question 13 based on the data shown below, calculate the correlation coefficient (to three decimal places)…

Explanation:

Step1: Calculate the means of x and y

Let (n = 8). (\bar{x}=\frac{4 + 5+6+7+8+9+10+11}{8}=\frac{60}{8}=7.5) (\bar{y}=\frac{95.2+93.4+92+86.8+85.2+77.6+78.8+73.4}{8}=\frac{692.4}{8}=86.55)

Step2: Calculate the numerator of the correlation - coefficient formula

[ \begin{align*} \sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})&=(4 - 7.5)(95.2-86.55)+(5 - 7.5)(93.4 - 86.55)+(6 - 7.5)(92-86.55)+(7 - 7.5)(86.8-86.55)+(8 - 7.5)(85.2-86.55)+(9 - 7.5)(77.6-86.55)+(10 - 7.5)(78.8-86.55)+(11 - 7.5)(73.4-86.55)\ &=(- 3.5)\times8.65+(-2.5)\times6.85+(-1.5)\times5.45+(-0.5)\times0.25+0.5\times(-1.35)+1.5\times(-8.95)+2.5\times(-7.75)+3.5\times(-13.15)\ &=-30.275-17.125 - 8.175-0.125-0.675-13.425-19.375-46.025\ &=-135.2 \end{align*} ]

Step3: Calculate the denominator of the correlation - coefficient formula

[ \begin{align*} \sum_{i = 1}^{n}(x_i-\bar{x})^2&=(4 - 7.5)^2+(5 - 7.5)^2+(6 - 7.5)^2+(7 - 7.5)^2+(8 - 7.5)^2+(9 - 7.5)^2+(10 - 7.5)^2+(11 - 7.5)^2\ &=(-3.5)^2+(-2.5)^2+(-1.5)^2+(-0.5)^2+0.5^2+1.5^2+2.5^2+3.5^2\ &=12.25 + 6.25+2.25+0.25+0.25+2.25+6.25+12.25\ &=42 \end{align*} ] [ \begin{align*} \sum_{i = 1}^{n}(y_i-\bar{y})^2&=(95.2-86.55)^2+(93.4 - 86.55)^2+(92-86.55)^2+(86.8-86.55)^2+(85.2-86.55)^2+(77.6-86.55)^2+(78.8-86.55)^2+(73.4-86.55)^2\ &=8.65^2+6.85^2+5.45^2+0.25^2+(-1.35)^2+(-8.95)^2+(-7.75)^2+(-13.15)^2\ &=74.8225+46.9225+29.7025+0.0625+1.8225+80.1025+60.0625+172.9225\ &=466.42 \end{align*} ] (\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}=\sqrt{42\times466.42}=\sqrt{19589.64}\approx139.963)

Step4: Calculate the correlation coefficient r

(r=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i-\bar{x})^2\sum_{i = 1}^{n}(y_i-\bar{y})^2}}=\frac{-135.2}{139.963}\approx - 0.966)

Answer:

(-0.966)