Fox Module 17: Hat values practice problem
**Exercise 17.1: Hat values
A statistician regresses the nine Y values on the nine X values.
Y | 6.87 | 6.58 | 7.69 | 6.96 | 7.39 | 14.14 | 7.90 | 15.62 | 13.23 |
X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
A. What are the hat values at each point?
B. What is the minimum hat value in any regression equation?
C. What is the maximum hat value in any regression equation?
Part A: The hat values are shown in the table below.
# Pts | 9 | Deviance | Deviance Squared | Hat Value |
values | 1 | -4 | 16 | 0.378 |
| 2 | -3 | 9 | 0.261 |
| 3 | -2 | 4 | 0.178 |
| 4 | -1 | 1 | 0.128 |
| 5 | 0 | 0 | 0.111 |
| 6 | 1 | 1 | 0.128 |
| 7 | 2 | 4 | 0.178 |
| 8 | 3 | 9 | 0.261 |
| 9 | 4 | 16 | 0.378 |
ave/tot | 5 | 0 | 60 | 0.222 |
Each hat value is: (Fox page 245) .
Part B: If xj = , the hat value is 1/n; this is the minimum.
Part C: Suppose the explanatory variable has N independent points, of which (N-1) are zero and 1 is N.
The points xj = 0 have hat values of 1/n + 1/n2 / ( (n-1)/n2 + (n-1)2/n2 ) = 1/n + 1 / ( (n-1) + (n-1)2 ).
As n ➝ ∞, the hat values ➝ 1/n.
The point xj = N has a hat value of 1/n + (n-1)2/n2 / ( (n-1)/n2 + (n-1)2/n2 ) = 1/n + (n-1)2 / ( (n-1) + (n-1)2 )
= 1/n + (n-1) / n = 1.
As n ➝ ∞, the hat value ➝ 1.