Chapter 10, Problem 6E

(a)

To determine

To find: the expected value of y when the value of x is 2.

Answer to Problem 6E

1.44

Explanation of Solution

Given:

  $\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8log}x$

Calculation:

Model is

  $\begin{array}{c}\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8log}x\\ =1.2+0.8\left(\text{log}2\right)\\ =1.2+0.8\left(0.3010\right)\\ =1.2+0.2408\\ =1.4408\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 1.44

(b)

To determine

To find: the expected value of y when the value of x is 2.

Answer to Problem 6E

630.96

Explanation of Solution

Given:

  $\text{log <mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8}x$

Calculation:

  $\begin{array}{c}\text{log <mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8}x\\ =1.2+0.8\left(2\right)\\ =1.2+1.6\\ =2.8\\ \widehat{y}=e^{2.8}\\ =630.957\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 630.96

(c)

To determine

To find: the expected value of y when the value of x is 2.

Answer to Problem 6E

5.78

Explanation of Solution

Given:

  $\text{In <mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8}\text{In}x$

Calculation:

Model is

  $\begin{array}{c}\text{In <mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8}\text{In}x\\ =1.2+0.8\left(\text{In}2\right)\\ =1.2+0.8\left(0.693\right)\\ =1.2+0.5545\\ =1.7545\\ \widehat{y}=e^{2.8}\\ =5.78\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 5.78.

(d)

To determine

To find: the expected value of y when the value of x is 2.

Answer to Problem 6E

1.67

Explanation of Solution

Given:

  ${\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover>}}^2\text{=1}\text{.2 + 0}\text{.8}x$

Calculation:

Model is

  $\begin{array}{c}{\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover>}}^2\text{=1}\text{.2 + 0}\text{.8}x\\ =1.2+0.8\left(2\right)\\ =1.2+1.6\\ =2.8\\ \widehat{y}=\sqrt{2.8}\\ =1.67\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 1.67

(e)

To determine

To find: the expected value of y when the value of x is 2.

Answer to Problem 6E

0.128

Explanation of Solution

Given:

  $\frac{1}{{\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover>}}^2}\text{=1}\text{.2 + 0}\text{.8}x$

Calculation:

  $\begin{array}{c}\frac{1}{{\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover>}}^2}\text{=1}\text{.2 + 0}\text{.8}x\\ =1.2+0.8\left(2\right)\\ =1.2+1.6\\ =2.8\\ \widehat{y}=\frac{1}{{\left(2.8\right)}^2}\\ =\frac{1}{7.84}\\ =0.128\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 0.128.