By NEAS - 6/30/2024 6:35:28 PM
MS Module 19: Correlation – practice problems
(The attached PDF file has better formatting.)
Exercise 19.1: Correlation
A regression model Yj = β0 + β1 Xj + εj has N = 11 observations. The sample correlation between X and Y is 0.60. We test the null hypothesis H0: ρ = 0 (the true correlation between the X and Y variables is zero).
A. What is the t value to test the null hypothesis?
B. What is the p value to test the null hypothesis?
Part A: The t value is R √(n-2) / √(1– R2) = 0.6 × (11 – 2)0.5 / (1 – 0.62)0.5 = 2.25000
Part B: The t distribution has n-2 degrees of freedom, so the p value for a two-tailed test is 0.051 (table look-up or spread-sheet function).
Question: The sample correlation is 0.60, which is much different from zero, yet the p value is 5.1%, which does not satisfy even a 5% significance level.
Answer: The scenario has only 11 observations. Even if the true correlation is zero, a sample with only a few observations often shows a high sample correlation.
Exercise 19.2: Correlation and β1
A linear regression with 11 data points has an estimated β1 of 4.5 and a sample correlation between the X and Y values of 0.60.
A. What is the t value to test the null hypothesis that the correlation ρ is zero?
B. What is the t value to test the null hypothesis that β1 is zero?
C. What is the standard deviation of the estimate of β1?
Part A: The t value to test the null hypothesis that the correlation ρ is zero is r √(n-2) / √(1– r2) =
0.6 × (11 – 2)0.5 / (1 – 0.62)0.5 = 2.25000
Part B: The t value to test the null hypothesis that β1 is zero is 1 / σ(1) = 4.5 / σ(1), where
● 1 is the estimate of β1.
● s(1) is the standard deviation of the estimate of β1.
Part C: The two tests are the same: ρ = 0 implies β1 = 0 and vice versa.
4.5 / s(1) = 2.25 ➾ s(1) = 4.5 / 2.25 = 2.
|
|