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
$$ \hat{y}_{x=95} = -0.8355+ (0.9869)*15 = 92.923 $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators