Answer to Question #213192 in Statistics and Probability for Manasa Reddyrajula

Question

Find coefficient of correlation of the following data. Use this data to estimate y when x =

12.

X

3

6

7

10

Y

12

18

25

22

Answer 1

The given data is

Regression coefficient of y on x

$b_{yx}=\dfrac{n\sum xy-\sum x.\sum y}{n\sum x^2-(\sum x)^2}$

$=\dfrac{4\times 539-26\times77}{4\times 194-(26)^2}\\[9pt]=\dfrac{154}{100}=1.54$

Regression coefficient of x on y

$b_{xy}=\dfrac{n\sum xy-\sum x.\sum y}{n\sum y^2-(\sum y)^2}$

$=\dfrac{4\times 539-26\times 77}{4\times 1577-(77)^2}\\[9pt]=\dfrac{154}{379}=0.406$

Correlation coefficient $r=\sqrt{b_{yx}\times b_{xy}}=\sqrt{1.54\times 0.406}=\sqrt{0.62}=0.787$

Here, $\bar{x}=\dfrac{\sum x}{n}=\dfrac{26}{4}=6.5$

$\bar{y}=\dfrac{\sum y}{n}=\dfrac{77}{4}=19.25$

Regression line of y on x is

$y-\bar{y}=b_{yx}(x-\bar{x})$

$\Rightarrow y-19.25=1.54(x-6.5)$

Value of y when x=12 is

$y-19.25=1.54(12-6.5)\\\Rightarrow y=8.47+19.25=27.72$