what is the correlation coefficient, r, for this data?\n|x|y|\n|35|7|\n|40|8|\n|46|4|\n|46|9|\n|47|10|\nround…

Explanation:

Step1: Calculate the means

Let $x = [35,40,46,46,47]$, $y=[7,8,4,9,10]$. $\bar{x}=\frac{35 + 40+46+46+47}{5}=\frac{214}{5}=42.8$ $\bar{y}=\frac{7 + 8+4+9+10}{5}=\frac{38}{5}=7.6$

Step2: Calculate the numerator and denominator components

Let $n = 5$. For the numerator: [ \begin{align*} \sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})&=(35 - 42.8)(7-7.6)+(40 - 42.8)(8 - 7.6)+(46-42.8)(4 - 7.6)+(46-42.8)(9 - 7.6)+(47-42.8)(10 - 7.6)\ &=(-7.8)\times(-0.6)+(-2.8)\times0.4 + 3.2\times(-3.6)+3.2\times1.4+4.2\times2.4\ &=4.68-1.12-11.52 + 4.48+10.08\ &=6.6 \end{align*} ] For the denominator: [ \begin{align*} \sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\sqrt{\sum_{i = 1}^{n}(y_i - \bar{y})^2}&=\sqrt{(35 - 42.8)^2+(40 - 42.8)^2+(46-42.8)^2+(46-42.8)^2+(47-42.8)^2}\sqrt{(7 - 7.6)^2+(8 - 7.6)^2+(4 - 7.6)^2+(9 - 7.6)^2+(10 - 7.6)^2}\ &=\sqrt{(-7.8)^2+(-2.8)^2+3.2^2+3.2^2+4.2^2}\sqrt{(-0.6)^2+0.4^2+(-3.6)^2+1.4^2+2.4^2}\ &=\sqrt{60.84 + 7.84+10.24+10.24+17.64}\sqrt{0.36+0.16 + 12.96+1.96+5.76}\ &=\sqrt{106.8}\sqrt{21.2}\ &\approx10.33\times4.60\ &\approx47.52 \end{align*} ]

Step3: Calculate the correlation coefficient

$r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\sqrt{\sum_{i = 1}^{n}(y_i - \bar{y})^2}}=\frac{6.6}{47.52}\approx0.1$

Answer:

$0.1$