Anger | 6 | 7 | 5 | 21 | 13 | 5 | 13 | 14 |

Vigor | 30 | 23 | 29 | 22 | 19 | 19 | 28 | 19 |

Calculate Spearman’s rank correlation coefficient.

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Login Now**Solution:**

Anger(x) |
Vigor(y) |
Rank of x(R_{1}) |
Rank of x(R_{2}) |
d = R_{1} + R_{2} |
d^{2} |

6 | 30 | 3 | 8 | -5 | 25 |

7 | 23 | 4 | 5 | -1 | 1 |

5 | 29 | 1.5 | 7 | -5.5 | 30.25 |

21 | 22 | 8 | 4 | 4 | 16 |

13 | 19 | 5.5 | 2 | 3.5 | 12.25 |

5 | 19 | 1.5 | 2 | -0.5 | 0.25 |

13 | 28 | 5.5 | 6 | -0.5 | 0.25 |

14 | 19 | 7 | 2 | 5 | 25 |

Σd = 0 | Σd^{2} = 110 |

Now,

\(R = 1-\frac{6\left \{ \sum d^2 + \frac{m_1(m_1^2 – 1)}{12} + \frac{m_2(m_2^2 – 1)}{12} + \frac{m_3(m_3^2 – 1)}{12} \right \}}{n(n^2 – 1)}\)

Here,

m_{1} = 2 (5 is repeated 2 times)

m_{2} = 2 (13 is repeated 2 times)

m_{3} = 3 (19 is repeated 3 times)

\(= 1-\frac{6\left \{ 110 + \frac{2(2^2 – 1)}{12} + \frac{2(2^2 – 1)}{12} + \frac{3(3^2 – 1)}{12} \right \}}{8(8^2 – 1)}\)

\(= 1-\frac{6\left \{ 110 + 0.5 + 0.5 + 23 \right \}}{504}\)

= -0.35

Hence, Sperman’s rank correlation cofficient is -0.35

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