)(. 83),
^

X-n p
^
^
~ N (0,1), (we assume that n is large enough for normality approximat ion)
n p ( 1- p )
^
^
^
^
^
^


 P p  z / 2 p ( 1- p ) / n  p  p  z / 2 p ( 1- p ) / n   1  




P .17  1.96 (0.17)(. 83) / 100  p  .17  1.96 (.17)(. 83) / 100  0.95
p  (0.096, 0.244) with 95% confidence
Part II #5,6
(0.5)(0.5)
1.96 2  (0.5)(0.5)
 z0.25
 0.03,
n
2
n
(0.03)
n  1067.11,1068 is the necessary sample size
if the servey result is 0.3
the margin of error  1.96
(0.3)(0.7)
 0.0274  2.75%, lightly decreasing
1068