Question

consider the following small data set.
subject x y
1 7 27
2 13 24
3 6 30
4 5 24
5 9 20
find the linear correlation coefficient.
r = 0.401
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Explanation:

Step1: Calculate the means of \(x\) and \(y\)

Let \(x_1 = 7,x_2=13,x_3 = 6,x_4=5,x_5 = 9\); \(y_1 = 27,y_2=24,y_3 = 30,y_4=24,y_5 = 20\).
\(\bar{x}=\frac{7 + 13+6+5+9}{5}=\frac{40}{5}=8\)
\(\bar{y}=\frac{27+24 + 30+24+20}{5}=\frac{125}{5}=25\)

Step2: Calculate the numerator of the correlation - coefficient formula

\[

$$\begin{align*} &\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})\\ =&(7 - 8)(27-25)+(13 - 8)(24 - 25)+(6 - 8)(30 - 25)+(5 - 8)(24 - 25)+(9 - 8)(20 - 25)\\ =&(-1)\times2+5\times(-1)+(-2)\times5+(-3)\times(-1)+1\times(-5)\\ =&-2-5 - 10 + 3-5\\ =&-19 \end{align*}$$

\]

Step3: Calculate the denominator of the correlation - coefficient formula

\[

$$\begin{align*} &\sqrt{\sum_{i = 1}^{5}(x_i-\bar{x})^2\sum_{i = 1}^{5}(y_i-\bar{y})^2}\\ \sum_{i = 1}^{5}(x_i-\bar{x})^2&=(7 - 8)^2+(13 - 8)^2+(6 - 8)^2+(5 - 8)^2+(9 - 8)^2\\ &=(-1)^2+5^2+(-2)^2+(-3)^2+1^2\\ &=1 + 25+4 + 9+1\\ &=40 \end{align*}$$

\]
\[

$$\begin{align*} \sum_{i = 1}^{5}(y_i-\bar{y})^2&=(27 - 25)^2+(24 - 25)^2+(30 - 25)^2+(24 - 25)^2+(20 - 25)^2\\ &=2^2+(-1)^2+5^2+(-1)^2+(-5)^2\\ &=4 + 1+25+1+25\\ &=56 \end{align*}$$

\]
\(\sqrt{\sum_{i = 1}^{5}(x_i-\bar{x})^2\sum_{i = 1}^{5}(y_i-\bar{y})^2}=\sqrt{40\times56}=\sqrt{2240}\approx47.33\)

Step4: Calculate the correlation coefficient \(r\)

\(r=\frac{\sum_{i = 1}^{5}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{5}(x_i-\bar{x})^2\sum_{i = 1}^{5}(y_i-\bar{y})^2}}=\frac{-19}{47.33}\approx - 0.401\)

Answer:

\(-0.401\)