Obtain the expression E ( n i − n j ) .

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

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Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

Answer 6

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

Answer 7

Chapter 14.3, Problem 6E

a.

To determine

Obtain the expression E(ni−nj).

Answer 8

a.

Expert Solution

Explanation of Solution

In this context, it is assumed that the assumptions associated with a multinomial experiment are all satisfied. Based on Section 5.9, each of the ni's, i=1,2,...,k follows binomial distribution with parameters n and pi. Further, Cov(ni,nj)=−npipj if i≠j.

The value of the expression E(ni−nj) is given below:

E(ni−nj)=E(ni)−E(nj)=npi−npj=n(pi−pj)

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

Answer 13

b.

To determine

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

Answer 14

b.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

E(ni−njn)=1nE(ni−nj)=1n(E(ni)−E(nj))=1n(npi−npj)=pi−pj

An unbiased estimator for θ=pi−pj is given below:

θ^=ni−njn

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

Answer 19

c.

To determine

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

Answer 20

c.

Expert Solution

Explanation of Solution

It is known that Cov(ni,nj)=−npipj if i≠j. Also, the variance of a linear combination is given below:

V(aX+bY)=V(aX)+V(bY)+2Cov(aX,bY)=a2V(X)+b2V(Y)+2abCov(X,Y)

Now,

V(ni−nj)=V(ni)+V(−nj)+2Cov(ni,−nj)=V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)=V(ni)+V(nj)−2Cov(ni,nj)=npi(1−pi)+npj(1−pj)−2(−npipj)=npi(1−pi)+npj(1−pj)+2npipj=n[pi(1−pi)+pj(1−pj)+2pipj]

Hence, it has been verified that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

Answer 21

c.

Expert Solution

Answer 22

c.

Expert Solution

Answer 23

Expert Solution

Answer 24

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

Answer 25

d.

To determine

Obtain the variance of unbiased estimator with reference to Part (c).

Answer 26

d.

Expert Solution

Explanation of Solution

From Part (a), E(ni−nj)=n(pi−pj).

Then,

V(ni−njn)=1n2V(ni−nj)=1n2[V(ni)+V(−nj)+2Cov(ni,−nj)]=1n2[V(ni)+(−12)V(nj)+2(−1)Cov(ni,nj)]=1n2[V(ni)+V(nj)−2Cov(ni,nj)]=1n2[npi(1−pi)+npj(1−pj)+2npipj]=1n2[n[pi(1−pi)+pj(1−pj)+2pipj]]=1n[pi(1−pi)+pj(1−pj)+2pipj]

Answer 27

d.

Expert Solution

Answer 28

d.

Expert Solution

Answer 29

Expert Solution

Answer 30

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

Answer 31

e.

To determine

Obtain the consistent estimator of n−1V(ni−nj).

Answer 32

e.

Expert Solution

Explanation of Solution

A consistent estimator for n−1V(ni−nj) is p^i−p^j because it is an unbiased estimator and variance reaches at 0 as n increases.

Answer 33

e.

Expert Solution

Answer 34

e.

Expert Solution

Answer 35

Expert Solution

Answer 36

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

Answer 37

f.

To determine

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

Answer 38

f.

Expert Solution

Explanation of Solution

It is known that the variable Hn=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has normal distribution.

Since pi and pj are consistent estimators, then σp^i−p^jσ^p^i−p^j tends to 0.

Now,

Hnσp^i−p^jσ^p^i−p^j=p^i−p^j−pi+pj1n[pi(1−pi)+pj(1−pj)+2pipj] has the limiting standard deviation.

In this context, it has been showed that a transformation of Hn has a standard normal distribution.

Therefore, (1−α)100% confidence interval for pi−pj is given by: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn

Answer 39

f.

Expert Solution

Answer 40

f.

Expert Solution

Answer 41

Expert Solution

Answer 42

Ch. 14.3 - Historically, the proportions of all Caucasians in...Ch. 14.3 - Prob. 2E Ch. 14.3 - Prob. 3E Ch. 14.3 - Prob. 4E Ch. 14.3 - Prob. 5E Ch. 14.3 - Prob. 6E Ch. 14.3 - Prob. 7E Ch. 14.3 - Prob. 8E Ch. 14.3 - Prob. 9E Ch. 14.3 - Prob. 10E

Obtain the expression E ( n i − n j ) .

Videos

Obtain the expression E(ni−nj).

Explanation of Solution

Provide the unbiased estimator of (pi−pj) with reference to Part (a).

Explanation of Solution

Verify that V(ni−nj)=n[pi(1−pi)+pj(1−pj)+2pipj].

Explanation of Solution

Obtain the variance of unbiased estimator with reference to Part (c).

Explanation of Solution

Obtain the consistent estimator of n−1V(ni−nj).

Explanation of Solution

Verify that a large sample (1−α)100% confidence interval for pi−pj is given below: p^i−p^j±zα2pi(1−pi)+pj(1−pj)+2pipjn.

Explanation of Solution

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