# Solved (Free): In an experiment to determine the relationship between frequency and the inductive reactance of an electrical circuit

#### ByDr. Raju Chaudhari

Mar 13, 2021

In an experiment to determine the relationship between frequency and the inductive reactance of an electrical circuit, the following results were obtained :

Frequency (Hertz) Inductive reactance (Ohms)
50 30
100 65
150 90
200 130
250 150
300 190
350 200

a. Determine the equation of the regression line of inductive reactance on frequency, assuming a linear relationship.
b. Determine the product moment correlation.

#### Solution

Let $x$ denote the frequency and $y$ denote the inductive reactance in Hertz.

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 scatter diagram is

$x$ $y$ $x^2$ $y^2$ $xy$
1 50 30 2500 900 1500
2 100 65 10000 4225 6500
3 150 90 22500 8100 13500
4 200 130 40000 16900 26000
5 250 150 62500 22500 37500
6 300 190 90000 36100 57000
7 350 200 122500 40000 70000
Total 1400 855 350000 128725 212000

The sample mean of $x$ is

 \begin{aligned} \overline{x}&=\frac{1}{n} \sum_{i=1}^n x_i\\ &=\frac{1400}{7}\\ &=200 \end{aligned}

The sample mean of $y$ is

 \begin{aligned} \overline{y}&=\frac{1}{n} \sum_{i=1}^n y_i\\ &=\frac{855}{7}\\ &=122.1429 \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{7*212000-(1400)(855)}{7*(350000)-(1400)^2}\\ &= \frac{287000}{490000}\\ &= 0.5857. \end{aligned}

The estimate of intercept is

 \begin{aligned} b_0&=\overline{y}-b_1\overline{x}\\ &=122.1429-(0.586)*200\\ &=5.0029. \end{aligned}

The best fitted simple linear regression model to predict inductive reactance from frequency is

 \begin{aligned} \hat{y} &= 5.0029+ (0.5857)*x \end{aligned}

b) Product moment 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}{7 -1}\big(350000-\frac{1400^2}{7}\big)\\ &= 11666.6667. \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}{7 -1}\big(128725-\frac{855^2}{7}\big)\\ &= 4048.8095. \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}{7-1}\big(212000 - \frac{1400\times 855}{7} \big)\\ &=6833.3333. \end{aligned}

The product moment correlation coefficient is

 $$\begin{eqnarray*} r &=& \frac{Cov(X,Y)}{\sqrt{V(x)*V(Y)}} \\ &=&\frac{s_{xy}}{\sqrt{s_x^2\times s_y^2}}\\ &=& \frac{6833.3333}{\sqrt{11666.6667* 4048.8095}}\\ & = & 0.9942. \end{eqnarray*}$$

There is a strong positive relation between the inductive reactance and frequency.