In: Statistics and Probability
Problem (multiple regression). Following data are taken from textbook (p.614)
Student |
Math-proficiency X1 |
SAT-Math X2 |
Calculus final (college 1st year) Y |
1 |
72 |
462 |
71 |
2 |
96 |
545 |
92 |
3 |
68 |
585 |
72 |
4 |
86 |
580 |
82 |
5 |
70 |
592 |
74 |
6 |
73 |
516 |
71 |
7 |
91 |
638 |
100 |
8 |
75 |
615 |
87 |
9 |
76 |
596 |
81 |
V= 56331/9 = 6259 S=sqrt6259 = 79.1
The least-square equation is Y=-26.6+0.777X1+0.082 X2, that is :
Calculus final=-26.6+0.777 *math-proficiency +0.082* Sat-Math
95% Confidence Interval | |||
X1 | X2 | lower | upper |
70 | 592 | 71.359 | 81.206 |
SS | |
Residual | 97.3230 |
Total | 848.8889 |
The regression equation is:
y = -26.6155 + 0.7763*x1 + 0.0820*x2
R2 = 0.885
88.5% of the variability in the model is explained.
Source | SS | df | MS | F | p-value |
Regression | 751.5659 | 2 | 375.7829 | 23.17 | .0015 |
Residual | 97.3230 | 6 | 16.2205 | ||
Total | 848.8889 | 8 |
The hypothesis being tested is:
H0: β1 = β2 = 0
H1: At least one βi ≠ 0
The p-value is 0.0015.
Since the p-value (0.0015) is less than the significance level (0.05), we can reject the null hypothesis.
Therefore, we can conclude that the model is significant.
The calculations are:
R² | 0.885 | |||||
Adjusted R² | 0.847 | |||||
R | 0.941 | |||||
Std. Error | 4.027 | |||||
n | 9 | |||||
k | 2 | |||||
Dep. Var. | Y | |||||
ANOVA table | ||||||
Source | SS | df | MS | F | p-value | |
Regression | 751.5659 | 2 | 375.7829 | 23.17 | .0015 | |
Residual | 97.3230 | 6 | 16.2205 | |||
Total | 848.8889 | 8 | ||||
Regression output | confidence interval | |||||
variables | coefficients | std. error | t (df=6) | p-value | 95% lower | 95% upper |
Intercept | -26.6155 | |||||
X1 | 0.7763 | 0.1465 | 5.300 | .0018 | 0.4179 | 1.1347 |
X2 | 0.0820 | 0.0270 | 3.039 | .0228 | 0.0160 | 0.1481 |
Predicted values for: Y | ||||||
95% Confidence Interval | 95% Prediction Interval | |||||
X1 | X2 | lower | upper | lower | upper | |
70 | 592 | 71.359 | 81.206 | 65.266 | 87.299 |