Calculating the Regression Line of y on x

The question asks us to find the line of regression of y on x using the provided data for 10 observations on price (x) and supply (y). The line of regression of y on x is typically represented by the equation:

\(y = a + bx\)

where:

• \(y\) is the dependent variable (supply).
• \(x\) is the independent variable (price).
• \(b\) is the regression coefficient of y on x, representing the change in y for a one-unit change in x.
• \(a\) is the y-intercept, representing the expected value of y when x is 0.

We are given the following summary statistics for n = 10 observations:

• \(\sum x = 130\)
• \(\sum y = 220\)
• \(\sum x^2 = 2288\)
• \(\sum y^2 = 5506\)
• \(\sum xy = 3467\)

To find the line of regression \(y = a + bx\), we first need to calculate the regression coefficient \(b\) and then the intercept \(a\).

Calculating the Regression Coefficient (b)

The formula for the regression coefficient \(b\) of y on x is:

\(b = \frac{n \sum xy - (\sum x)(\sum y)}{n \sum x^2 - (\sum x)^2}\)

Let's plug in the given values:

\(b = \frac{(10)(3467) - (130)(220)}{(10)(2288) - (130)^2}\)

\(b = \frac{34670 - 28600}{22880 - 16900}\)

\(b = \frac{6070}{5980}\)

\(b = \frac{607}{598}\)

Now, let's calculate the decimal value for \(b\):

\(b \approx 1.01505\)

Rounding this to two decimal places gives approximately \(1.02\).

Calculating the Intercept (a)

The formula for the intercept \(a\) is:

\(a = \bar{y} - b\bar{x}\)

First, we need to find the means of x and y:

\(\bar{x} = \frac{\sum x}{n} = \frac{130}{10} = 13\)

\(\bar{y} = \frac{\sum y}{n} = \frac{220}{10} = 22\)

Now, using the value of \(b\) we calculated (we will use the rounded value \(b \approx 1.02\) to match the options provided, as is often done in practice):

\(a = 22 - (1.02)(13)\)

\(a = 22 - 13.26\)

\(a = 8.74\)

Forming the Regression Line Equation

Now that we have the values for \(a\) and \(b\), we can write the equation for the line of regression of y on x:

\(y = a + bx\)

\(y = 8.74 + 1.02x\)

This can also be written as:

\(y = 1.02x + 8.74\)

Comparing with Options

Let's compare our calculated regression line with the given options:

Our calculated line \(y = 1.02x + 8.74\) matches Option 2 exactly. The slight difference in the exact calculated \(b\) (\(\approx 1.01505\)) compared to the option's value (\(1.02\)) is due to standard rounding. When \(b\) is rounded to two decimal places (\(1.02\)), calculating \(a\) using this rounded \(b\) value yields \(8.74\), which perfectly matches the option.

Revision Table: Key Concepts

Additional Information on Linear Regression

Linear regression is a statistical method used to model the linear relationship between a dependent variable (y) and one or more independent variables (x). When there is only one independent variable, it is called simple linear regression.

• Purpose: The primary goals of simple linear regression are to model the relationship between two variables, predict the value of the dependent variable based on the independent variable, and understand the strength and direction of the relationship.
• Assumptions: Linear regression relies on several assumptions, including linearity (the relationship is linear), independence of errors, homoscedasticity (constant variance of errors), and normality of errors.
• Interpretation of Coefficients: The slope (\(b\)) indicates how much the dependent variable is expected to change when the independent variable increases by one unit. The intercept (\(a\)) is the expected value of the dependent variable when the independent variable is zero. Note that interpreting the intercept only makes sense if zero is a meaningful value for the independent variable and falls within the range of the data.
• Coefficient of Determination (\(R^2\)): This value measures the proportion of the variance in the dependent variable that is predictable from the independent variable. It ranges from 0 to 1. A higher \(R^2\) indicates that the model explains a larger portion of the variance.

The regression line is also known as the "line of best fit" because it minimizes the sum of the squared vertical distances (residuals) between the observed data points and the line.