Answer to Question #317037 in Statistics and Probability for Kiddo

Question

The table given below is for scores in Management Accounting M.A) and Quantitative Techniques (Q.T).

Student

A

B

C

D

E

F

G

H

M.A

86

77

68

71

67

90

78

71

Q.T

80

82

73

69

72

85

84

65

Test for existence of linear relationship at 5% level of significance.

Answer 1

$Pearson?s\,\,correlation\,\,coefficient:\\n=8\\\sum{x_i}=522\\\sum{y_i}=610\\\sum{{x_i}^2}=39308\\\sum{{y_i}^2}=46904\\\sum{x_iy_i}=39818\\r=\frac{n\sum{x_iy_i}-\sum{x_i}\sum{y_i}}{\sqrt{\left( n\sum{{x_i}^2}-\left( \sum{x_i} \right) ^2 \right) \left( n\sum{{y_i}^2}-\left( \sum{y_i} \right) ^2 \right)}}=\\=\frac{8\cdot 39818-522\cdot 610}{\sqrt{\left( 8\cdot 39308-522^2 \right) \left( 8\cdot 46904-610^2 \right)}}=0.0108141\\T=r\sqrt{\frac{n-2}{1-r^2}}=0.0108141\sqrt{\frac{8-2}{1-0.0108141^2}}=0.0264905\sim t_{n-2}=t_6\\P-value:\\P\left( \left| T \right|>0.0265 \right) =2F_{t,6}\left( -0.0265 \right) =2\cdot 0.48986=0.97972>0.05\Rightarrow \\\Rightarrow insignificant$