Question

Use the data in Exercises \(7-8\) to calculate the coefficient of determination, \(r^{2} .\) What information does this value give about the usefulness of the linear model? $$ \begin{array}{r|rrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$

Short answer

To summarize, the given dataset has a coefficient of determination (\(r^2\)) of 0.5041, which means that approximately 50.41% of the variability in the y values can be explained by the linear model using x values. Although the linear model might be useful in capturing the relationship between x and y, further investigation and model refinement would be beneficial as there is still a significant amount of variability unexplained.

Step-by-step solution

Calculate the means of x and y values

First, we need to find the means of the x and y values in our dataset. To do this, we add up the values of x and y separately and then divide by the number of data points (5 in our case). For x values: \(\bar{x} = \frac{-2 + (-1) + 0 + 1 + 2}{5} = 0\) For y values: \(\bar{y} = \frac{1+1+3+5+5}{5} = 3\)

Calculate the covariance and variances for x and y

Next, we need to calculate the covariance of x and y, which is given by the formula: \(Cov(x,y) = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{n}\) We also need the variances of x and y to calculate r later on: For x: \(Var(x) = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n}\) For y: \(Var(y) = \frac{\sum_{i=1}^n (y_i - \bar{y})^2}{n}\) We can calculate each of these values using the given dataset: \(Cov(x,y) = \frac{(-2)(-2)+(-1)(-1)+0(0)+1(2)+2(2)}{5} = \frac{10}{5} = 2\) \(Var(x) = \frac{(-2)^2+(-1)^2+0^2+1^2+2^2}{5} = \frac{10}{5} = 2\) \(Var(y) = \frac{(1-3)^2+(1-3)^2+(3-3)^2+(5-3)^2+(5-3)^2}{5} = \frac{20}{5} = 4\)

Calculate the correlation coefficient (r)

Now we have all the information needed to calculate the correlation coefficient (r) using the formula: \(r = \frac{Cov(x,y)}{\sqrt{Var(x) \cdot Var(y)}}\) \(r = \frac{2}{\sqrt{2(4)}} = \frac{2}{\sqrt{8}} = \frac{1}{\sqrt{2}} = 0.71\)

Calculate the coefficient of determination (r^2)

Now, we can find the coefficient of determination (r^2) by simply squaring the value of r. \(r^2 = (0.71)^2 = 0.5041\)

Interpret the coefficient of determination

The coefficient of determination (\(r^2\)) is 0.5041, which lies between 0 and 1. It tells us that approximately 50.41% of the variability in the y values is explained by the linear model using x values. Since the value is above 0.5, it suggests the linear model might be useful in capturing the relationship between x and y, but there is still a considerable amount of variability unexplained. Further investigation and model refinement might be needed to better capture the relationship between the variables.

Key concepts

Linear Regression

Linear regression is a statistical method that models the relationship between two variables by fitting a linear equation to observed data. One variable is considered dependent and the other independent. The goal is to use the linear relationship estimated from the data to predict values of the dependent variable from the independent variable.

Let’s look at a simple example. In the given exercise, we have pairs of x (independent variable) and y (dependent variable) values. These points can be plotted on a graph, and linear regression aims to draw a straight line—known as the regression line—that best fits these points. Visually, this line will minimize the sum of the distances between the points and the line itself.

Mathematically, the regression line is typically expressed in the form of an equation:
\( y = \beta_0 + \beta_1 x \) where \( \beta_0 \) is the y-intercept and \( \beta_1 \) is the slope of the line. These coefficients are calculated using methods such as the Least Squares method, which minimizes the sum of the squared differences between the observed values and the values predicted by the model.

The coefficient of determination, denoted as \( r^2 \), quantifies the quality of the regression model by measuring the proportion of variance in the dependent variable that is predictable from the independent variable. As seen in the solution, \( r^2 \) gives us invaluable insight into how useful the linear model is in explaining the relationship between x and y.

Correlation Coefficient

The correlation coefficient, often denoted as \( r \), measures the strength and direction of a linear relationship between two variables. Its value ranges from -1 to 1, where 1 indicates a perfect positive linear correlation, -1 indicates a perfect negative linear correlation, and 0 indicates no linear correlation.

In simple terms, if \( r \) is close to 1, it means that as one variable increases, the other one also increases in a linear pattern. Conversely, if \( r \) is close to -1, it implies that as one variable increases, the other one decreases. An \( r \) value close to 0 would suggest that there is little to no linear relationship between the variables.

\( r = \frac{Cov(x,y)}{\text{SD}(x) \times \text{SD}(y)} = \frac{Cov(x,y)}{\text{sqrt}{[Var(x) \times Var(y)]}} \)

The correlation coefficient \( r \) itself is derived from the covariance of the variables normalized by their standard deviations. In the exercise, the calculation of \( r \) has led to a value of approximately 0.71, indicating a moderate positive linear relationship between x and y.

Variance and Covariance

Variance and covariance are two fundamental concepts in statistics that describe the spread and the relationship of data, respectively.

Variance, denoted as \( Var(x) \) for a variable \( x \) and \( Var(y) \) for a variable \( y \), measures how much the values in a dataset spread out around the mean. If the variance is high, the data points are spread out widely from their mean, indicating great variability. If the variance is low, the data points are closer to the mean, indicating less variability.

Covariance, on the other hand, measures how two variables vary together. A positive covariance implies that as one variable increases, the other variable tends to increase as well. A negative covariance indicates that as one variable increases, the other variable tends to decrease.

In the context of the exercise, the calculation of variance provided the necessary values to understand the spread of both x and y variables, while the computation of covariance allowed for the understanding of their relationship. Knowing both variance and covariance was essential to calculate the correlation coefficient and the coefficient of determination, ultimately gauging the performance of the linear regression model.