The following results were obtained experimentally when verifying Hooke's Law.

Load (N) | Extension (mm) |
---|---|

2 | 2 |

5 | 23 |

8 | 62 |

11 | 119 |

15 | 223 |

a. Create a scatter plot.

b. Produce the best-fit straight line and determine its equation.

c. Determine value, degree and nature of the correlation of the data.

#### Solution

Let $x$ denote the Load and $y$ denote the Extension in mm.

a. The scatter diagram is

$x$ | $y$ | $x^2$ | $y^2$ | $xy$ | |
---|---|---|---|---|---|

1 | 2 | 2 | 4 | 4 | 4 |

2 | 5 | 23 | 25 | 529 | 115 |

3 | 8 | 62 | 64 | 3844 | 496 |

4 | 11 | 119 | 121 | 14161 | 1309 |

5 | 15 | 223 | 225 | 49729 | 3345 |

Total | 41 | 429 | 439 | 68267 | 5269 |

b. Let the simple linear regression model of $Y$ on $X$ is

$$y=\beta_0 + \beta_1x +e$$

By the method of least square, the estimates of $\beta_1$ and $\beta_0$ are respectively

` $$ \begin{aligned} \hat{\beta}_1 & = \frac{n \sum xy - (\sum x)(\sum y)}{n(\sum x^2) -(\sum x)^2} \end{aligned} $$ `

and

` $$ \begin{aligned} \hat{\beta}_0&=\overline{y}-\hat{\beta}_1\overline{x} \end{aligned} $$ `

The sample mean of $x$ is

` $$ \begin{aligned} \overline{x}&=\frac{1}{n} \sum_{i=1}^n x_i\\ &=\frac{41}{5}\\ &=8.2 \end{aligned} $$ `

The sample mean of $y$ is

` $$ \begin{aligned} \overline{y}&=\frac{1}{n} \sum_{i=1}^n y_i\\ &=\frac{429}{5}\\ &=85.8 \end{aligned} $$ `

The estimate of $\beta_1$ is given by

` $$ \begin{aligned} b_1 & = \frac{n \sum xy - (\sum x)(\sum y)}{n(\sum x^2) -(\sum x)^2}\\ & = \frac{5*5269-(41)(429)}{5*(439)-(41)^2}\\ &= \frac{8756}{514}\\ &= 17.035. \end{aligned} $$ `

The estimate of intercept is

` $$ \begin{aligned} b_0&=\overline{y}-b_1\overline{x}\\ &=85.8-(17.035)*8.2\\ &=-53.887. \end{aligned} $$ `

The best fitted simple linear regression model to predict Extension from Load is

` $$ \begin{aligned} \hat{y} &= -53.887+ (17.035)*x \end{aligned} $$ `

c. Correlation coefficient:

The sample variance of $X$ is

` $$ \begin{aligned} s_{x}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_i^2 - \frac{(\sum_{i=1}^n x_i)^2}{n}\bigg)\\ &= \frac{1}{5 -1}\big(439-\frac{41^2}{5}\big)\\ &= 25.7. \end{aligned} $$ `

The sample variance of $Y$ is

` $$ \begin{aligned} s_{y}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(y_i -\overline{y})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n y_i^2 - \frac{(\sum_{i=1}^n y_i)^2}{n}\bigg)\\ &= \frac{1}{5 -1}\big(68267-\frac{429^2}{5}\big)\\ &= 7864.7. \end{aligned} $$ `

The covariance between $X$ and $Y$ is

` $$ \begin{aligned} s_{xy}&=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})(y_i-\overline{y})\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_iy_i - \frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}\bigg)\\ &=\frac{1}{5-1}\big(5269 - \frac{41\times 429}{5} \big)\\ &=437.8. \end{aligned} $$ `

The product moment correlation coefficient is

` $$ \begin{eqnarray*} r &=& \frac{Cov(X,Y)}{\sqrt{V(x)*V(Y)}} \\ &=&\frac{s_{xy}}{\sqrt{s_x^2\times s_y^2}}\\ &=& \frac{437.8}{\sqrt{25.7* 7864.7}}\\ & = & 0.9738. \end{eqnarray*} $$ `

There is a strong positive relation between Load and Extension.

#### Further Reading

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