Answer to Question #274993 in Statistics and Probability for V.kathiravan

Question

Find the least square regression for the following.

X: 0 1 2 3 4

Y: 2 3 5 4 6

Answer 1

From the given data of x & y

CASE 1 :

Let us consider that y is the dependent variable and x is an independent variable then the formula for the regression line is given by,

y = a + bx

where,

a = y intercept

b = slope of the line

The formula for a and b is given by,

$b = \frac{n\sum(xy)- \sum x\sum y}{n\sum(x)^2-(\sum x)^2} \ \ \ \ \ \ \ and\ \ \ \ \ a = \frac{\sum y}{n} - b\frac{\sum x}{n}$

where n = sample size = 6

from the above table we get the values of a and b as under

$b = \frac{n\sum(xy)- \sum x\sum y}{n\sum(x^2)-(\sum x)^2} = \frac{6*174 - 21*42}{6*91 - 21^2} = 1.5428$

$a = \frac{\sum y}{n} - b\frac{\sum x}{n} = \frac{42}{6} - 1.542857*\frac{21}{6} = 1.6$

Hence the equation of the regression line becomes y = 1.6 + 1.54x

So the value of y, when x = 2.5 can be found using above equation

y = 1.6 + 1.54*2.5 = 5.457

(X₁,Y₁) = (2.5 , 5.457)

CASE 2 :

Let us consider that x is the dependent variable and y is an independent variable then the formula for the regression line is given by,

x = a + by

where,

a = x intercept

b = slope of the line

The formula for a and b is given by,

$b = \frac{n\sum(xy)- \sum x\sum y}{n\sum(y)^2-(\sum y)^2} \ \ \ \ \ \ \ and\ \ \ \ \ a = \frac{\sum x}{n} - b\frac{\sum y}{n}$

where n = sample size = 6

from the above table we get the values of a and b as under

$b = \frac{n\sum(xy)- \sum x\sum y}{n\sum(y)^2-(\sum y)^2} = \frac{6*174 - 21*42}{6*336-42^2} = \frac{9}{14} = 0.64285$

$a = \frac{\sum x}{n} - b\frac{\sum y}{n} = \frac{21}{6} - 0.64285*\frac{42}{6} = -1$

Hence the equation of the regression line becomes x = -1 + 0.64y