Statistics-And-Probability Question 608
Question: For 10 observations on price (x) and supply (y), the following data was obtained: $ \sum{x=130,\sum{y=220,}} $ $ \sum{x^{2}=2288,\sum{y^{2}=5506}} $ and $ \sum{xy=3467} $ What is line of regression of y on x?
Options:
A) $ y=0.91x+8.74 $
B) $ y=1.02x+8.74 $
C) $ y=1.02x-7.02 $
D) $ y=0.91x-7.02 $
Show Answer
Answer:
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}-{{( \frac{\Sigma y}{n} )}^{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 $
BETA
