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