Question

Consider the following five data points:
x -1.0 0.0 1.0 2.0 3.0
Y -1.0 1.0 1.0 2.5 3.5
a. Use regression analysis to calculate by hand the estimated coefficients of the equation Y = B + aX.
b. Compute the coefficient of determination.
c. What is the predicted value of Y for X = 1.0? For X = 3.5?

Answer 1

Answer on Question #73784 – Math – Statistics and Probability

Question

Consider the following five data points:

a. Use regression analysis to calculate by hand the estimated coefficients of the equation

$$y = b + ax$$

.

b. Compute the coefficient of determination.

c. What is the predicted value of $y$ for $x = 1.0$? For $x = 3.5$?

Solution

a. Regression equation of $y$ on $x$:

$$y = \mu_y + \frac{\operatorname{Cov}(x, y)}{\sigma_x^2} (x - \mu_x).$$

where $\mu_x$ and $\mu_y$ are mean values of $x$ and $y$, $\sigma_x^2$ is a variance of $x$ and $\operatorname{Cov}(x, y)$ is a covariance between $x$ and $y$.

Let us calculate the necessary values:

$$\mu_x = E[X] = \frac{-1.0 + 0.0 + 1.0 + 2.0 + 3.0}{5} = 1.0$$ $$\mu_y = E[Y] = \frac{-1.0 + 1.0 + 1.0 + 2.5 + 3.5}{5} = 1.4$$ $$\begin{array}{l} \operatorname{Cov}(x, y) = E[XY] - E[X]E[Y] \\ = \frac{(-1.0) \cdot (-1.0) + 0.0 \cdot 1.0 + 1.0 \cdot 1.0 + 2.0 \cdot 2.5 + 3.0 \cdot 3.5}{5} - 1.0 \cdot 1.4 = 2.1 \\ \end{array}$$ $$\begin{array}{l} \sigma_x^2 = Var[X] = \frac{(-1.0 - 1.0)^2 + (0.0 - 1.0)^2 + (1.0 - 1.0)^2 + (2.0 - 1.0)^2 + (3.0 - 1.0)^2}{5} \\ = 2.0 \\ \end{array}$$

Therefore, the regression equation is

$$y = 1.4 + \frac{2.1}{2.0} (x - 1.0)$$

or

$$y = 1.05x + 0.35$$

b. Let us first calculate the predicted values $\hat{y}_i$ of dependent variable by formula

$$\hat{y}_i = 1.05x_i + 0.35$$

The results we put in the Table 1:

The coefficient of determination $R^2$ is defined as

$$R^2 = 1 - \frac{SS_{res}}{SS_{tot}},$$

where

$$\begin{aligned} SS_{res} = & \sum_{i} (y_i - \hat{y}_i)^2 \\ & = (-1.0 + 0.7)^2 + (1.0 - 0.35)^2 + (1.0 - 1.4)^2 + (2.5 - 2.45)^2 + (3.5 - 3.5)^2 \\ & = 0.675; \end{aligned}$$ $$\begin{aligned} SS_{tot} = & \sum_{i} (y_i - \mu_y)^2 \\ & = (-1.0 - 1.4)^2 + (1.0 - 1.4)^2 + (1.0 - 1.4)^2 + (2.5 - 1.4)^2 + (3.5 - 1.4)^2 \\ & = 11.7 \end{aligned}$$

Therefore,

$$R^2 = 1 - \frac{0.675}{11.7} = 0.94$$

c. The predicted values of $y$ is calculated in Table 1. So the predicted value of $y$ for $x = 1.0$ is equal to $1.05 \cdot 1.0 + 0.35 = 1.4$ and the predicted value of $y$ for $x = 3.5$ is equal to $1.05 \cdot 3.5 + 0.35 = 4.025$

Answer: a. $y = 1.05x + 0.35, b = 0.35, a = 1.05$; b. $R^2 = 0.94$; c. 1.4 and 4.025.

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