Chapter 10, Problem 5E

(a)

To determine

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

Answer to Problem 5E

16.44

Explanation of Solution

Given:

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

Calculation:

Model is

  $\begin{array}{c}\text{In <mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8}x\\ =1.2+0.8\left(2\right)\\ =1.2+1.6\\ =2.8\\ \widehat{y}=e^{2.8}\\ =16.44\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 16.44.

(b)

To determine

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

Answer to Problem 5E

7.84

Explanation of Solution

Given:

  $\sqrt{\widehat{y}}=1.2+0.8x$

Calculation:

Model is

  $\begin{array}{c}\sqrt{\widehat{y}}=1.2+0.8x\\ =1.2+0.8\left(2\right)\\ =1.2+1.6\\ =2.8\\ \widehat{y}={\left(2.8\right)}^2\\ =7.84\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 7.84.

(c)

To determine

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

Answer to Problem 5E

0.357

Explanation of Solution

Given:

  $\frac{1}{\widehat{y}}=1.2+0.8x$

Calculation:

Model is

  $\begin{array}{c}\frac{1}{\widehat{y}}=1.2+0.8x\\ =1.2+0.8\left(2\right)\\ =1.2+1.6\\ =2.8\\ \widehat{y}=\frac{1}{\left(2.8\right)}\\ =0.357\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 0.357.

(d)

To determine

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

Answer to Problem 5E

1.75

Explanation of Solution

Given:

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

Calculation:

Model is

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

Therefore, expected value of $\widehat{y}$ for $x=2$ is 1.75.

(e)

To determine

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

Answer to Problem 5E

27.59

Explanation of Solution

Given:

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

Calculation:

Mode is

  $\begin{array}{c}\text{Log <mover accent="true"> <mo>y </mo><mo>^</mo></mover> =1}\text{.2 + 0}\text{.8Log}x\\ =1.2+0.8\left(\text{Log}2\right)\\ =1.2+0.8\left(0.3010\right)\text{<mover accent="true"> <mo>y </mo><mo>^</mo></mover> }={10}^{1.4408}\\ =1.2+0.2408=27.59\\ =1.4408\end{array}$

Therefore, expected value of $\widehat{y}$ for $x=2$ is 27.59.