Question

(a) draw a scatter diagram of the data. choose the correct graph below.
(b) by hand, compute the correlation coefficient.
the correlation coefficient is ( r = 0.977 ). (round to three decimal places as needed.)
(c) determine whether there is a linear relation between ( x ) and ( y ).
because the correlation coefficient is positive and the absolute value of the correlation coefficient, 0.977, is greater than the critical value for this data set, linear relation exists between ( x ) and ( y ).
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the means

Let \(x = [2,4,6,6,7]\) and \(y=[4,8,10,14,20]\)
The mean of \(x\), \(\bar{x}=\frac{2 + 4+6+6+7}{5}=\frac{25}{5} = 5\)
The mean of \(y\), \(\bar{y}=\frac{4+8+10+14+20}{5}=\frac{56}{5}=11.2\)

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 \(i = 1\): \((2 - 5)(4-11.2)=(-3)\times(-7.2) = 21.6\)
• For \(i = 2\): \((4 - 5)(8 - 11.2)=(-1)\times(-3.2)=3.2\)
• For \(i = 3\): \((6 - 5)(10 - 11.2)=(1)\times(-1.2)=-1.2\)
• For \(i = 4\): \((6 - 5)(14 - 11.2)=(1)\times(2.8)=2.8\)
• For \(i = 5\): \((7 - 5)(20 - 11.2)=(2)\times(8.8)=17.6\)
• \(\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})=21.6+3.2-1.2 + 2.8+17.6=44\)
• Calculate \((x_{i}-\bar{x})^{2}\):
• For \(i = 1\): \((2 - 5)^{2}=9\)
• For \(i = 2\): \((4 - 5)^{2}=1\)
• For \(i = 3\): \((6 - 5)^{2}=1\)
• For \(i = 4\): \((6 - 5)^{2}=1\)
• For \(i = 5\): \((7 - 5)^{2}=4\)
• \(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=9 + 1+1+1+4=16\)
• Calculate \((y_{i}-\bar{y})^{2}\):
• For \(i = 1\): \((4 - 11.2)^{2}=(-7.2)^{2}=51.84\)
• For \(i = 2\): \((8 - 11.2)^{2}=(-3.2)^{2}=10.24\)
• For \(i = 3\): \((10 - 11.2)^{2}=(-1.2)^{2}=1.44\)
• For \(i = 4\): \((14 - 11.2)^{2}=(2.8)^{2}=7.84\)
• For \(i = 5\): \((20 - 11.2)^{2}=(8.8)^{2}=77.44\)
• \(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=51.84+10.24 + 1.44+7.84+77.44=148.8\)

Step3: Calculate the correlation coefficient \(r\)

\(r=\frac{44}{\sqrt{16\times148.8}}=\frac{44}{\sqrt{2380.8}}\approx\frac{44}{48.8}\approx0.902\) (Rounded to three decimal places)

Answer:

\(r = 0.902\)