Question

1. the table below gives the amount of time students in a class studied for a test and their test scores.

hours studied	2	5	1	0	4	2	3
test score	77	92	70	63	90	75	84

(a) (3pts) linear regression equation:
(b) (4pts) graph the data on a scatter plot, find the line of best fit, and write the equation for the line you draw.
(c) (3pts) determine whether linear relation exist or not (explain by critical value):
(d) (3pts) using the linear regression equation predict a student test score if they studied for 6hrs.

Explanation:

Step1: Calculate means

Let $x$ be hours studied and $y$ be test - score.
$n = 7$
$\bar{x}=\frac{2 + 5+1+0+4+2+3}{7}=\frac{17}{7}\approx2.43$
$\bar{y}=\frac{77 + 92+70+63+90+75+84}{7}=\frac{551}{7}=78.71$

Step2: Calculate numerator and denominator for slope $b_1$

$\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(2 - 2.43)(77 - 78.71)+(5 - 2.43)(92 - 78.71)+(1 - 2.43)(70 - 78.71)+(0 - 2.43)(63 - 78.71)+(4 - 2.43)(90 - 78.71)+(2 - 2.43)(75 - 78.71)+(3 - 2.43)(84 - 78.71)$
$=(-0.43)(-1.71)+(2.57)(13.29)+(-1.43)(-8.71)+(-2.43)(-15.71)+(1.57)(11.29)+(-0.43)(-3.71)+(0.57)(5.29)$
$=0.7353+34.1553 + 12.4553+38.1753+17.7253+1.5953+3.0153$
$=107.8571$
$\sum_{i = 1}^{n}(x_i-\bar{x})^2=(2 - 2.43)^2+(5 - 2.43)^2+(1 - 2.43)^2+(0 - 2.43)^2+(4 - 2.43)^2+(2 - 2.43)^2+(3 - 2.43)^2$
$=(-0.43)^2+(2.57)^2+(-1.43)^2+(-2.43)^2+(1.57)^2+(-0.43)^2+(0.57)^2$
$=0.1849+6.6049+2.0449+5.9049+2.4649+0.1849+0.3249$
$=17.7143$
$b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}=\frac{107.8571}{17.7143}\approx6.09$

Step3: Calculate $b_0$

$b_0=\bar{y}-b_1\bar{x}=78.71-6.09\times2.43$
$=78.71 - 14.8087=63.9013$
The linear - regression equation is $y = 63.90+6.09x$

Step4: For part (b)

To graph the scatter - plot, we plot the points $(2,77),(5,92),(1,70),(0,63),(4,90),(2,75),(3,84)$ on a coordinate plane. The line of best fit is $y = 63.90+6.09x$

Step5: Calculate correlation coefficient $r$

$r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i - \bar{y})^2}}$
First, calculate $\sum_{i = 1}^{n}(y_i - \bar{y})^2=(77 - 78.71)^2+(92 - 78.71)^2+(70 - 78.71)^2+(63 - 78.71)^2+(90 - 78.71)^2+(75 - 78.71)^2+(84 - 78.71)^2$
$=(-1.71)^2+(13.29)^2+(-8.71)^2+(-15.71)^2+(11.29)^2+(-3.71)^2+(5.29)^2$
$=2.9241+176.6241+75.8641+246.8041+127.4641+13.7641+27.9841$
$=671.4327$
$r=\frac{107.8571}{\sqrt{17.7143\times671.4327}}=\frac{107.8571}{\sqrt{11894.39}}=\frac{107.8571}{109.0614}\approx0.99$
For $n = 7$ and a significance level of $\alpha=0.05$, the critical value $r_{crit}\approx0.754$. Since $|r| = 0.99>0.754$, a linear relation exists.

Step6: Predict for $x = 6$

$y=63.90+6.09\times6=63.90 + 36.54=100.44$

Answer:

(a) $y = 63.90+6.09x$
(b) Scatter - plot with points $(2,77),(5,92),(1,70),(0,63),(4,90),(2,75),(3,84)$ and line of best fit $y = 63.90+6.09x$
(c) A linear relation exists because $r\approx0.99$ and the critical value for $n = 7$ and $\alpha = 0.05$ is approximately $0.754$ and $|r|>r_{crit}$
(d) $100.44$