Solved (Free): A STAT 200 instructor believes that the average quiz score is a good predictor of final exam score

ByDr. Raju Chaudhari

Mar 15, 2021

A STAT 200 instructor believes that the average quiz score is a good predictor of final exam score. A random sample of 10 students produced the following data where x is the average quiz score and y is the final exam score.

x 80 93 50 60 100 40 85 70 75 85
y 70 96 50 63 96 38 83 60 77 87

(a) Find an equation of the least squares regression line. Show all work; writing the correct equation, without supporting work, will receive no credit.
(b) Based on the equation from part (a), what is the predicted final exam score if the average quiz score is 95? Show all work and justify your answer.

Solution

a. Let $x$ denote average quiz score and $y$ denote final exam score.

$x$ $y$ $x^2$ $y^2$ $xy$
80 70 6400 4900 5600
93 96 8649 9216 8928
50 50 2500 2500 2500
60 63 3600 3969 3780
100 96 10000 9216 9600
40 38 1600 1444 1520
85 83 7225 6889 7055
70 60 4900 3600 4200
75 77 5625 5929 5775
85 87 7225 7569 7395
738 720 57724 55232 56353

The last row is the total.

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{10*56353-(738)(720)}{10*(57724)-(738)^2}\\ &= \frac{32170}{32596}\\ &= 0.9869. \end{aligned}

The estimate of intercept is

 \begin{aligned} b_0&=\overline{y}-b_1\overline{x}\\ &=\bigg(\frac{720}{10}\bigg)-(0.9869)*\bigg(\frac{738}{10}\bigg)\\ &=-0.8355013 \end{aligned}

The best fitted linear regression equation is

 $$\hat{y} = -0.8355+ (0.9869)*x$$

b. The predicted final exam score if the average quiz
score is $x=95$ is

 $$\hat{y}_{x=95} = -0.8355+ (0.9869)*15 = 92.923$$