ENM 500
Regression Exercises
0. Use regression techniques to fit a line to the points (0, 0) and (3, 4) using the format
Y
0
4
x
0
3
x2
xy
Yhat
Y-Yhat (Y-Yhat)2

Write the normal equations.
1. Given the following
X = 15
Y = 17
X2 = 55 Y2 = 77
XY = 54 n = 5
Find Sxx Sxy Syy A B SSerror s2 t( = 0) R2 SSexplained
Write y-hat and X-hat equations. B = 3/19.2 A = 3 – 3/19.2 * 17/5 = 2.46875
X-hat = 2.46875 + 0.15625y
Y-hat = 2.5 + 0.3x
2. Sketch a scatterplot and fit regression line to it using the data : x  1 3 5; y  4 9 8.
3. Use multiple linear regression techniques to fit a plane between the points given below.
Y
5
6
0
x1
0
4
1
x2
0
3
-1
x12
x22
x1x2
x1Y

Solve your linear equations and write the equation of the plane.
4. Fit a polynomial of degree 2 between three points below.
x
x2
x2
x4
x3
xY
x2 Y
2
2
Y
x1
x2
x1
x2 x1x2 x1Y x2Y
5
0
6
4
0
1

Solve your equations and write the polynomial.
x2Y
5. Demo of Example 8.25 (ex8.25)
(mapcar #' R-sq (list x1 x2 x3 x4 x5) (list-of 5 y))
(mapcar #' MSE-r (list x1 x2 x3 x4 x5) (list-of 5 y))
1. Solution for #1.X = 15 Y = 17 X2 = 55 Y2 = 77 XY = 54
Find Sxx Sxy Syy A B SSerror s2 t( = 0) R2 SSexplained
Y-Hat = 2.500 + 0.300X
Sxx = 10 Sxy = 3
SSerror = 18.3 sý = 6.1
Explained Variation =
0.9
Syy =
19.2 R-sq = 0.0469
Sum of Y-squared =
77
For ß = 0, (t = 0.384111 P-VALUE = 0.726505)
For  = 0, (t = 0.965114 P-VALUE = 0.40568)
Sa = 2.590 Sb = 0.781
(-2.168, 2.768) or 0.300 +- 2.468
95% confidence interval for Beta NIL
(-5.686, 10.686) or 2.500 +- 8.186
95% confidence interval for Alpha NIL
Residuals are:
-0.800
0.900
-1.400
3.300
-2.
The yhats are (2.8 3.1 3.4 3.7 4.0)
The b coefficients in Y's are -0.2 -0.1 0 0.1 0.2
The a coefficients in Y's are 0.8 0.5 0.2 -0.1 -0.4
(-0.091 6.891 ) or 3.400 +- 3.491
95% confidence interval for Y-Hat at x = x-bar NIL
(-5.15026 11.95026) or 3.400 +- 8.550
95% confidence interval for Y-Predict at x = x-bar NIL
F-ratio = Explain/Error/3 = 0.1475.
Source
Regression
Residual Error
Total
Analysis of Variance
SS
df
MS
0.9
1
0.9
18.3
3
6.1
19.2
4
F
P-value
0.148
0.7265