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

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

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

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.

Further Reading

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