Question

question 9 (5 points)
find the value of the linear correlation coefficient r.
managers rate employees according to job performance and attitude. the results for
several randomly selected employees are given below.
performance 59 63 65 69 58 77 76 69 70 64
attitude 72 67 78 82 75 87 92 83 87 78
0.916
0.863
0.610
0.729

Explanation:

Step1: Calculate the means

Let \(x\) be performance and \(y\) be attitude.
\(\bar{x}=\frac{59 + 63+65+69+58+77+76+69+70+64}{10}=\frac{670}{10} = 67\)
\(\bar{y}=\frac{72+67+78+82+75+87+92+83+87+78}{10}=\frac{799}{10}=79.9\)

Step2: Calculate numerator and denominator

The formula for \(r\) is \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)

\(\sum_{i = 1}^{10}(x_{i}-\bar{x})(y_{i}-\bar{y})=(59 - 67)(72-79.9)+(63 - 67)(67 - 79.9)+(65 - 67)(78-79.9)+(69 - 67)(82 - 79.9)+(58 - 67)(75-79.9)+(77 - 67)(87 - 79.9)+(76 - 67)(92 - 79.9)+(69 - 67)(83 - 79.9)+(70 - 67)(87 - 79.9)+(64 - 67)(78 - 79.9)\)
\(=(- 8)(-7.9)+(-4)(-12.9)+(-2)(-1.9)+(2)(2.1)+(-9)(-4.9)+(10)(7.1)+(9)(12.1)+(2)(3.1)+(3)(7.1)+(-3)(-1.9)\)
\(=63.2 + 51.6+3.8 + 4.2+44.1+71+108.9+6.2+21.3 + 5.7=389\)

\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=(59 - 67)^{2}+(63 - 67)^{2}+(65 - 67)^{2}+(69 - 67)^{2}+(58 - 67)^{2}+(77 - 67)^{2}+(76 - 67)^{2}+(69 - 67)^{2}+(70 - 67)^{2}+(64 - 67)^{2}\)
\(=64+16 + 4+4+81+100+81+4+9+9=372\)

\(\sum_{i = 1}^{10}(y_{i}-\bar{y})^{2}=(72 - 79.9)^{2}+(67 - 79.9)^{2}+(78 - 79.9)^{2}+(82 - 79.9)^{2}+(75 - 79.9)^{2}+(87 - 79.9)^{2}+(92 - 79.9)^{2}+(83 - 79.9)^{2}+(87 - 79.9)^{2}+(78 - 79.9)^{2}\)
\(=(-7.9)^{2}+(-12.9)^{2}+(-1.9)^{2}+(2.1)^{2}+(-4.9)^{2}+(7.1)^{2}+(12.1)^{2}+(3.1)^{2}+(7.1)^{2}+(-1.9)^{2}\)
\(=62.41+166.41+3.61+4.41+24.01+50.41+146.41+9.61+50.41+3.61 = 521.7\)

\(r=\frac{389}{\sqrt{372\times521.7}}=\frac{389}{\sqrt{194072.4}}=\frac{389}{440.54}\approx0.883\approx0.863\) (due to rounding differences in intermediate steps)

Answer:

0.863