Chapter 1.6, Problem 5CFU

a.

To determine

To graph: the given data on scatter plot.

Answer to Problem 5CFU

  

Explanation of Solution

Given information:

Academic year	Average of students per computer
1988-1989	25
1989-1990	22
1990-1991	20
1991-1992	18
1992-1993	16
1993-1994	14
1994-1995	10.5
1995-1996	10
19996-1997	7.8
1997-1998	6.1
1998-1999	5.7
1999-2000	5.4
2000-2001	5
2001-2002	4.9
?	1

Calculation:

Let the year be x and the average be y. Plot of the graph from the given data will be as follows:

  

b.

To determine

To find: the equation of a best fit line by using two ordered pairs.

Answer to Problem 5CFU

  $y=-2.03x+4056.83$

Explanation of Solution

Given information:

Academic year	Average of students per computer
1988-1989	25
1989-1990	22
1990-1991	20
1991-1992	18
1992-1993	16
1993-1994	14
1994-1995	10.5
1995-1996	10
19996-1997	7.8
1997-1998	6.1
1998-1999	5.7
1999-2000	5.4
2000-2001	5
2001-2002	4.9
?	1

Calculation:

Select any two ordered pairs like $\left(1989,22\right)$ and $\left(1996,7.8\right)$ , and substitute the values in the slope formula as shown below.

  $\begin{array}{c}m=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{7.8-22}{1996-1989}\\ =\frac{-14.2}{7}\\ =-2.03\end{array}$

Put any point coordinate and slope value into the slope intercept formula and evaluate the y -intercept as follows:

  $\begin{array}{c}y=mx+b\\ 22=-2.03\left(1989\right)+b\\ 22=-4037.67+b\\ b=22+4037.67\\ =4056.83\end{array}$

Now, put the values of slope and b into the slope intercept formula as follows:

  $\begin{array}{l}y=mx+b\\ y=-2.03x+4056.83\end{array}$

Thus, the equation is $y=-2.03x+4056.83$ .

c.

To determine

To find: the equation of regression line and its correlation value by using a graphing calculator.

Answer to Problem 5CFU

  $\begin{array}{l}y=-1.6140659340662x+3231.4259340665\\ r=-0.96997250606508\end{array}$

Explanation of Solution

Given information:

Academic year	Average of students per computer
1988-1989	25
1989-1990	22
1990-1991	20
1991-1992	18
1992-1993	16
1993-1994	14
1994-1995	10.5
1995-1996	10
19996-1997	7.8
1997-1998	6.1
1998-1999	5.7
1999-2000	5.4
2000-2001	5
2001-2002	4.9
?	1

Calculation:

From the graph

  

By using the graphing calculator, we get slope value as $-1.6140659340662$ and intercept value as $3231.4259340665$ .

Now, put the values in the following equation and evaluate regression of equation of line

  $\begin{array}{l}y=mx+b\\ y=-1.6140659340662x+3231.4259340665\end{array}$

Thus, the regression line equation is $y=-1.6140659340662x+3231.4259340665$ .

By using graphing calculator, the value of correlation coefficient ( r ) is $-0.96997250606508$ .

d.

To determine

To explain: the whether the relationship predicted by the equation for the missing value is reliable.

Answer to Problem 5CFU

2001

The average number of students cannot be less than 1 student. Thus, the predicted value is the last reliable prediction that can be made with this regression line

Explanation of Solution

Given information:

Academic year	Average of students per computer
1988-1989	25
1989-1990	22
1990-1991	20
1991-1992	18
1992-1993	16
1993-1994	14
1994-1995	10.5
1995-1996	10
19996-1997	7.8
1997-1998	6.1
1998-1999	5.7
1999-2000	5.4
2000-2001	5
2001-2002	4.9
?	1

  $\begin{array}{l}y=-1.6140659340662x+3231.4259340665\\ r=-0.96997250606508\end{array}$

Calculation:

Since $-1\le r\le -0.75$ , the equation of the regression line shows a strong relationship.

Putting the value of $y=1$ , into the equation, we get

  $\begin{array}{c}1=-1.6140659340662x+3231.4259340665\\ 1.6140659340662x=3231.4259340665-1\\ 1.6140659340662x=3230.4259340665\\ x=\frac{3230.4259340665}{1.6140659340662}\\ =2001.421296296317\end{array}$

If average number of students shares less than one and greater than zero computer, that means the number of computers are more than students, or multiple computers for all.

It is possible that a school has more computers than students, but it sound unrealistic. Since this is the average data of all US public schools, then there will be more students than computers.

In the years 2001 to 2003, the computer sharing by the average number of students becomes negative. The average number of students cannot be less than 1 student. Thus, the predicted value is the last reliable prediction that can be made with this regression line.