For 4 data points of two correlated variables x and y, it is given that ∑ x = 24, ∑ y = 11, ∑ x 2= 202, ∑ xy = 84, ∑ y 2= 39 Fit a least squares line to this data using x as independent variable.
Understanding Least Squares Regression
Least squares regression is a method used to find the best-fitting line (or curve) through a set of data points. When we fit a straight line, it is called linear regression. The goal is to find a line \(y = a + bx\) such that the sum of the squared differences between the observed y-values and the y-values predicted by the line is minimized.
In this problem, we are given data for two correlated variables, x and y, with x as the independent variable and y as the dependent variable. We need to fit the least squares line \(y = a + bx\).
Given Data
We are given the following information for \(n=4\) data points:
- Sum of x values, ∑ x = 24
- Sum of y values, ∑ y = 11
- Sum of x squared values, ∑ x2 = 202
- Sum of xy values, ∑ xy = 84
- Sum of y squared values, ∑ y2 = 39 (Note: This value is not needed for finding the regression line of y on x).
Normal Equations for Least Squares Line \(y = a + bx\)
The coefficients \(a\) (y-intercept) and \(b\) (slope) of the least squares line \(y = a + bx\) are determined by solving the following system of linear equations, known as the normal equations:
- ∑ y = na + b ∑ x
- ∑ xy = a ∑ x + b ∑ x2
Substituting Given Values
Substitute the given values (\(n=4\), ∑ x = 24, ∑ y = 11, ∑ x2 = 202, ∑ xy = 84) into the normal equations:
- 11 = 4a + b(24) → 4a + 24b = 11 (Equation 1)
- 84 = a(24) + b(202) → 24a + 202b = 84 (Equation 2)
We now have a system of two linear equations:
\[ \begin{align*} 4a + 24b &= 11 \quad &(1) \\ 24a + 202b &= 84 \quad &(2) \end{align*} \]
Solving the System of Equations
We can solve this system for \(a\) and \(b\). Let's eliminate \(a\). Multiply Equation (1) by 6:
\[ 6 \times (4a + 24b) = 6 \times 11 \]
\[ 24a + 144b = 66 \quad &(3) \]
Now, subtract Equation (3) from Equation (2):
\[ (24a + 202b) - (24a + 144b) = 84 - 66 \]
\[ 202b - 144b = 18 \]
\[ 58b = 18 \]
\[ b = \frac{18}{58} = \frac{9}{29} \]
Now substitute the value of \(b = \frac{9}{29}\) back into Equation (1) to find \(a\):
\[ 4a + 24\left(\frac{9}{29}\right) = 11 \]
\[ 4a + \frac{216}{29} = 11 \]
\[ 4a = 11 - \frac{216}{29} \]
\[ 4a = \frac{11 \times 29 - 216}{29} \]
\[ 4a = \frac{319 - 216}{29} \]
\[ 4a = \frac{103}{29} \]
\[ a = \frac{103}{4 \times 29} = \frac{103}{116} \]
The Least Squares Line Equation
With \(a = \frac{103}{116}\) and \(b = \frac{9}{29}\), the equation of the least squares line \(y = a + bx\) is:
\[ y = \frac{103}{116} + \frac{9}{29}x \]
To express this in a form similar to the options, we can find a common denominator:
\[ y = \frac{103}{116} + \frac{9 \times 4}{29 \times 4}x \]
\[ y = \frac{103}{116} + \frac{36}{116}x \]
\[ y = \frac{1}{116}(103 + 36x) \]
Comparing with Options
Let's compare our derived equation with the given options:
- Option 1: \(y=\dfrac{1}{116}(36+103x)\)
- Option 2: \(x=\dfrac{1}{116}(36+103y)\) (This is a line of x on y)
- Option 3: \(x=\dfrac{1}{116}(103+36y)\) (This is a line of x on y)
- Option 4: \(y=\dfrac{1}{116}(103+36x)\)
Our calculated equation matches Option 4.
Least Squares Regression Revision
| Concept | Description |
|---|---|
| Least Squares Method | A statistical method to find the best-fitting line by minimizing the sum of the squared vertical distances (residuals) between the observed data points and the line. |
| Regression Line (y on x) | A line of the form \(y = a + bx\) where x is the independent variable and y is the dependent variable. Used to predict y based on x. |
| Normal Equations | A set of linear equations derived from the principle of least squares, used to solve for the coefficients (a and b) of the regression line. For \(y = a + bx\), the equations are ∑ y = na + b ∑ x and ∑ xy = a ∑ x + b ∑ x2. |
| Independent Variable (x) | The variable that is manipulated or observed to see its effect on the dependent variable. Plotted on the horizontal axis. |
| Dependent Variable (y) | The variable that is being predicted or explained. Its value depends on the independent variable. Plotted on the vertical axis. |
Additional Information on Regression Analysis
Regression analysis is a powerful statistical tool used to understand the relationship between variables. Linear regression, which fits a straight line, is the simplest form. The slope (b) represents the change in the dependent variable (y) for a one-unit increase in the independent variable (x). The y-intercept (a) is the predicted value of y when x is zero.
It's important to note the distinction between the regression line of y on x (\(y = a + bx\)) and the regression line of x on y (\(x = c + dy\)). They are generally different lines, except in the case of perfect correlation. The choice of which line to fit depends on which variable is considered independent and which is dependent (i.e., which variable you are trying to predict).
The goodness of fit of a regression line can be assessed using metrics like the coefficient of determination (R2), which measures the proportion of the variance in the dependent variable that is predictable from the independent variable.
Important Questions from Correlation and Regression
Variable Y regresses with variable X with the conditions that \(\overline X = 5.50\) , \(\overline Y = 3.50\) and b = 1.50 in the linear regression model (Y = a + bX), where \(\overline Y\) and \(\overline X\) are means of the respective variables and b refers to gradient of line of Y w. r. t X. Which one of the following values of parameter 'a' of the model is correct?
Given below are two statements:
Statement l: One of the assumptions under OLS method states that the regression model is linear in the parameters, though it may or may not be linear in the variables.
Statement ll: The variance of the error, or disturbance, term is not the same regardless of the value of the explanatory variable under OLS.
In light of the above statements, choose the most appropriate answer from the options given below
If two regression coefficients are -0.8 and -0.2, then the value of coefficient of correlation is
Which of the following statements relating to Correlation and Regression are true?
(a) The Coefficient of Correlation is independent of change of origin and scale.
(b) The Coefficient of Correlation between the two variables is the arithmetic average of the two Regression Coefficients.
(c) The probable error of the Coefficient Correlation is 0.6745 times its standard error.
(d) Coefficient of Correlation multiplied by the ratio between the standard deviations of the two variables denotes the slope of the regression line.
Code:
If two regression coefficients are 0.8 and 1.2, which one of the following is the value of coefficient of correlation?
