Question

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

x	y
5	93.4
6	92
7	86.8
8	85.2
9	77.6
10	78.8
11	73.4

r =
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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:

\(-0.966\)