Question

what is the r - value of the following data, to three decimal places?
a. - 0.816
b. 0.816
c. - 0.903
d. 0.903

Explanation:

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

The mean of \(x\) values: \(\bar{x}=\frac{4 + 5+8+9+13}{5}=\frac{39}{5} = 7.8\)
The mean of \(y\) values: \(\bar{y}=\frac{23+12 + 10+9+2}{5}=\frac{56}{5}=11.2\)

Step2: Calculate the numerator and denominator of the correlation formula

The numerator \(n\sum xy-\sum x\sum y\):
\(\sum xy=(4\times23)+(5\times12)+(8\times10)+(9\times9)+(13\times2)=92 + 60+80+81+26=339\)
\(n\sum xy-\sum x\sum y=5\times339-(39\times56)=1695 - 2184=-489\)

The denominator \(\sqrt{(n\sum x^{2}-(\sum x)^{2})(n\sum y^{2}-(\sum y)^{2})}\)
\(\sum x^{2}=4^{2}+5^{2}+8^{2}+9^{2}+13^{2}=16 + 25+64+81+169=355\)
\(n\sum x^{2}-(\sum x)^{2}=5\times355-(39)^{2}=1775-1521 = 254\)
\(\sum y^{2}=23^{2}+12^{2}+10^{2}+9^{2}+2^{2}=529+144+100+81+4=858\)
\(n\sum y^{2}-(\sum y)^{2}=5\times858-(56)^{2}=4290 - 3136=1154\)
The denominator \(\sqrt{254\times1154}=\sqrt{293116}\approx541.4\)

Step3: Calculate the correlation coefficient \(r\)

\(r=\frac{-489}{541.4}\approx - 0.903\)

Answer:

C. -0.903