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

#### ByDr. Raju Chaudhari

Oct 4, 2020

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).