Question

The mean unemployment rate in March of each year from 1992 to 2002 is x̄ = 5.45, with a standard deviation of s_x = 1.1. The mean unemployment rate in August of each year for the same time frame is ȳ = 5.72, with a standard deviation of s_y = 1.4. The correlation coefficient is r = 0.95.

Part A: Find the equation of the least-squares regression line for predicting August unemployment rate from March's unemployment rate. Show your work.
Part B: Use the regression line to predict the unemployment rate if March's rate is 5.2. Show your work.
Part C: Find and interpret r-squared.

Answer 1

Part A:

Slope (a) is given by:

a = r X S_y/S_x

= 0.95 X 1.4/1.1

= 1.2091

Y Intercept (b) is given by:

b = - (a X )

= 5.72 - (1.2091 X 5.45)

= 0.8695

Thus, the equation of the least-squares regression line for predicting August unemployment rate from March's unemployment rate is given by:

y = 1.2091 x + 0.8695

Part B:
For x = 5.2, we get:

y = (1.2091 X 5.2) + 0.8695

= 7.15682

So,

Answer is:

7.15682

Part C:

(i)

Given:

Correlation Coefficient (r) = 0.95

So,

Coefficient of Determination (R²) = 0.9025

So,

Answer is:

0.9025

(ii)

R² = 0.9025 = 90.25% is a statistical measure of how close the data are to the fitted regression line. R² = 90.25 % is the percentage of response variable explained by the regressor in the model.