Tom Moore asked...
----- Forwarded message from Thomas L. Moore -----
Hello,
Does anyone know of a good example of cubic regression that you'd be
willing to share?
Thanks.
----- End of forwarded message from Thomas L. Moore -----
I don't know if this is what Tom had in mind, but it is one of my
favorite datasets. I dug it out and reran my analysis for Tom, and
thought I'd share it with others as well. (The software is Minitab 11.)
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
Univariate Data Well Fit by a Cubic
MTB > notitles
MTB > gstd
* NOTE * Standard Graphics are enabled.
Professional Graphics are disabled.
Use the GPRO command to enable Professional Graphics.
MTB > note Turn back the clock to Minitab 5.1
MTB > retr 'e:\stats\minitab8\stats1a\smt15.16'
Retrieving worksheet from file: e:\stats\minitab8\stats1a\smt15.16
Worksheet was saved on 11/15/1996
MTB > print c1 c2
Row height age
1 86.500 2
2 95.500 3
3 103.000 4
4 109.800 5
5 116.400 6
6 122.400 7
7 128.200 8
8 133.800 9
9 139.600 10
10 145.000 11
MTB > note Girl's "typical" heights in cm from 1980 World Almanac
MTB > plot c1 c2
-
- *
140+ *
-
height - *
- *
- *
120+
- *
-
- *
- *
100+
- *
-
- *
-
------+---------+---------+---------+---------+---------+age
2.0 4.0 6.0 8.0 10.0 12.0
MTB > correlation c1 c2
Correlation of height and age = 0.997
MTB > note Relationship appears fairly linear
MTB > regress c1 1 c2;
SUBC> residuals in c3.
The regression equation is
height = 76.6 + 6.37 age
Predictor Coef StDev T P
Constant 76.641 1.188 64.52 0.000
age 6.3661 0.1672 38.08 0.000
S = 1.518 R-Sq = 99.5% R-Sq(adj) = 99.4%
Analysis of Variance
Source DF SS MS F P
Regression 1 3343.5 3343.5 1450.45 0.000
Error 8 18.4 2.3
Total 9 3361.9
Unusual Observations
Obs age height Fit StDev Fit Residual St Resid
1 2.0 86.500 89.373 0.892 -2.873 -2.34R
R denotes an observation with a large standardized residual
MTB > note Is this a good model? Why? Let's look at the residulals.
MTB > plot c3 c2
-
1.5+ *
- * *
C3 - *
- *
-
0.0+ *
- *
- *
-
-
-1.5+
- *
-
-
-
-3.0+ *
------+---------+---------+---------+---------+---------+age
2.0 4.0 6.0 8.0 10.0 12.0
MTB > note Looks curved to me.
MTB > note Data used and analyzed to this point in Siegel and Morgan,
MTB > note Statistics and Data Analysis: An Introduction, Wiley, 1996, pp.554-556
MTB > note This is the best example I've ever seen of the power of residual plots!-)
MTB > note However, I hate to let sleeping dogs lay (or lie).
MTB > note Would a quadratic term help?
MTB > let c4=c2*c2
MTB > name c4 'age-sqr'
MTB > regress c1 2 c2 c4;
SUBC> residuals in c5.
The regression equation is
height = 70.6 + 8.67 age - 0.177 age-sqr
Predictor Coef StDev T P
Constant 70.6133 0.8601 82.10 0.000
age 8.6706 0.2962 29.28 0.000
age-sqr -0.17727 0.02236 -7.93 0.000
S = 0.5138 R-Sq = 99.9% R-Sq(adj) = 99.9%
Analysis of Variance
Source DF SS MS F P
Regression 2 3360.0 1680.0 6362.89 0.000
Error 7 1.8 0.3
Total 9 3361.9
Source DF Seq SS
age 1 3343.5
age-sqr 1 16.6
Unusual Observations
Obs age height Fit StDev Fit Residual St Resid
1 2.0 86.500 87.245 0.404 -0.745 -2.35R
R denotes an observation with a large standardized residual
MTB > plot c5 vs. c2
-
-
0.50+ * * *
-
C5 - *
-
- *
0.00+ *
-
- *
-
- *
-0.50+ *
-
- *
-
-
------+---------+---------+---------+---------+---------+age
2.0 4.0 6.0 8.0 10.0 12.0
MTB > note Oy! It's still curved, but not a parabola.
MTB > let c6=c2*c4
MTB > name c6 'age-cube'
MTB > regress c1 3 c2 c4 c6;
SUBC> residuals in c7.
The regression equation is
height = 66.5 + 11.3 age - 0.629 age-sqr + 0.0232 age-cube
Predictor Coef StDev T P
Constant 66.4594 0.6508 102.12 0.000
age 11.2662 0.3755 30.01 0.000
age-sqr -0.62879 0.06328 -9.94 0.000
age-cube 0.023155 0.003221 7.19 0.000
S = 0.1790 R-Sq = 100.0% R-Sq(adj) = 100.0%
Analysis of Variance
Source DF SS MS F P
Regression 3 3361.7 1120.6 34976.76 0.000
Error 6 0.2 0.0
Total 9 3361.9
Source DF Seq SS
age 1 3343.5
age-sqr 1 16.6
age-cube 1 1.7
Unusual Observations
Obs age height Fit StDev Fit Residual St Resid
1 2.0 86.500 86.662 0.162 -0.162 -2.16R
R denotes an observation with a large standardized residual
MTB > plot c7 c2
0.30+
- *
C7 -
- *
-
0.15+
-
-
- *
-
0.00+ * *
- *
- *
-
- *
-0.15+ *
- *
------+---------+---------+---------+---------+---------+age
2.0 4.0 6.0 8.0 10.0 12.0
Looks much more random now.
MTB > print c1-c7
Row height age C3 age-sqr C5 age-cube C7
1 86.500 2 -2.87273 4 -0.745455 8 -0.161958
2 95.500 3 -0.23879 9 0.470303 27 0.275804
3 103.000 4 0.89515 16 0.540605 64 0.054358
4 109.800 5 1.32909 25 0.265457 125 -0.165219
5 116.400 6 1.56303 36 0.144849 216 -0.021864
6 122.400 7 1.19697 49 -0.221212 343 -0.054499
7 128.200 8 0.63090 64 -0.432732 512 -0.002056
8 133.800 9 -0.13515 81 -0.489696 729 -0.003449
9 139.600 10 -0.70121 100 0.007883 1000 0.202382
10 145.000 11 -1.66728 121 0.459998 1331 -0.123499
MTB > let c8=abs(c7)
MTB > average c8
Mean of C8 = 0.10651
MTB > note Data is given to nearest tenth, at best, so I'll stop here.
_
| | Robert W. Hayden
| | Work: Department of Mathematics
/ | Plymouth State College MSC#29
| | Plymouth, New Hampshire 03264 USA
| * | fax (603) 535-2943
/ | Home: 82 River Street (use this in the summer)
| ) Ashland, NH 03217
L_____/ (603) 968-9914 (use this year-round)
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