Chapter 4.4, Problem 19PFA

To determine

To find: the best equation which represents the data in the table.

Answer to Problem 19PFA

The correct approximate equation is $s=60+6h$ .

Explanation of Solution

Given information:

Given table,

Given options are,

A.$S=40+15h$

B.$S=60+6h$

C.$S=55+10h$

D.$S=80+3h$

Calculation:

Given the table of the values Study hours (h) = 0, 1, 2, 3, 4, 5, 6,

Exam score (s) = 55, 70, 83, 76, 87, 92, 90

Sum of the exam score (s)	Sum of the Study hours (h) mean	Sum of the exam score (s) mean	Product of (sum of study hours (h) mean)(sum of the exam score (s)mean)	(study hours(h) mean) $\wedge$ 2
55	$\left(0-3\right)=-3$	$\left(55-79\right)=-24$	$-3\times -24=72$	$\left(0-3\right)\wedge 2$
70	$\left(1-3\right)=-2$	$\left(70-79\right)=-9$	$-2\times -9=18$	$\left(1-3\right)\wedge 2$
83	$\left(2-3\right)=-1$	$\left(83-79\right)=4$	$-1\times 4=-4$	$\left(2-3\right)\wedge 2$
76	$\left(3-3\right)=0$	$\left(76-79\right)=-3$	$0\times -3=0$	$\left(3-3\right)\wedge 2$
87	$\left(4-3\right)=1$	$\left(87-79\right)=8$	$1\times 8=8$	$\left(4-3\right)\wedge 2$
92	$\left(5-3\right)=2$	$\left(92-79\right)=13$	$2\times 13=26$	$\left(5-3\right)\wedge 2$
90	$\left(6-3\right)=3$	$\left(90-79\right)=11$	$3\times 11=33$	$\left(6-3\right)\wedge 2$
Sum = 553			Sum of product = 153	Sum = 28
Mean = (553/7)=79

Regression equation $y=bx+a$

Now,

  $\begin{array}{c}b=\frac{153}{28}\\ =5.46429\\ \approx 6\end{array}$

Now,

  $\begin{array}{c}a=79-\left(5.46\times 3\right)\\ =62.60714\\ \approx 60\end{array}$

And then,

  $s=5.46429h+62.60714$

Therefore, the correct approximate equation is $s=60+6h$ .