Question

for the following data set, (a) draw a scatter diagram, (b) compute the correlation coefficient, and (c) comment on the type of relation that appears to exist between x and y.

(a) which of the following scatter diagrams corresponds to the given data?

(b) compute the correlation coefficient.
r ≈ (round to three decimal places as needed.)

Explanation:

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

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\).
Given \(x = [2,5,7,7,9]\), \(n = 5\). \(\sum_{i=1}^{5}x_{i}=2 + 5+7+7+9=30\), so \(\bar{x}=\frac{30}{5}=6\).
Given \(y = [1.3,1.6,1.8,2.0,2.4]\), \(\sum_{i = 1}^{5}y_{i}=1.3 + 1.6+1.8+2.0+2.4 = 9.1\), so \(\bar{y}=\frac{9.1}{5}=1.82\).

Step2: Calculate the numerator \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\)

\((2 - 6)(1.3 - 1.82)=(-4)\times(-0.52) = 2.08\)
\((5 - 6)(1.6 - 1.82)=(-1)\times(-0.22)=0.22\)
\((7 - 6)(1.8 - 1.82)=(1)\times(-0.02)=-0.02\)
\((7 - 6)(2.0 - 1.82)=(1)\times(0.18)=0.18\)
\((9 - 6)(2.4 - 1.82)=(3)\times(0.58)=1.74\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=2.08 + 0.22-0.02 + 0.18+1.74=4.2\)

Step3: Calculate the denominator \(\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}\)

\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(2 - 6)^{2}+(5 - 6)^{2}+(7 - 6)^{2}+(7 - 6)^{2}+(9 - 6)^{2}=16 + 1+1+1+9 = 28\)
\(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=(1.3 - 1.82)^{2}+(1.6 - 1.82)^{2}+(1.8 - 1.82)^{2}+(2.0 - 1.82)^{2}+(2.4 - 1.82)^{2}\)
\(=0.2704+0.0484 + 0.0004+0.0324+0.3364=0.688\)
\(\sqrt{28\times0.688}=\sqrt{19.264}\approx4.39\)

Step4: Calculate the correlation coefficient \(r\)

The formula for the correlation coefficient \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)
\(r=\frac{4.2}{4.39}\approx0.957\)

Answer:

\(r\approx0.957\)