Chapter 13.3, Problem 24E

a.

To determine

To Find: the least square regression line for the data.

Answer to Problem 24E

The least square regression line is $y=0.138x+22.1$ .

Explanation of Solution

Given:

The relationship between the amount of fertilizer x (in pounds per acre) and the wheat yield y (in bushels per acre)

  $\begin{array}{ccccc}\text{Fertilizer,}x& 100& 150& 200& 25\\ \text{Yield,}y& 35& 44& 50& 56\end{array}$

Concept Used:

The least square regression line for the set of points $\left(x,y\right)$ can be calculated as $f\left(x\right)=ax+b$

Where $a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\left({\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}\right)}{n{\displaystyle \sum _{i=1}^n{x}_i{}^2}-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}$ and $b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{i=1}^n{x}_i}\right)$

Calculation:

Here number of observations (n) is 4

Construct a table as shown below:

  $\begin{array}{cccc}x& y& xy& x^2\\ 100& 35& 3500& 10000\\ 150& 44& 6600& 22500\\ 200& 50& 10000& 40000\\ 250& 56& 14000& 62500\\ {\displaystyle \sum _{i=1}^4{x}_i}=700& {\displaystyle \sum _{i=1}^4{y}_i}=185& {\displaystyle \sum _{i=1}^4{x}_i{y}_i}=34100& {\displaystyle \sum _{i=1}^4{x}_i{}^2}=135000\end{array}$

Now for the regression line

  $\begin{array}{c}a=\frac{n{\displaystyle \sum _{i=1}^n{x}_i{y}_i}-\left({\displaystyle \sum _{i=1}^n{x}_i}{\displaystyle \sum _{i=1}^n{y}_i}\right)}{n{\displaystyle \sum _{i=1}^n{x}_i{}^2-{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2}}\\ =\frac{4\left(34100\right)-\left(\left(700\right)\left(185\right)\right)}{4\left(135000\right)-{\left(700\right)}^2}\\ =0.138\\ b=\frac{1}{n}\left({\displaystyle \sum _{i=1}^n{y}_i}-a{\displaystyle \sum _{i=1}^n{x}_i}\right)\\ =\frac{1}{4}\left(185-a\left(700\right)\right)\\ =\frac{1}{4}\left(185-\left(0.138\right)\left(700\right)\right)\left(\because a=0.138\right)\\ =22.1\end{array}$

Hence the least square regression line formed will be $y=0.138x+22.1$

b.

To determine

To Estimate: the yield for a fertilizer application of 160 pounds per acre.

Answer to Problem 24E

The estimated yield for a fertilizer application of 160 pounds per acre is 44 bushels per acre.

Explanation of Solution

Given:

The relationship between the amount of fertilizer x (in pounds per acre) and the wheat yield y (in bushels per acre)

  $\begin{array}{ccccc}\text{Fertilizer,}x& 100& 150& 200& 25\\ \text{Yield,}y& 35& 44& 50& 56\end{array}$

The least square regression line is $y=0.138x+22.1$

Concept Used:

Put $x=160$ in the given least square regression line.

Calculation:

At $x=160,y=0.138\left(160\right)+22.1$

On evaluating:

  $\begin{array}{c}y=0.138\left(160\right)+22.1\\ =22.08+22.1\\ =44.18\approx 44\text{bushels}\text{per}\text{acre}\end{array}$

c.

To determine

To Estimate: the amount of fertilizer needed to yield of 70 bushels per acre.

Answer to Problem 24E

The amount of fertilizer needed to yield of 70 bushels per acre is 347 pounds per acre.

Explanation of Solution

Given:

The relationship between the amount of fertilizer x (in pounds per acre) and the wheat yield y (in bushels per acre)

  $\begin{array}{ccccc}\text{Fertilizer,}x& 100& 150& 200& 25\\ \text{Yield,}y& 35& 44& 50& 56\end{array}$

The least square regression line is $y=0.138x+22.1$

Concept Used:

Put $y=70$ in the given least square regression line and find the value of x.

Calculation:

At $y=70$ , x will be

  $\begin{array}{l}70=0.138\left(x\right)+22.1\\ 0.138x=70-22.1\\ 0.138x=47.9\\ x=\frac{92.1}{0.138}\\ x=347.1014\\ x\approx 347\text{pounds}\text{per}\text{acre}\end{array}$