Statistics Assignment 9

This document contains solutions for Statistics Assignment 9. It includes confidence intervals, hypothesis testing, and probability distributions. Subject: Statistics, Course Code: 630, Course Name: Not mentioned, College/University: Not mentioned

Statistics 630 - Assignment 9
(due Thursday, April 4, 2019, 8am CDT)
Name ________________________________________________
Email Address ___________________________________________________________
Statistics 630 - Assignment 9
(due Thursday, April 4, 2019, 8am CDT)
6.3.1)
Based on given data
Sample size = n = 10
Sample mean = [4.7+ 5.5+ 4.4 + 3.3+4.6 +5.3+5.2+ 4.8+ 5.7+ 5.3]/10 = 4.88
Sample standard deviation
X [x-mean]2
4.7 0.0324
5.5 0.3844
4.4 0.2304
3.3 2.4964
4.6 0.0784
5.3 0.1764
5.2 0.1024
4.8 0.0064
5.7 0.6724
5.3 0.1764
Total = 44.8 Total = 4.356
Mean = 44.8/10 = 4.48
Variance = 4.356/9 = 0.484
s.d = √0.484 = 0.6957
To test
H0: μ=5
H1: μ ≠5

a) 95% confidence interval for μ
x Z∗
√ s . d
n− ¿
+¿¿ ¿
4.88 1.96∗ √ 0.5
10− ¿
+ ¿¿ ¿
= 4.88 0.4383− ¿+ ¿¿ ¿
4.4417 < μ < 5.3183
b) for σ unknown
95% confidence interval for μ is
x t∗s . d
√n− ¿
+¿¿ ¿
4.88 2.2622∗0.6957
√10− ¿
+¿ ¿ ¿
4.88 0.4977− ¿+¿ ¿ ¿
4.3823 < μ < 5.3777
6.3.8)
To test:
Test statistic:
z= ^θ−θ 0
√ θ 0( 1−θ 0)
n
N (0 , 1)͠
H 0
Where, n: sample size
^θ : Sample proportion
θ0: Hypothesize population proportion
We reject the null hypothesis or H0 if the observed value of |z| > Zα/2, where, ∝=0.10
Now, ^θ=0.62 ,θ0 = 0.65, n = 250
The observed value of the test statistic is: z = -0.99
The critical value: Zα/2 = 1.645

Now, z = -0.99 < 1.645
Thus we accept the null hypothesis or H0.
Hence we do not have sufficient evidence to suggest that the proportion of voters in a given
population is not equal to 0.65.
The 90 % confidence interval for the population proportion is:
^θ z¿
( √ ^θ(1− ^θ)
n )− ¿
+ ¿¿ ¿
,
Where, is the multiplier
Here, z¿=1.645 , ^θ=0.62 , n=250
Lower limit: 0.5695
Upper limit: 0.6705
The 90 % confidence interval for the population proportion θ is: (0.5695, 0.6705)
6.5.1)
N( μ0, σ 2 ¿ where μ0 is known and σ 2 is known and θϵ ( 0 , ∞ ) isunknown
Therefore, X N( μ0, σ 2 ¿
The PDF is given as f ( x , μ 0 , σ2 )= 1
σ √ 2 π ∗e
−1
2 ( x−μ 0
σ )
2
, ∞ <x−μ 0< ∞, σ 2> 0 …….eq1
To find fisher information I (σ2 )
I ( σ2 ) =f ( x , σ 2 ) =E [ −∂2 logf ( x , σ2 )
∂2 σ2 ]
Taking log on both side by eq(1), we get
logf ( x , σ2 )=−log √ 2 π−logσ − 1
2 ( x−μ 0
σ )
2
Differentiating σ 2 we get
∂logf ( x , σ 2 )
∂ σ2 =−0− 1
σ −( x−μ 0
σ )2
(−2 σ−3 )
∂logf ( x , σ 2 )
∂ σ2 =−1
σ + ( x−μ 0 ) 2
σ3

Differentiating w.r, to σ 2 , we get
∂2 logf ( x , σ2 )
∂2 σ2 =− ( 1
σ2 ) +( x−μ 0)2 ( −3 σ−4 )
∂2 logf ( x , σ2 )
∂2 σ2 = ( 1
σ2 ) −3 ( x−μ 0 ) 2
σ4
Now, I ( σ2 )=E [ ∂2 logf ( x , σ2 )
∂2 σ2 ]
= E [− ( 1
σ2 −3 ( x−μ 0 )2
σ 4 ) ]
= −E ( 1
σ 2 )+ 3
σ4 E ( x−μ 0 )2
= −1
σ2 + 3
σ 4 ∗σ2
E ( x−μ 0 )2=σ2
= −1
σ2 + 3
σ 4 = 2
σ2
Therefore,
I ( σ2 )= 2
σ 2
6.5.2)
X1,…, xn is a sample γ ¿0, θ ¿ distribution
α0 is known and θϵ ( 0 , ∞ ) isunknown
Fisher information I ( θ )=?
x γ ¿0, θ ¿
f ( x )= θα 0
⌈ α 0 ¿ ¿

L ( x :θ ) = θnα 0
⌈ α 0 ¿ ¿
Log L = c + n α 0log θ−θ∑ xI + ¿0-1)∑
i=1
n
logxi
∂logL
∂ θ = nα 0
θ −∑ xI = 0
nα 0
θ =∑ xi
^θ= nα 0
∑ xi = nα 0
n x
Therefore,
^θ= α 0
x is MLE
∂2 logL
∂ θ2 =−nα 0
θ2
−E ( ∂2 logL
∂ θ2 ) =−E ( −nα 0
θ2 )
= n ∝0
θ2
I ( θ )= 1
−E ( ∂2 logL
∂θ2 )
I ( θ ) = θ2
nα 0
6.5.3)
From Pareto distribution PDF it is given by
f ( x , α ) =αx0αx-α – 1, x≥ x0

Taking log on both side gives
I ( α ) =−E [ ∂2
∂ α2 f ( x , α ) ]
= 1
α2
Therefore,
In(α ¿=¿ (α)
= n
α2
6.5.4)
The pivot for, λ= λ−λ
√ λ
√ λ
Poisson distribution mean = λ=x
= 1
n ∑ x
= 1
20 (1.93)
= 9.65
The estimated mean, λ=λ=9.65
The standard deviation = √ λ= √ 9.65 = 3.106
Construction of the 95% confidence interval
[ λ−z∗( λ
√ n ) , λ+ z∗( λ
√ n ) ]
Z*0.05 = 1.645
95% confidence interval based on pivot sample
= [ 9.65−1.645 ( 3.106
√ 20 ) , 9.65+ 1.645 ( 3.106
√ 20 ) ]

= (9.65 – 1.1143, 9.65 + 1.143)
= (8.507, 10.793)
6.5.5)
Based theorem of central limit, Let Xi, I = 1, 2, … n be a sequence of i.i.d distributed random
variable with mean μ= 1
λ and variance σ 2= 1
λ2
For large n, Y = ∑ xI
N
n
λ , n
λ2
Therefore, Y, which is Gamma(n, λ ¿ , will have a normal approximation for large n value
Here estimating ^λ= 1
x
Where, x= 1
n ∑ xI = 43991.75
27 =1627.4722
λ= 1
λ from the method of moment estimator
^λ= 1
1627.4722 = 0.0006145
Gamma(α , λ ¿ N ( α
λ , α
λ2 )
N( 2
0.0006145 , 2
0.00061452 ¿
90% confidence interval for mean
0.90 = (-1.6449 < N(0, 1)) < 1.6449)
0.90(-1.6449 < μ−μ
σ / √n <1.6949¿

0.90 = (-1.6449 < 3254.94−μ
442.9 <1.6449 ¿
2526.34 < μ<3983.53
6.5.6)
When α =1
Therefore,
Gamma(1, λ) N ( 1
λ , 1
λ2 )
Therefore,
N( 1
0.0006145 , 1
0.00061452 ¿
90% confidence interval for mean
0.90 = (-1.6449 < N(0, 1)) < 1.6449)
0.90(-1.6449 < μ−μ
σ / √n <1.6949¿
0.90 = (-1.6449 < 1627.47−μ
313.2071 <1.6449 ¿
1112.27 < μ<2142.66
6.2.5)
X1,…, xn is a random sample Gamma¿0, θ ¿ distribution
α0 > 0 is known and θϵ ( 0 , ∞ ) isunknown
f ( x :α 0 , β )= θα 0
θα 0 ⌈ α 0 ¿ ¿ , x > 0
The likelihood function is

L ( θ ) = 1
θnα 0 [ ⌈ α 0¿ ¿ ¿ ¿n ¿∏
i=1
n
{xα 0−1
}e
−∑ xi
θ
Log L = - n α0logθ−nlog ⌈ ( α 0)¿ ¿I + ¿0-1)∑
i=1
n
logxI - ∑ xi
θ
∂logL(θ)
∂ θ =−nα 0
θ −
∑
i=1
n
xi
θ2
= 0
θ nα0 = ∑
i=1
n
xi
^θ=
1
α 0 ∗1
n ∑
i=1
n
xi
^θ= x
α 0 , where x= 1
n ∑
i=1
n
xi
Hence θ is consistence
6.2.12)
X1, X2,…., Xn N ( μ . σ2 )
i .i . d , μ=μ 0=known
Unbiased = 1
σ √ 2 π ∗e
−1
2 σ2 ∑
i=1
n
( xi−μ 0 ) 2
, 1 - α < x<∞
lnL = -nlnσ −nln √2 π − 1
2 σ2 ∑
i=1
n
( xi−μ 0 )2
∂lnL
∂ σ =−n
σ + 1
σ 3 ∑
i=1
n
( xi−μ 0 )2
^σ 2= 1
n ∑
i=1
n
( xi−μ 0 )2 = MLE of σ 2
Change in MLE in location scale model
Unbiased = 1
σ √ 2 π ∗e
−1
2 σ2 ∑
i=1
n
( xi−μ 0 ) 2
, 1 - α < x<∞

μ=unknown
lnL = -nlnσ −nln √2 π − 1
2 σ2 ∑
i=1
n
( xi−μ 0 )2
∂lnL
∂ μ = 1
2 σ 2∗2∑
i=1
n
( xi−μ )=0
^μ= 1
n ∑
i=1
n
xi=x = MLE of μ
Putting ^μ in place of μ
∂lnL
∂ σ =−n
σ + 1
σ 3 ∑
i=1
n
( xi−x )2
^σ 2= 1
n ∑
i=1
n
( xi−x ) 2 = MLE of σ 2
Hence σ 2 is consistence
In problem 6.2.12
Asymptotic distribution of MLE is normal
As γlogL
γ σ2 isconsistent solution of likelihood function and is asymptotically distributed with mean
0 and variance = 1
I ( θ ) = 1
n ∑
i=1
n
( xi−μ 0 )2
So,
σ 2 N ( 0 , 1
n ∑
i=1
n
( xi−μ 0 ) 2
)