Question

compute the correlation coefficient.

x	20	31	23	29	7	28	-6
y	30	4	-2	18	3	36	9

send data to excel
use the ti - 84 plus calculator as needed.
the correlation coefficient is r = . round the answer to at least three decimal places.

Explanation:

Step1: Calculate the means of x and y

Let \(x = [20,31,23,29,7,28, - 6]\) and \(y=[30,4, - 2,18,3,36,9]\).
The mean of \(x\), \(\bar{x}=\frac{20 + 31+23+29+7+28+( - 6)}{7}=\frac{132}{7}\approx18.857\).
The mean of \(y\), \(\bar{y}=\frac{30 + 4+( - 2)+18+3+36+9}{7}=\frac{98}{7} = 14\).

Step2: Calculate the numerator of the correlation - coefficient formula

The numerator \(S_{xy}=\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})\).
\((20 - 18.857)(30 - 14)+(31 - 18.857)(4 - 14)+(23 - 18.857)( - 2 - 14)+(29 - 18.857)(18 - 14)+(7 - 18.857)(3 - 14)+(28 - 18.857)(36 - 14)+( - 6 - 18.857)(9 - 14)\)
\(=1.143\times16+12.143\times( - 10)+4.143\times( - 16)+10.143\times4+( - 11.857)\times( - 11)+9.143\times22+( - 24.857)\times( - 5)\)
\(=18.288-121.43-66.288 + 40.572+130.427+201.146 + 124.285\)
\(=327\).

Step3: Calculate the denominator of the correlation - coefficient formula

The standard - deviation of \(x\), \(S_{x}=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(20 - 18.857)^{2}+(31 - 18.857)^{2}+(23 - 18.857)^{2}+(29 - 18.857)^{2}+(7 - 18.857)^{2}+(28 - 18.857)^{2}+( - 6 - 18.857)^{2}\)
\(=1.143^{2}+12.143^{2}+4.143^{2}+10.143^{2}+( - 11.857)^{2}+9.143^{2}+( - 24.857)^{2}\)
\(=1.306+147.45+17.164+102.88+140.59+83.59+617.87\)
\(=1000.85\).
\(S_{x}=\sqrt{\frac{1000.85}{6}}\approx12.91\).
The standard - deviation of \(y\), \(S_{y}=\sqrt{\frac{\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}{n - 1}}\).
\(\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}=(30 - 14)^{2}+(4 - 14)^{2}+( - 2 - 14)^{2}+(18 - 14)^{2}+(3 - 14)^{2}+(36 - 14)^{2}+(9 - 14)^{2}\)
\(=16^{2}+( - 10)^{2}+( - 16)^{2}+4^{2}+( - 11)^{2}+22^{2}+( - 5)^{2}\)
\(=256 + 100+256+16+121+484+25\)
\(=1258\).
\(S_{y}=\sqrt{\frac{1258}{6}}\approx14.47\).
The denominator \(S_{x}S_{y}=12.91\times14.47\approx186.81\).

Step4: Calculate the correlation coefficient

The correlation coefficient \(r=\frac{S_{xy}}{S_{x}S_{y}}=\frac{327}{186.81}\approx0.547\).

Answer:

\(0.547\)