Question

find the correlation coefficient $r$ of the data below. round your answer to the nearest hundredth. show your work here

Explanation:

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

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\) and \(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}\).
For \(x\) - values: \(x_1 = 2.8,x_2=-5.5,x_3=-0.2,x_4=-6.4,x_5=-4.6,x_6=-7.7\)
\(\sum_{i=1}^{6}x_{i}=2.8+( - 5.5)+(-0.2)+(-6.4)+(-4.6)+(-7.7)=2.8-5.5 - 0.2-6.4-4.6-7.7=-21.6\)
\(\bar{x}=\frac{-21.6}{6}=-3.6\)
For \(y\) - values: \(y_1=-9.8,y_2=-3.3,y_3=-2.7,y_4=-2.7,y_5 = 12.5,y_6=4.7\)
\(\sum_{i=1}^{6}y_{i}=-9.8+( - 3.3)+(-2.7)+(-2.7)+12.5 + 4.7=-18.5 + 17.2=-1.3\)
\(\bar{y}=\frac{-1.3}{6}\approx - 0.22\)

Step2: Calculate the numerator and denominator of the correlation coefficient formula

The formula for the correlation coefficient \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)
Calculate \((x_{i}-\bar{x})(y_{i}-\bar{y})\):
For \(i = 1\): \((2.8-( - 3.6))(-9.8-( - 0.22))=(6.4)(-9.58)=-61.312\)
For \(i = 2\): \((-5.5-( - 3.6))(-3.3-( - 0.22))=(-1.9)(-3.08) = 5.852\)
For \(i = 3\): \((-0.2-( - 3.6))(-2.7-( - 0.22))=(3.4)(-2.48)=-8.432\)
For \(i = 4\): \((-6.4-( - 3.6))(-2.7-( - 0.22))=(-2.8)(-2.48)=6.944\)
For \(i = 5\): \((-4.6-( - 3.6))(12.5-( - 0.22))=(-1)(12.72)=-12.72\)
For \(i = 6\): \((-7.7-( - 3.6))(4.7-( - 0.22))=(-4.1)(4.92)=-20.172\)
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})(y_{i}-\bar{y})=-61.312 + 5.852-8.432 + 6.944-12.72-20.172=-90.84\)

Calculate \((x_{i}-\bar{x})^{2}\):
For \(i = 1\): \((2.8-( - 3.6))^{2}=(6.4)^{2}=40.96\)
For \(i = 2\): \((-5.5-( - 3.6))^{2}=(-1.9)^{2}=3.61\)
For \(i = 3\): \((-0.2-( - 3.6))^{2}=(3.4)^{2}=11.56\)
For \(i = 4\): \((-6.4-( - 3.6))^{2}=(-2.8)^{2}=7.84\)
For \(i = 5\): \((-4.6-( - 3.6))^{2}=(-1)^{2}=1\)
For \(i = 6\): \((-7.7-( - 3.6))^{2}=(-4.1)^{2}=16.81\)
\(\sum_{i = 1}^{6}(x_{i}-\bar{x})^{2}=40.96+3.61 + 11.56+7.84+1+16.81=81.78\)

Calculate \((y_{i}-\bar{y})^{2}\):
For \(i = 1\): \((-9.8-( - 0.22))^{2}=(-9.58)^{2}\approx91.7764\)
For \(i = 2\): \((-3.3-( - 0.22))^{2}=(-3.08)^{2}\approx9.4864\)
For \(i = 3\): \((-2.7-( - 0.22))^{2}=(-2.48)^{2}\approx6.1504\)
For \(i = 4\): \((-2.7-( - 0.22))^{2}=(-2.48)^{2}\approx6.1504\)
For \(i = 5\): \((12.5-( - 0.22))^{2}=(12.72)^{2}\approx161.7984\)
For \(i = 6\): \((4.7-( - 0.22))^{2}=(4.92)^{2}\approx24.2064\)
\(\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}=91.7764+9.4864+6.1504+6.1504+161.7984+24.2064=299.5684\)

Step3: Calculate the correlation coefficient \(r\)

\(r=\frac{-90.84}{\sqrt{81.78\times299.5684}}\)
\(81.78\times299.5684\approx81.78\times300 = 24534\) (approximate for calculation, actual \(81.78\times299.5684=81.78\times(300 - 0.4316)=24534-81.78\times0.4316\approx24534 - 35.33=24498.67\))
\(\sqrt{24498.67}\approx156.52\)
\(r=\frac{-90.84}{156.52}\approx - 0.58\)

Answer:

\(-0.58\)