# Solved (Free): A calculus instructor asked a random sample of 8 students to record their study times per lesson in calculus course

#### ByDr. Raju Chaudhari

Mar 15, 2021

A calculus instructor asked a random sample of 8 students to record their study times per lesson in calculus course. The following table represents the data for total study times over the two weeks and test scores at the end of two weeks.

Study time (X) 10 15 12 20 8 16 14 22
Grade (y) 92 81 84 74 85 80 84 80

a. Determine the equation of the regression line for the data.
b. Use the regression line to predict the grade of a student who studies for 15 hours.

#### Solution

a. Let $x$ denote study time and $y$ denote grade.

Sr. No. $x$ $y$ $x^2$ $y^2$ $xy$
1 10 92 100 8464 920
2 15 81 225 6561 1215
3 12 84 144 7056 1008
4 20 74 400 5476 1480
5 8 85 64 7225 680
6 16 80 256 6400 1280
7 14 84 196 7056 1176
8 22 80 484 6400 1760
Total 117 660 1869 54638 9519

The slope $b_1$ is given by
 \begin{aligned} b_1 & = \frac{n \sum xy - (\sum x)(\sum y)}{n(\sum x^2) -(\sum x)^2}\\ & = \frac{8*9519-(117)(660)}{8*(1869)-(117)^2}\\ &= \frac{-1068}{1263}\\ &= -0.8456. \end{aligned}

The estimate of intercept is
 \begin{aligned} b_0&=\overline{y}-b_1\overline{x}\\ &=\bigg(\frac{660}{8}\bigg)-(-0.8456)*\bigg(\frac{117}{8}\bigg)\\ &=94.8669834 \end{aligned}

The best fitted linear regression equation is
 $$\hat{y} = 94.867+ (-0.8456)*x$$

b. Predicticted value of the grade of a student who studies for $x=15$ hours is

 $$\hat{y}_{x=15} = 94.867+ (-0.8456)*15 = 82.183$$