Question

Sir Francis Galton, in the late 1800s, was the first to introduce the statistical concepts of regression and correlation. He studied the relationships between pairs of variables such as the size of parents and the size of their offspring Data similar to that which he studied are given below, with the variable x denoting the height (in centimeters) of a human father and the variable y denoting the height at maturity (in centimeters) of the father's oldest son. The data are given in tabular form and also displayed in the Figure 1 scatter plot. Height of father, Height of son, y (in centimeters) (i centimeters) 161.0 183.2 172.6 176.5 189.0 175.4 201.0 172.9 161.1 201.7 182.7 155.6 187.1 191.4 191.6 190.7 174.1 176.4 178.1 192.3 181.3 166.5 189.5 188.6 175.3 187.5 195.4 189.3 185.9 180 160 150 160 1 180 190 200 210 Figure 1 The least-squares regression line for these data has a slope of approximately 0.50 Answer the following. Carry your intermediate computations to at least four decimal places, and round your answers as specified below. (If necessary, consult a list of formulas.) What is the value of the y-intercept of the least-squares regression line for these data? Round your answer to at least two decimal places What is the value of the sample correlation coefficient for these data? Round your answer to at least three decimal places

Answer 1

Given in the question

Slope of the equation =0.5

Intercept of line can be calculated as

a = (summation(Y) - b*Summation(X))/n

Here summation (Y) = 2900.5

Summation(X) = 2893.5

a= (2900.5-(2893.5*0.5))/16=(2900.5-1446.75)/16 = 1453.75/16 = 90.85

So regression equation line is

Y=90.85+0.5*X

Coefficient of correlation = (N*summation(XY) - summation (X)*Summation(Y) / Sqrt(((n*Summation(X^2) -(summation(x^2))((n*summaion(y^2) - (summation(y))^2))

= (16*525983.06)-(2893.5*2900.5)/Sqrt((16*526151.99)-(2893.5*2393.5))((16*526920.59)-(2900.5*2900.5)))

=23132.21/28666.008 = 0.807