Chapter 4.5, Problem 27E

(a)

To determine

To find: The equation of the line of best fit and correlation coefficient

Answer to Problem 27E

The equation of the line of best fit is $y=-0.08x+3.8$

Explanation of Solution

Given:

Hours, x	Grade point average, y
$10$ $5$ $3$ $12$ $20$ $15$ $8$ $4$ $16$	$3.0$ $3.4$ $3.5$ $2.7$ $2.1$ $2.8$ $3.0$ $3.7$ $2.5$

Calculation:

Consider the table shows the grade point average $y$ of several students and the numbers $x$ of hours they spend watching television each week

To use the linear regression feature of the $TI-83$ calculator, let us first press $STAT$ the key, then $ENTER$ to go to the statistics menu

  

Now press the $1$ key and go to the edit menu and then write the data containing values of $x$ in $L1$ and data containing the value of $y$ in $L2$

  

Now to calculate the linear regression, that is $ax+b$ , first turn the diagnostics on, from the home screen , press $2nd$ key and then press $CATALOG$ . In catalog select “ Diagnostic on” and then press $ENTER$ twice. Finally press the $STAT$ key, go to right arrow in the $CALC$ menu, select $Lin\mathrm{Re}g(ax+b)$ and then press $ENTER$

In order to input the list to use while finding the linear regression, press $2nd$ key, then press $1$ to select $L1$ , then use comma $(,)$ o separate and finally press $2nd$ key and the $L2$ (by selecting) , press comma $(,)$

  $VARS$ ,right arrow, then $ENTER$ twice

  

Therefore, the equation of the line of best fit is $y=-0.08x+3.8$

To make the graph using graphing calculator, press the $ZOOM$ key, select $"Zoomstat"$ and then press $ENTER$

The graph obtained is shown below

  

Conclusion:

The equation of the line of best fit is $y=-0.08x+3.8$

(b)

To determine

To find: The slope and $y$ intercept of the line of best fit

Answer to Problem 27E

The slope and $y$ intercept is

  $\begin{array}{l}Slope=-0.08\\ y-\mathrm{int}ercept=3.8\end{array}$

Explanation of Solution

Given:

Hours, x	Grade point average, y
$10$ $5$ $3$ $12$ $20$ $15$ $8$ $4$ $16$	$3.0$ $3.4$ $3.5$ $2.7$ $2.1$ $2.8$ $3.0$ $3.7$ $2.5$

Calculation:

The slope of the line is $-0.08$ . This means the grade point average is decrease as the hours is increase by $0.08$ times

The $y-\mathrm{int}ercept$ is $3.8$

No meaning in this context because the time cannot be zero

Conclusion:

There may be no causal relationship between year and the number of text messages

(c)

To determine

To find: The students grade point average

Answer to Problem 27E

The grade point average of the student is $2.7$

Explanation of Solution

Given:

Hours, x	Grade point average, y
$10$ $5$ $3$ $12$ $20$ $15$ $8$ $4$ $16$	$3.0$ $3.4$ $3.5$ $2.7$ $2.1$ $2.8$ $3.0$ $3.7$ $2.5$

Calculation:

Consider another student watches about $14$ hours of television each week

Put the value $x=14$ in the equation

  $\begin{array}{l}y=-0.08x+3.8\\ =-1.12+3.8\\ =2.68\end{array}$

The grade point average of the student is $2.7$

Conclusion:

Therefore, the grade point average of the student is $2.7$

(d)

To determine

To find: Whether there is a casual relationship

Answer to Problem 27E

There is no causal relationship as the time consumed to watch the television student does not determine the average grade

Explanation of Solution

Given:

Hours, x	Grade point average, y
$10$ $5$ $3$ $12$ $20$ $15$ $8$ $4$ $16$	$3.0$ $3.4$ $3.5$ $2.7$ $2.1$ $2.8$ $3.0$ $3.7$ $2.5$

Calculation:

There is no causal relationship between time spent watching television and grade point average. As the time consumed to watch the television student does not determine the average grade that is GPA

Conclusion:

Therefore, there is no causal relationship between time spent watching television and grade point average.