早稲田大学 創造理工学研究科 経営システム工学専攻 2021年8月実施 概率统计

标签：

• Waseda-University
• Independence-of-Random-Variables
• Simple-Linear-Regression

Author

思齐塾, 祭音Myyura

Description

次の2変数のデータに基づき各設問に解答せよ。

$$\begin{array}{|c|c|c|c|c|c|} \hline No & 1 & 2 & 3 & 4 & 5 \\ \hline x & 1 & 2 & 2 & 2 & 3 \\ \hline y & 4 & 4 & 2 & 0 & 0 \\ \hline \end{array}$$

(1) 相関係数 (correlation coefficient) を求めよ。なお、計算結果に根号を残してもよい。

(2) yを目的変数 (dependent variable), xを説明変数 (independent variable) として、単回帰分析 (single regression analysis) を行って直線をあてはめる。最小2乗法 (least squares method) による直線の推定式 (estimated equation) を求めよ。

题目描述

根据以下两变量数据回答问题：

$$\begin{array}{|c|c|c|c|c|c|} \hline \mathrm{No.}&1&2&3&4&5\\ \hline x&1&2&2&2&3\\ \hline y&4&4&2&0&0\\ \hline \end{array}$$

1. 求相关系数，结果中可以保留根号。
2. 以 $y$ 为因变量、$x$ 为自变量作简单线性回归，求最小二乘法拟合直线的估计方程。

Kai

(1) Let's calculate the correlation coefficient $r$ .

First, calculate the means of $x$ and $y$ :

$$\bar{x} = \frac{1+2+2+2+3}{5} = \frac{10}{5} = 2$$

$$\bar{y} = \frac{4+4+2+0+0}{5} = \frac{10}{5} = 2$$

Next, calculate the standard deviations of $x$ and $y$ :

$$s_x = \sqrt{\frac{(1-2)^2 + (2-2)^2 + (2-2)^2 + (2-2)^2 + (3-2)^2}{5-1}} = \sqrt{\frac{1+0+0+0+1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

$$s_y = \sqrt{\frac{(4-2)^2 + (4-2)^2 + (2-2)^2 + (0-2)^2 + (0-2)^2}{5-1}} = \sqrt{\frac{4+4+0+4+4}{4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$$

Now, calculate the covariance of $x$ and $y$ :

$$cov(x, y) = \frac{\sum_{i=1}^5 (x_i - \bar{x})(y_i - \bar{y})}{5-1} = \frac{(1-2)(4-2) + (2-2)(4-2) + (2-2)(2-2) + (2-2)(0-2) + (3-2)(0-2)}{4} = \frac{(-1)(2) + (0)(2) + (0)(0) + (0)(-2) + (1)(-2)}{4} = \frac{-2 + 0 + 0 + 0 - 2}{4} = \frac{-4}{4} = -1$$

Finally, calculate the correlation coefficient:

$$r = \frac{cov(x, y)}{s_x s_y} = \frac{-1}{\frac{1}{\sqrt{2}} \cdot 2} = \frac{-1}{\frac{2}{\sqrt{2}}} = \frac{-\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}$$

(2) To find the estimated equation $\hat{y} = a + bx$ using the least squares method, we need to calculate $a$ and $b$ .

The formula for $b$ is:

$$b = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2}$$

We already calculated the numerator in part (1) as $cov(x, y) \times (n-1)$ , so $\sum_{i=1}^5 (x_i - \bar{x})(y_i - \bar{y}) = -4$ .

Now, let's calculate the denominator:

$$\sum_{i=1}^5 (x_i - \bar{x})^2 = (1-2)^2 + (2-2)^2 + (2-2)^2 + (2-2)^2 + (3-2)^2 = 1 + 0 + 0 + 0 + 1 = 2$$

So, $b = \frac{-4}{2} = -2$ .

The formula for $a$ is:

$$a = \bar{y} - b\bar{x}$$

We have $\bar{x} = 2$ , $\bar{y} = 2$ , and $b = -2$ , so

$$a = 2 - (-2)(2) = 2 + 4 = 6$$

Therefore, the estimated equation is:

$$\hat{y} = 6 - 2x$$