A) \[y=0.91x+8.74\]
B) \[y=1.02x+8.74\]
C) \[y=1.02x-7.02\]
D) \[y=0.91x-7.02\]
Correct Answer: B
Solution :
[b] Line of regression of y on x is: |
\[y-\bar{y}={{b}_{yx}}(x-\bar{x})\] |
\[\bar{y}=\frac{\Sigma y}{n};\bar{x}\frac{\Sigma x}{n}\Rightarrow \bar{y}=\frac{220}{10}=22;\bar{x}=\frac{130}{10}=13\] |
\[{{b}_{yx}}=r.\frac{{{\sigma }_{y}}}{{{\sigma }_{x}}}\] |
\[r=\frac{n\Sigma xy-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma {{x}^{2}}-{{(\Sigma x)}^{2}}][n\Sigma {{y}^{2}}-{{(\Sigma y)}^{2}}]}}\] |
\[=\frac{10(3467)-(130)(220)}{\sqrt{[(10\times 2288)-{{130}^{2}}][(10\times 5506)-({{220}^{2}})]}}\] |
\[{{\sigma }_{y}}=\sqrt{\frac{\Sigma {{y}^{2}}}{n}-{{\left( \frac{\Sigma y}{n} \right)}^{2}}}\Rightarrow {{\sigma }_{y}}=8.2;{{\sigma }_{x}}=7.73.\] |
\[\Rightarrow {{b}_{xy}}=0.962\times \frac{8.2}{7.73}=1.02\] |
\[\Rightarrow \]Line of regression of y on x is; |
\[y-22=1.02(x-13)\] |
\[\Rightarrow y=1.02x+8.74\] |
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