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

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

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators