Chapter 7.1, Problem 46E

Solution

a.

To calculate: The linear regression models for the populations of both cities using the given data.

Time (years)	San Antonio (thousands)	San Diego ( thousands)
1980	786	876
1990	935	1111
2000	1145	1223
2010	1327	1307
2015	1480	1395

The obtained matrix is $y\approx 19.60x+762.18$ and $y\approx 13.75x+921.10$ .

Given Information:

The table is defined,

Time (years)	San Antonio (thousands)	San Diego ( thousands)
1980	786	876
1990	935	1111
2000	1145	1223
2010	1327	1307
2015	1480	1395

Calculation:

Consider the given table,

Time (years)	San Antonio (thousands)	San Diego ( thousands)
1980	786	876
1990	935	1111
2000	1145	1223
2010	1327	1307
2015	1480	1395

Use a graphing calculator to find the linear rgeression.

Step 1. Insert the table in calculator by using the table feature and take as 0 for 1980.

  

Step 2. Use the LinReg feature with first two list ${\text{L}}_1$ and ${\text{L}}_2$ o obtain the linear regression model for the population of San Antonio.

  

And,

  

Step 3. Use the LinReg feature with first two list ${\text{L}}_1$ and ${\text{L}}_3$ o obtain the linear regression model for the population of San Diego.

  

And,

  

Step 4. Use the Graph feature to plot the both the graph as scatter graph.

  

Therefore, the obtained equation is $y\approx 19.60x+762.18$ and $y\approx 13.75x+921.10$ .

b.

To calculate: The time when the populations of the two cities were the same as well as the common population of the two cities.

The obtained result is $\text{1291}\text{.38 thousand}$ .

Given Information:

The equations from the previous part is $y\approx 19.60x+762.18$ and $y\approx 13.75x+921.10$ .

Calculation:

Consider the given information,

Equate both the equations $y\approx 19.60x+762.18$ and $y\approx 13.75x+921.10$ to find that time.

  $\begin{array}{c}19.60x+762.18=13.75x+921.10\\ 19.60x-13.75x=921.10-762.18\\ \text{5}\text{.85}x=\text{158}\text{.92}\\ x\approx 27\end{array}$

Add the obtained value in 1980.

  $1980+27=2007$

The corresponding year is 2007.

Now, substitute 27 for $x$ in any equation.

  $\begin{array}{c}y\approx 19.60\left(27\right)+762.18\\ =\text{529}\text{.2}+762.18\\ =\text{1291}\text{.38 thousand}\end{array}$

Hence, the obtained result is $\text{1291}\text{.38 thousand}$ .