[From Bruce Abbott (950117.1145 EST)]

Just to make Rick Marken happy, here is a Minitab multiple regression analysis

of the THREECV1 data. The data analyzed are from the same run as in my first

post but consist of every fifth row (as opposed to the first 1800 rows used

before). This sampling was done to provide sufficient workspace for Minitab

to conduct the regression analysis. Here are the results:

The regression equation is

H = - 150 + 0.426 C1 + 0.501 C2 - 0.456 C3

Predictor Coef Stdev t-ratio p

Constant -150.13 74.42 -2.02 0.044

C1 0.42566 0.01829 23.27 0.000

C2 0.50106 0.01919 26.12 0.000

C3 -0.4558 0.2334 -1.95 0.051

s = 33.13 R-sq = 69.9% R-sq(adj) = 69.7%

Thus, movement of the three cursors can account for nearly 70% of the

variation in handle position. Interestingly, the three cursors "loaded" about

equally (at least in the raw weights).

What follows is provided to allow comparison of the results using every fifth

row of the full data set to those using the first 1800 rows as was presented

in the previous post.

Here is the correlation matrix:

C1 C2 C3

C2 0.264

C3 0.026 0.073

H 0.641 0.685 0.013

And here is the regression of handle position on cursor 3 position:

The regression equation is

H = - 48 + 0.152 C3

Predictor Coef Stdev t-ratio p

Constant -48.3 135.2 -0.36 0.721

C3 0.1516 0.4234 0.36 0.720

s = 60.25 R-sq = 0.0% R-sq(adj) = 0.0%

This regression shows that essentially none of the variation in handle

position can be accounted for by variation in cursor 3 position.

The disturbance acting on C3 was computed by subtracting the handle position;

handle position was then regressed on the disturbance:

The regression equation is

H = 317 - 0.993 D3

Predictor Coef Stdev t-ratio p

Constant 317.129 1.062 298.48 0.000

D3 -0.993445 0.003271 -303.69 0.000

s = 5.296 R-sq = 99.2% R-sq(adj) = 99.2%

The correlation between H and D3 was -0.996.

And here are the descriptive stats:

N MEAN MEDIAN TRMEAN STDEV SEMEAN

C1 720 319.01 306.00 320.20 70.04 2.61

C2 720 319.13 316.00 319.89 66.93 2.49

C3 720 319.22 320.00 319.42 5.31 0.20

H 720 0.08 1.00 -0.99 60.21 2.24

D3 720 319.14 320.00 320.16 60.37 2.25

FULL DATA SET VERSUS EVERY FIFTH OBSERVATION

The PCT analysis on the full data set was compared to the same analysis on the

set consisting of every fifth row, with the following result:

FULL FIFTH FULL FIFTH

H-C: C1 0.6417 0.6414 K 0.0881 0.3319

C2 0.6849 0.6850 RMS 3.9 4.0

C3 0.0122 0.0134 H-M 0.9979 0.9978

The only parameter to change much was K. Obviously there is a lot of

redundancy in consecutive points; every fifth point gives nearly identical

results (except for K) to those obtained with the full data set.

Regards,

Bruce