Question

Sample Covariance. For a set of n data points, the sample covariance, $s_{xy+}$is given by

The sample covariance can be used as an alternative method for tinding the slope and y-intercept of a regression line. The formulas are

$b_1=s_v/x_k^2\text{and}{b}_0=\hat{y}-b_1{\hat{i}}_n$

where $s_i$denotes the sample standard deviation of the x-values.

a. Use Equation (4.1) to determine the sample covariance of the data points in Exercise 4,45.

b. Use Equation (4.2) and your answer from part (a) to find the regression equation. Compare your result to that found in Exercise 4.57.

Short answer

a) Sample covariance is $\frac{-15}{4}$

b) The regression equation is$\hat{y}=2.875-0.625x$

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that

$$\begin{gathered} n=5 \\ \sum x_i=15 \\ \sum y_i=5 \\ \sum x_i{y}_i=0 \\ \sum x_i^2=69 \end{gathered}$$

Part (a) Step 2: Explanation

The calculation of sample covariance is showed below,

$$\begin{gathered} \text{Sample covariance}=\frac{\sum x_i{y}_i-\left(\sum x_i\right)\frac{\left(\sum y_i\right)}{n}}{n-1} \\ =\frac{0-\left(15\right)\frac{\left(5\right)}{5}}{5} \\ =\frac{-15}{4} \end{gathered}$$

Part (b) Step 1: Given information

Given in the question that

$$\begin{gathered} n=5 \\ \sum x_i=15 \\ \sum y_i=5 \\ \sum x_i{y}_i=0 \\ \sum x_i^2=69 \end{gathered}$$

Part (b) Step 2: Explanation

Here the value of the $S_{xx}$will be,

$$\begin{gathered} S_{xx}=\frac{\sum x_i^2-\frac{\sum {\left(x_i\right)}^2}{n}}{n-1} \\ {S}_{xx}=\frac{69-\frac{(15{)}^2}{5}}{4} \\ {S}_{xx}=\frac{69-45}{4} \\ {S}_{xx}=6 \end{gathered}$$

Then the slop will be,

$$\begin{gathered} b_1=\frac{S_{xy}}{S_{xx}} \\ {b}_1=\frac{\left(-\frac{15}{4}\right)}{6} \\ {b}_1=-0.625 \end{gathered}$$

Now we need to calculate the y-intercept

$$\begin{gathered} b_0=\overline{y}-b_1\overline{x} \\ {b}_0=1-(-0.625)3 \\ {b}_0=1+1.875 \\ {b}_0=2.875 \end{gathered}$$

Therefore we can represent the regression equation as

$$\begin{gathered} \widehat{y}=b_0+b_{1}x \\ \widehat{y}=2.875-0.625x \end{gathered}$$