Question

Consider the following five data points:
x -1. 0 1 2 3
Y -1 1 2 4 5
a) use regression analysis to calculate by hand the estimated coefficients of the equation Y = B + aX
(b) compute the standard error and the t-statistics for the coefficient of determination of X

Answer 1

Answer on Question #73785 – Math – Statistics and Probability

Consider the following five data points:

$$\begin{array}{l} X \quad - 1 \quad 0 \quad 1 \quad 2 \quad 3 \\ Y \quad - 1 \quad 1 \quad 2 \quad 4 \quad 5 \\ \end{array}$$

a) use regression analysis to calculate by hand the estimated coefficients of the equation $Y = B + AX$

Solution

$$\begin{array}{l} \text{mean of } x = \bar{x} = \frac{\sum x_i}{n} \\ \text{mean of } y = \bar{y} = \frac{\sum y_i}{n} \\ \end{array}$$ $$\begin{array}{l} \text{trend line: } Y = B + AX, A = \frac{S_{xy}}{S_{xx}}, B = \bar{y} - A\bar{x} \\ S_{xx} = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{\sum (x_i)^2}{n} - (\bar{x})^2 \\ S_{yy} = \frac{\sum (y_i - \bar{y})^2}{n} = \frac{\sum (y_i)^2}{n} - (\bar{y})^2 \\ S_{xy} = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n} = \frac{\sum x_i \cdot y_i}{n} - \bar{x} \cdot \bar{y} \\ \end{array}$$ $$\begin{array}{l} \text{mean of } x = \bar{x} = \frac{-1 + 0 + 1 + 2 + 3}{5} = 1 \\ \text{mean of } y = \bar{y} = \frac{-1 + 1 + 2 + 4 + 5}{5} = 2.2 \\ S_{xx} = \frac{(-1)^2 + (0)^2 + (1)^2 + (2)^2 + (3)^2}{5} - (1)^2 = 2 \\ S_{yy} = \frac{(-1)^2 + (1)^2 + (2)^2 + (4)^2 + (5)^2}{5} - (2.2)^2 = 4.56 \\ S_{xy} = \frac{(-1)(-1) + (0)(1) + (1)(2) + (2)(4) + (3)(5)}{5} - (1)(2.2) = 3 \\ A = \frac{3}{2} = 1.5 \\ B = 2.2 - 1.5(1) = 0.7 \\ Y = 0.7 + 1.5X \\ \end{array}$$

(b) compute the standard error and the t-statistics for the coefficient of determination of $X$

**Solution**

$$r = \frac {S _ {x y}}{\sqrt {S _ {x x} \sqrt {S _ {y y}}}}$$

the coefficient of determination of $X$

$$r ^ {2} = \frac {\left(S _ {x y}\right) ^ {2}}{S _ {x x} S _ {y y}}$$ $$r ^ {2} = \frac {(3) ^ {2}}{2 (4 . 5 6)} = 0. 9 8 6 8 4 2 1 0 5$$

$r = 0.993399267$

$$S S T = \sum_ {i} (y _ {i} - \bar {y}) ^ {2} = 2 2. 8$$ $$S S R = \sum_ {i} (\hat {y} _ {i} - \bar {y}) ^ {2} = 2 2. 5$$ $$r ^ {2} = \frac {S S R}{S S T} = \frac {2 2 . 5}{2 2 . 8} \approx 0. 9 8 6 8 4 2 1 0 5$$ $$S E E = \sqrt {\frac {\sum (\hat {y} - y) ^ {2}}{n - 2}}$$ $$S E E = \sqrt {\frac {0 . 0 4 + 0 . 0 9 + 0 . 0 4 + 0 . 0 9 + 0 . 0 4}{5 - 2}} = \sqrt {0 . 1} \approx 0. 3 1 6 2 2 7 7 6 6$$

Standard error of slope

$$S E _ {A} = \frac {S E E}{\sqrt {\sum (x - \bar {x}) ^ {2}}}$$ $$S E _ {A} = \frac {\sqrt {0 . 1}}{\sqrt {(- 1 - 1) ^ {2} + (0 - 1) ^ {2} + (1 - 1) ^ {2} + (2 - 1) ^ {2} + (3 - 1) ^ {2}}} = 0. 1$$

The test for the significance of regression for the data in the table

The test is carried out using the $t -$ test on the coefficient $A$

$$H _ {0}: A = 0$$ $$H _ {1}: A \neq 0$$

The test statistic can be calculated using the following equation:

$$\text{Test Statistic: } t_0 = \frac{A - 0}{SE_A} = \frac{1.5}{0.1} = 15$$

This statistic based on the $t$ distribution with 3 ($n - 2 = 5 - 2 = 3$) degrees of freedom can be obtained as follows:

$$\alpha = 0.01$$ $$\alpha/2 = 0.005$$ $$df = n - 2 = 5 - 2 = 3$$ $$\text{Critical value: } = 5.841$$

Statistical conclusion:

Reject $H_0$, there is evidence that $A$ is not equal to zero and that a significant relationship exists between $y$ and $x$.

The coefficient of non-determination was used in the $t$-test to see if there was significant linear correlation.

The standard error of the estimate is the square root of the coefficient of non-determination divided by its degrees of freedom.

$$s_e = \sqrt{\frac{1 - r^2}{n - 2}}$$ $$s_e = \sqrt{\frac{1 - 0.986842105}{5 - 2}} \approx 0.066226618$$

The test is carried out using the $t$-test on the coefficient $1 - r^2$.

$$H_0: 1 - r^2 = 1$$ $$H_1: 1 - r^2 < 0$$

The test statistic can be calculated using the following equation:

$$\text{Test Statistic: } t_0 = \frac{(1 - r^2) - 1}{s_e} = \frac{-0.986842105}{0.066226618} \approx -14.90098898$$

This statistic based on the $t$ distribution with 3 ($n - 2 = 5 - 2 = 3$) degrees of freedom can be obtained as follows:

$$\alpha = 0.01$$ $$df = n - 2 = 5 - 2 = 3$$ $$\text{Critical value: } = 4.54070$$

Statistical conclusion:

Reject $H_0$, there is evidence that $(1 - r^2)$ is not equal to 1 and that a significant relationship exists between $y$ and $x$.

Answer provided by https://www.AssignmentExpert.com