Statistics Assignment 9

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