Let X have a Weibull distribution with parameters α and β , so E ( X ) = β ⋅ Γ ( 1 + 1/ α ) V ( X ) = β 2 { Γ ( 1 + 2/ α ) - [ Γ (1 + 1/ α )] 2 } a. Based on a random sample X 1 ,.., X n , write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function. b. If n = 20, x ¯ = 28.0, and ∑ x i 2 = 16,500 , compute the estimates. [ Hint: [ Γ (1.2)] 2 / Γ (1.4) = .95.]

Question

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Answer 1

Textbook Question

Chapter 6.2, Problem 21E

Let X have a Weibull distribution with parameters α and β, so

$E ( X ) = β ⋅ Γ ( 1 + 1/ α ) V ( X ) = β 2 { Γ ( 1 + 2/ α ) - [ Γ (1 + 1/ α )] 2 }$

a. Based on a random sample X₁,.., X_n, write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function.
b. If n = 20, $x ¯$ = 28.0, and $∑ x i 2 = 16,500$ , compute the estimates. [Hint: [Γ(1.2)]²/ Γ(1.4) = .95.]

a.

Expert Solution

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

b.

Expert Solution

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

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To evaluate the success of a 1-year experimental program designed to increase the mathematical achievement of underprivileged high school seniors, a random sample of participants in the program will be selected and their mathematics scores will be compared with the previous year’s statewide average of 525 for underprivileged seniors. The researchers want to determine whether the experimental program has increased the mean achievement level over the previous year’s statewide average. If alpha=.05, what sample size is needed to have a probability of Type II error of at most .025 if the actual mean is increased to 550? From previous results, sigma=80.

Answer 2

Textbook Question

Chapter 6.2, Problem 21E

Let X have a Weibull distribution with parameters α and β, so

$E ( X ) = β ⋅ Γ ( 1 + 1/ α ) V ( X ) = β 2 { Γ ( 1 + 2/ α ) - [ Γ (1 + 1/ α )] 2 }$

a. Based on a random sample X₁,.., X_n, write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function.
b. If n = 20, $x ¯$ = 28.0, and $∑ x i 2 = 16,500$ , compute the estimates. [Hint: [Γ(1.2)]²/ Γ(1.4) = .95.]

a.

Expert Solution

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

b.

Expert Solution

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

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Answer 3

Textbook Question

Chapter 6.2, Problem 21E

Let X have a Weibull distribution with parameters α and β, so

$E ( X ) = β ⋅ Γ ( 1 + 1/ α ) V ( X ) = β 2 { Γ ( 1 + 2/ α ) - [ Γ (1 + 1/ α )] 2 }$

a. Based on a random sample X₁,.., X_n, write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function.
b. If n = 20, $x ¯$ = 28.0, and $∑ x i 2 = 16,500$ , compute the estimates. [Hint: [Γ(1.2)]²/ Γ(1.4) = .95.]

a.

Expert Solution

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

b.

Expert Solution

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

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Answer 4

Textbook Question

Chapter 6.2, Problem 21E

Let X have a Weibull distribution with parameters α and β, so

$E ( X ) = β ⋅ Γ ( 1 + 1/ α ) V ( X ) = β 2 { Γ ( 1 + 2/ α ) - [ Γ (1 + 1/ α )] 2 }$

a. Based on a random sample X₁,.., X_n, write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function.
b. If n = 20, $x ¯$ = 28.0, and $∑ x i 2 = 16,500$ , compute the estimates. [Hint: [Γ(1.2)]²/ Γ(1.4) = .95.]

a.

Expert Solution

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

b.

Expert Solution

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

b.

Expert Solution

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

Answer 8

a.

Expert Solution

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Write the equations for the method of moments estimators of $β$ and $α$ .

Show that the estimate of $β$ can be obtained from a table of gamma function after obtaining the estimate of $α$ and the estimate of $α$ is the solution to a complicated equation involving the gamma function.

Answer 13

Answer to Problem 21E

The equations for the method of moments estimators of $β$ and $α$ are:

$β^=X¯Γ(1+1α^)_$ , and

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_$ .

Answer 14

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $α$ and $β$ , such that $E(X)=β⋅Γ(1+1α)$ and $V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}$ . A random sample of n observations $X1,X2,...,Xn$ is obtained.

Calculation:

First, calculate $E(X2)$ for the population having Weibull distribution:

It is known that:

$V(X)=E(X2)−[E(X)]2E(X2)=V(X)+[E(X)]2$ .

Now, for a Weibull distribution,

$V(X)=β2{Γ(1+2α)−[Γ(1+1α)]2}=β2⋅Γ(1+2α)−β2⋅[Γ(1+1α)]2=β2⋅Γ(1+2α)−[β⋅Γ(1+1α)]2=β2⋅Γ(1+2α)−[E(X)]2 (from given information).$

Thus,

$V(X)+[E(X)]2=β2⋅Γ(1+2α)E(X2)=β2⋅Γ(1+2α).$

For a random sample of size n, the first sample raw moment is the sample mean, that is $X¯$ and the second sample raw moment is $1n∑i=1nXi2$ .

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$X¯=E(X)$ and $1n∑i=1nXi2=E(X2)$ .

Denote $α^$ and $β^$ as the method of moments estimators as $α$ and $β$ .

Thus, the method of moments estimators are obtained as follows:

$X¯=E(X)=β^⋅Γ(1+1α^)β^=X¯Γ(1+1α^)_.$

This equation involves the gamma function.

Again,

$1n∑i=1nXi2=E(X2)=β^2⋅Γ(1+2α^).$

Substitute the expression for $β^$ in this equation:

$1n∑i=1nXi2=[X¯Γ(1+1α^)]2⋅Γ(1+2α^)1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2_.$

This equation involves the gamma function.

This equation is independent of $β^$ and thus, can be solved to obtained $α^$ . An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $α_$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $α_$ , $α^_$ , can be substituted in the first equation to obtain $β^_$ , the estimate for $β_$ .

Thus,

Answer 15

b.

Expert Solution

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Compute the estimates when $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ .

Answer 20

Answer to Problem 21E

The estimate of $α^$ is 1.2. The estimate of $β^$ is $28Γ(1.2)_$ .

Answer 21

Explanation of Solution

Calculation:

First, put $n=20$ , $x¯=28.0$ and $∑xi2=16,500$ in:

$1n∑i=1nXi2X¯2=Γ(1+2α^)[Γ(1+1α^)]2Γ(1+2α^)[Γ(1+1α^)]2=16,50020(28.0)2=825784=1.0523.$

Now, taking reciprocal of both sides of the above equation,

$[Γ(1+1α^)]2Γ(1+2α^)=11.0523=0.9503≈0.95.$

It is given in hint that:

$[Γ(1.2)]2Γ(1.4)=0.95$ .

As a result,

$[Γ(1+1α^)]2Γ(1+2α^)=[Γ(1.2)]2Γ(1.4)$ .

The above equation is an identity, indicating that:

$[Γ(1+1α^)]2=[Γ(1.2)]2$ , and,

$Γ(1+2α^)=Γ(1.4)$ .

From the second equation,

$1+2α^=1.42α^=1.4−1=0.4α^=20.4$

$=5_$ .

Substitute $α^=5$ and $x¯=28.0$ in the expression for $β^$ :

$β^=X¯Γ(1+1α^)=28Γ(1+15)=28Γ(1+0.2)=28Γ(1.2)_.$

Answer 22

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