ECON 451 Homework 7: Confidence Intervals and Estimator Properties

This document provides solutions for ECON 451 Homework 7, focusing on confidence intervals and the properties of estimators. Part A covers confidence interval calculations, including finding a 95% confidence interval for student IQ scores, determining whether males and females metabolize CH3Hg+ at the same rate, and assessing the effect of a diet on serum cholesterol using confidence intervals. Part B delves into properties of estimators, such as unbiased estimators, variance, and the Cramer-Rao lower bound. The document also explores the efficiency of estimators and the concept of sufficient statistics. The solutions involve statistical formulas, calculations, and explanations, providing a comprehensive understanding of the concepts. The assignment also references to the books by Larsen and Marx (2018) and Wakerly, Mendenhall, and Scheaffer (2008).

Running Head: HOMEWORK 7 1
ECON 451 Intermediate Introduction to Stats. and Econometrics I
Student’s Name
Institutional Affiliation

HOME WORK 7 2
ECON 451 PART A: CONFIDENCE INTERVAL
1. A commonly used IQ test is scaled to have a mean of 100 and a standard deviation
of 15. A school counselor was curious about the average IQ of the students in her
school and took a random sample of 50 students’ IQ scores. The average was y =
107:9. Find a 95% confidence interval for the student IQ in the school. Assume 50 is
a large enough sample. Ans: (103.7, 112.1)
Confidence interval is expressed by formulae
 ∓ z
 where Standard Error (
) = σ
√ n .
The value of Z score at 95% confidence interval is 1.96. The n is the sample size and
sigma the standard deviation.
Standard Error (
) = σ
√ n .=15/SQRT(50)=2.12
The value of z
= 1.96*2.12=4.158
Therefore Confidence interval Limit is Y
 107.9∓ (4.158) = [103.7,112.1]2. Mercury pollution is widely recognized as a serious ecological problems. Fish can
ingest and absorbs methylmercury (CH3Hg+), and then they are eaten by humans.
Men and women, however, may not metabolized CH3Hg+ the same way.
For men, the amount of CH3Hg+ half-lives in their system is believed to be 80 days.
For females, researchers gave 6 women were given a known amount of CH3Hg+ and
had a CH3Hg+ half-lives of {52, 69, 73, 88, 87, 56}
Assume for both genders, CH3Hg+ half-lives are normally distributed with a
standard deviation of 8 days.
Construct a 95% CI for the true female CH3Hg+ half-life. Based on the data, is it
believable that males and females metabolizes CH3Hg+ at the same rate? Explain.
Ans: No
Confidence interval is expressed by formulae μ ∓ z
 where Standard Error (
) = σ
√ n .
The value of Z score at 95% confidence interval is 1.96. The n is the sample size and
sigma the standard deviation.
Standard Error (
) = σ
√ n .=8/SQRT(6)=3.266
The value of z
= 1.96*3.266=6.401
Mean for days for women is = (52+69+ 73+ 88+87+56)/6=70.83
Therefore Confidence interval Limit is
 70.83∓ (6.401) = [64.43,77.23]

HOME WORK 7 3
Based on the data therefore males and females do not metabolize CH3Hg+ at the same
rate, Since the boundary above is 64.43 to 77.23 does not include 80 days as seen in
males. Hence answer is No
3. A physician who has a group of 38 female patients age 18-24 is on a special diet to
estimate the effect of the diet on total serum cholesterol. The average total serum
cholesterol for this group is 188.4. Based on a government study, it is known that
total serum cholesterol has a normal distribution with = 40.7. Given that the
national average total serum cholesterol for women aged 18-24 is 192, does the diet
seem to have any effect on women’ serum cholesterol in the study? Use 95% CI to
explain.
Standard Error (
) = σ
√ n .=40/SQRT(38)=6.489
The value of z
= 1.96*6.489=12.72
Therefore Confidence interval Limit is Y
 188.40∓ (12.72) = [175.68, 201.12]
Since the national average total serum cholesterol for women aged 18-24 is 192, the
national average lies within this boundary. This implies that the diet seem to have effect
on the womens’ serum cholesterol in the study.4. Researchers need to decide how many patients n to sample for their project. It is
known that the sample is drawn from a normal distribution with = 9.7. How large
does n have to be to guarantee that the length of the 99% CI for will be less than 2.
Ans: More than 626.
The formula is:
n = [(z*s)/E]2
The “s” is the standard deviation. The “E” is the desired margin of error. The Z is the
critical value from the z-table for a 95% CI.
n = [(zc*s)/E]2 =[(2.58*9.7)/1]^2=626.30
The sample size is therefore more than 626.5. A sample of size n = 16 is drawn from a normal distribution where = 10 but μ
is
unknown. If μ
= 20, what is the probability that the estimator μ
^ will lie between
19.0 and 21.0?Ans: 0.3108
Z = X−μ
σ / √n
Probability the μ estimate lies between 19 to 21
= 19−20
10/ √ 16 <z < 21−20
10/ √ 16

HOME WORK 7 4
= -0.4<Z<0.4
=P(19<X<21)=0.6554-0.3446=0.3108
ECON 451 PART B: PROPERTIES OF ESTIMATORS
1. What must c equal if the statistics c(Y1+2Y2) is to be an unbiased estimator for 1/ θ
Solve for c
E{c(y1+2y2)}= c{E(y1)+2E(y2)} = 3cE(y)
2. a). Show that ^θ3 is unbiased estimator for θ
E( ^θ3 ¿=αE ( ^θ1 ) +(1−α )E ¿
Therefore ^θ3 is unbiased estimator for θ
b.
For part b exploring the variance linear combination of the function would lead us to
Var( ^θ3 ¿=α 2 Var ( ^θ1 ) + ( 1−α ) 2 Var ¿
The minimum of this variance in a is attained at a*= δ 22
δ 12 +δ 22 considering the derivative
3
a. Are any of these estimators unbiased?
Where ^θ1= Y1, ^θ2= Y 1+2 Y 2
3 , and ^θ3=Y
For an exponential distribution, E[ ^θ]=θ
Therefore,
E ( ^θ1)=E(Y1)=θ
E¿=E ( Y 1+2Y 2
3 )= E ( Y 1 ) +2 E (Y 2 )
3 = θ+2 θ
3 =θ,
E¿=E ( Y 1+Y 2
2 )= E ( Y 1 ) + E ( Y 2 )
2 =θ +θ
2 =θ,
Therefore ^θ1, ^θ2 and ^θ3 are unbiased estimators

HOME WORK 7 5
b. Among the unbiased estimators, which has the smallest variance?
When finding out the variable with the smallest variance, there is need to compute. Consider
that the variance for an Exponential random variable is θ2
Therefore,
Var ( ^θ1)= Var(Y1)=θ2
Var ¿=Var ( Y 1+ 2Y 2
3 )= Var ( Y 1 ) + 4 Var ( Y 2 )
9 = θ2 + 4 θ2
9 =5 θ2
9
Var¿=Var ( Y 1+Y 2
2 )=Var ( Y 1 ) +var ( Y 2 )
4 =θ2+θ2
4 =θ2
2 ,
Thus ^θ3has the smallest variance. Probably a MVUE for θ .
4. Let Y1; Y2,.., Yn be a random sample .Compare the Cramer-Rao lower bound for the function
and compare to the variance of the ML estimator for θ, θestimate = y . Is y a best estimator
Solution
Using the Fisher Information in CRLB,
Therefore
And thus
Considering that E(Y)=θ then
CRLB is therefore
The next part is to find the Variance of Y.
It can be clearly seen that
Therefore Var(Y)=θ2
Var (Y ¿

HOME WORK 7 6
It can therefore be concluded that ^θ is the best estimator for θ
5 Consider the function
Show that mean is an efficient estimator for λ
It can be seen that CRLB is λ / n which is the variance of x. Hence x. Is an efficient estimator for
λ
6
a. P1=X1+2X2+3X3
This is to Prove that T(X)is sufficient for X by deriving the distribution of
X|T(X) =t.
Where X1=1,X2=1,X3=0
For the x1,x2,x3 implies P(X/t)=1/3
P1=(t=1)+2(t=1)+3(0)
P1=1/3+2/3=1
Hence is not sufficient
b. P2= X1 + X2 + X3 where x1=1,x2=1,x3=0
For the x1,x2,x3 implies P(X/t)=1/3
P1=(t=1)+(t=1)+(t=0)
P1=1/3+1/3=2/3
Hence is sufficient