Chapter 1.6, Problem 11E

a.

To determine

To graph: the data on the scatter plot.

Answer to Problem 11E

  

Explanation of Solution

Given information:

The table of values given is as follows:

Year (x)	Population in millions (y)
1	200
1650	500
1850	1000
1930	2000
1975	4000
2000	6050
2010	?

Calculation:

The graph for the given data is shown below.

  

b.

To determine

To find: best fit line using two ordered pairs.

Answer to Problem 11E

  $y=15.857x-25664$

Explanation of Solution

Given information:

The table of values given is as follows:

Year (x)	Population in millions (y)
1	200
1650	500
1850	1000
1930	2000
1975	4000
2000	6050
2010	?

Calculation:

Let’s select any two ordered pair like $\left(1650,500\right)$ and $\left(2000,6050\right)$ . Plug in the values in the slope equation and solve m as follows:

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{6050-500}{2000-1650}\\ =\frac{5550}{350}\\ =15.857\end{array}$

Now, plug the value of m and any one point coordinate into the slope intercept equation and evaluate b as follows:

  $\begin{array}{c}y=mx+b\\ 6050=\left(15.857\right)\left(2000\right)+b\\ 6050=31714+b\\ b=6050-31714\\ =-25664\end{array}$

Now, plug the values of m and b in the slope intercept formula as follows:

  $\begin{array}{l}y=mx+b\\ y=15.857x-25664\end{array}$

Thus, the line equation is $y=15.857x-25664$ .

c.

To determine

To find: the equation for regression line and correlation value using the graphing calculator.

Answer to Problem 11E

  $\begin{array}{c} y=1.6461803428259x-288.99538410342\\ r=0.55742877735414\end{array}$

Explanation of Solution

Given information:

The table of values given is as follows:

Year (x)	Population in millions (y)
1	200
1650	500
1850	1000
1930	2000
1975	4000
2000	6050
2010	?

Calculation:

The graph for the above equation is as follows:

  

Using the graphing calculator, the regression line equation is as follows:

  $y=1.6461803428259x-288.99538410342$

Using the graphing calculator, the correlation coefficient comes out to be, $r=0.55742877735414$

d.

To determine

To find: the missing value if regression line equation has strong or moderate relationship, and explains whether the prediction is reliable.

Answer to Problem 11E

  $\approx 3020$

Prediction is unreliable.

Explanation of Solution

Given information:

The table of values given is as follows:

Year (x)	Population in millions (y)
1	200
1650	500
1850	1000
1930	2000
1975	4000
2000	6050
2010	?

  $\begin{array}{c} y=1.6461803428259x-288.99538410342\\ r=0.55742877735414\end{array}$

Calculation:

The equation of regression line shows moderate relationship because $0.5\le r\le 0.75$ .

Plugging $x=2010$ into the regression formula and evaluate y as follows:

  $\begin{array}{c}y=1.6461803428259x-288.99538410342\\ y=1.6461803428259\left(2010\right)-288.99538410342\\ =\text{3308}\text{.822489080059}-288.99538410342\\ =3019.8271049766395\\ \approx 3020\end{array}$

The factors which make the regression line unreliable are as follows:

1. Regression line is on the low end of moderate, or moderately weak.
2. There is a 1650 year gap between two x -values. This gap affects the line heavily.

So, the predicted value is unreliable.