Question

professor g found a correlation between the number of hours spent studying in this class and the score in the weekly quiz score
number of hours spent and quiz score

number of hours spent studying	quiz score
8	4
15	9
18	9
17	10
7	7
5	4

what will be the quiz score for a student who studied 20 hours?
○ a. 10.94
○ b. 10
○ c.1.74
○ d. cannot determine the y value

Explanation:

Step1: Calculate means

Let $x$ be the number of hours studying and $y$ be the quiz - score.
$\bar{x}=\frac{2 + 8+15+18+17+7+5}{7}=\frac{72}{7}\approx10.29$
$\bar{y}=\frac{2 + 4+9+9+10+7+4}{7}=\frac{45}{7}\approx6.43$

Step2: Calculate slope ($b_1$)

$b_1=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}$

$x_i$	$y_i$	$x_i-\bar{x}$	$y_i-\bar{y}$	$(x_i - \bar{x})(y_i-\bar{y})$	$(x_i-\bar{x})^2$
8	4	$8 - 10.29=-2.29$	$4 - 6.43=-2.43$	$(-2.29)\times(-2.43)=5.57$	$(-2.29)^2 = 5.24$
15	9	$15 - 10.29 = 4.71$	$9 - 6.43=2.57$	$4.71\times2.57 = 12.10$	$4.71^2=22.18$
18	9	$18 - 10.29 = 7.71$	$9 - 6.43=2.57$	$7.71\times2.57 = 19.81$	$7.71^2=59.44$
17	10	$17 - 10.29 = 6.71$	$10 - 6.43 = 3.57$	$6.71\times3.57=23.95$	$6.71^2 = 45.02$
7	7	$7 - 10.29=-3.29$	$7 - 6.43 = 0.57$	$(-3.29)\times0.57=-1.87$	$(-3.29)^2 = 10.82$
5	4	$5 - 10.29=-5.29$	$4 - 6.43=-2.43$	$(-5.29)\times(-2.43)=12.86$	$(-5.29)^2 = 27.98$

$\sum_{i = 1}^{7}(x_i-\bar{x})(y_i - \bar{y})=36.72 + 5.57+12.10+19.81+23.95-1.87+12.86=108.14$
$\sum_{i = 1}^{7}(x_i-\bar{x})^2=68.72+5.24 + 22.18+59.44+45.02+10.82+27.98=239.4$

$b_1=\frac{108.14}{239.4}\approx0.452$

Step3: Calculate intercept ($b_0$)

$b_0=\bar{y}-b_1\bar{x}$
$b_0 = 6.43-0.452\times10.29=6.43 - 4.65=1.78$

Step4: Find the predicted value

The regression equation is $\hat{y}=b_0 + b_1x$.
When $x = 20$, $\hat{y}=1.78+0.452\times20=1.78 + 9.04=10.82\approx10.94$

Answer:

A. 10.94