Simple linear regression from summarize data
Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations.
The simple linear regression model of $Y$ on $X$ is
$$y_i=\beta_0 + \beta_1x_i +e_i$$ where,
- $y$ is a dependent variable,
- $x$ is an independent variable,
- $\beta_0$ is an intercept,
- $\beta_1$ is the slope,
- $e$ is the error term.
Formula
The simple linear regression model parameters $\beta_0$ and $\beta_1$ can be estimated using the method of least square.
The regression coefficients $\beta_1$ (slope) can be estimated as
$\hat{\beta}_1 = \frac{Cov(x,y)}{V(x)}=\dfrac{s_{xy}}{s_x^2}=r\dfrac{s_y}{s_x}$
The regression coefficients $\beta_0$ (intercept) can be estimated as
$\hat{\beta}_0=\overline{y}-\hat{\beta}_1\overline{x}$
where,
$\overline{x}=\dfrac{1}{n}\sum_{i=1}^n x_i$is the sample mean of $X$,$\overline{y}=\dfrac{1}{n}\sum_{i=1}^n y_i$is the sample mean of $Y$,$V(x) = s_x^2$is variance of $X$,$V(y) = s_y^2$is variance of $Y$,$Cov(x,y) = s_{xy}$is covariance between $X$ and $Y$,$r=\dfrac{Cov(x,y)}{\sqrt{V(x)V(y)}}$is the correlation coefficient between $X$ and $Y$,- $n$ is the number of data points.
Example
For 42 firms the number of employees $x$ and the amount of expenses for continuing education $x$ (in EUR) were recorded. The statistical summary of the data set is given by:
| Summary | Variable $x$ | Variable $y$ |
|---|---|---|
| Mean | 47 | 251 |
| Variance | 103 | 1960 |
The covariance between $X$ and $Y$ is equal to 377.4.
Estimate the expected amount of money spent for continuing education by a firm with 39 employees using least squares regression.
Solution
$X$ denote the number of employees and $Y$ denote amount of expenses for continuing education.
Given that $n=42$, mean of $X$ is $\overline{x}=47$, mean of $Y$ is $\overline{y}=251$, variance of $X$ is $V(x)=s_x^2=103$, variance of $Y$ is $V(y)=s_y^2=1960$ and the covariance between $X$ and $Y$ is $Cov(x,y) =s_{xy} = 377.6$.
The correlation coefficient between $X$ and $Y$ is
$$ \begin{aligned} r&= \frac{Cov(x,y)}{\sqrt{V(x)V(y)}}\\ &=\frac{377.6}{\sqrt{103\times 1960}}\\ &=\frac{377.6}{\sqrt{201880}}\\ &=\frac{377.6}{449.3106}\\ &=0.8404 \end{aligned} $$
The slope of the simple linear regression model is
$$ \begin{aligned} \hat{\beta}_1 &= r \cdot \frac{s_y}{s_x}\\ &= 0.8404 \cdot \sqrt{\frac{1960}{103}}\\ &= 3.666. \end{aligned} $$
OR
$$ \begin{aligned} \hat{\beta}_1 &=\frac{Cov(x,y)}{V(x)}\\ &=\frac{s_{xy}}{s_x^2}\\ &=\frac{377.6}{103}\\ &= 3.666. \end{aligned} $$
Intercept of the simple linear regression model is
$$ \begin{aligned} \hat{\beta}_0 &= \overline{y} - \hat{\beta}_1 \cdot \overline{x}\\ &=251 - 3.666 \cdot 47\\ &= 78.698. \end{aligned} $$
Using the estimates of $\beta_0$ and $\beta_1$, the best fitted simple linear regression model to predict amount of expenses for continuing education from number of employees is
$$ \begin{aligned} \hat{y} &= \hat{\beta}_0 + \hat{\beta}_1*x\\ &=78.698+ (3.666)*x \end{aligned} $$
The expected amount of expenses for continuing education when the number of employees $x=39$ is
$$ \begin{aligned} \hat{y}&=78.698 + (3.666)\times 39\\ &= 221.672\quad \text{ EUR} \end{aligned} $$
