The torque, T Nm, required to rotate shafts of different diameters, D mm, on a machine has been tested and recorded below.

D (mm) | 6 | 10 | 14 | 18 | 21 | 25 |
---|---|---|---|---|---|---|

T (Nm) | 5.5 | 7.0 | 9.5 | 12.5 | 13.5 | 16 |

a) Plot a scatter diagram of the results with diameter of the machine as the independent variable.

b) Use the method of least squares to determine the linear regression equation that relates the torque to the diameter.

c) Estimate the value of torque required when the diameter is 16 mm.

d) Calculate the coefficient of correlation between these variables and comment on your prediction obtained in (c).

#### Solution

a) Scatter diagram of the results with diameter of the machine as the independent variable

Sr. No. | $D$ (mm) | $T$ (Nm) | $D^2$ | $T^2$ | $D*T$ |
---|---|---|---|---|---|

1 | 6 | 5.5 | 36 | 30.25 | 33 |

2 | 10 | 7 | 100 | 49 | 70 |

3 | 14 | 9.5 | 196 | 90.25 | 133 |

4 | 18 | 12.5 | 324 | 156.25 | 225 |

5 | 21 | 13.5 | 441 | 182.25 | 283.5 |

6 | 25 | 16 | 625 | 256 | 400 |

Total | 94 | 64 | 1722 | 764 | 1144.5 |

b) $\overline{D} = \frac{\sum D}{n} = \frac{94}{6} = 15.6667$, $\overline{T} = \frac{\sum T}{n} = \frac{64}{6} = 10.6667$

` $$ \begin{eqnarray*} SS_{DD} &=& \sum D^2 -\frac{(\sum D)^2}{n} \\ &=& 1722 - \frac{(94)^2}{6}\\ & = & 249.333. \end{eqnarray*} $$ `

` $$ \begin{eqnarray*} SS_{TT} &=& \sum T^2 -\frac{(\sum T)^2}{n} \\ &=& 764 - \frac{(64)^2}{6}\\ & = & 81.333. \end{eqnarray*} $$ `

` $$ \begin{eqnarray*} SS_{DT} &=& \sum (D*T) -\frac{(\sum D)*(\sum T)}{n} \\ &=& 1144.5 - \frac{(94 * 64)}{6}\\ & = & 141.333 \end{eqnarray*} $$ `

So, `$\hat{b} = \frac{SS_{DT}}{SS_{DD}} = \frac{141.333}{249.333} = 0.56885$`

and

`$\hat{a} = \overline{T} - \hat{b} *\overline{D} = 10.6667 - 0.56889*15.6667 = 1.7539$`

.

So the linear regression equation that relates the torque to the diameter is

` $$ \begin{equation*} T = \hat{a} + \hat{b} *D = 1.7539 + 0.56885*D. \end{equation*} $$ `

c) When the diameter $D = 16$mm, the torque is

$\hat{T} = 1.7539 + 0.56885*16 = 10.8555$.

So the estimated torque for $D = 16$ mm is 10.8555 Nm.

d) The correlation coefficient is

` $$ \begin{eqnarray*} r &=& \frac{SS_{DT}}{\sqrt{SS_{DD}*SS_{TT}}} \\ &=& \frac{141.333}{\sqrt{249.333* 81.333}}\\ & = & 0.9960. \end{eqnarray*} $$ `

$r^2 = (0.9960)^2 = 0.9920$.

So 99% of the variation in Torque (T) is explained by Diameter (D).

#### Further Reading

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