# Regression Analysis – Standardized Residuals

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- This topic has 3 replies, 3 voices, and was last updated 15 years, 2 months ago by BeenThereDoneThat.

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- December 9, 2004 at 4:44 pm #37802
The company I work for purchases tubing from a third party vendor. This tubing is used on medical device sets, and therefore OD and ID dimensions are small (0.137 OD X 0.100 ID). This tubing is coiled in spools of 3500 ft each. The problem we are having is that there is a “feeling” by the mfg department and the planning department that we are not quite receiving 3500 ft in each spool.

I set out to determine if we could determine spool length indirectly through it’s weight. I cut the entire length of one spool into 1 foot sections (a VERY cumbersome job, by the way) and taped them together in groups of ten. Then I designed a randomized plan to obtain weights from predetermined lengths. Once the data was plotted in Minitab, the results were spectacular: there was an Rsq value of 100%! The p values for both the constant and the predictor (weight) was 0.0000. Everything looked peachy…that is until I looked at the residuals. The histogram of the standardized residuals resembles a lognormal or exponential distribution (definetely NOT normal).

Does this make sense? Here’s the Minitab output of the analysis:

Regression Analysis: Ft of Tubing versus Weight

The regression equation is

Ft of Tubing = – 16.7 + 271 Weight

Predictor Coef SE Coef T P

Constant -16.748 1.935 -8.65 0.000

Weight 271.239 0.296 916.67 0.000

S = 5.66740 R-Sq = 100.0% R-Sq(adj) = 100.0%

Analysis of Variance

Source DF SS MS F P

Regression 1 26989349 26989349 840280.60 0.000

Residual Error 42 1349 32

Total 43 26990698

Unusual Observations

Ft of

Obs Weight Tubing Fit SE Fit Residual St Resid

6 0.5 130.00 116.70 1.81 13.30 2.48R

7 0.2 50.00 33.97 1.89 16.03 3.00R

44 12.0 3260.00 3245.99 2.01 14.01 2.65R

R denotes an observation with a large standardized residual.0December 9, 2004 at 5:33 pm #112127ALMIR,The first thing that jumps out at me is the equation you got:

Ft of Tubing = – 16.7 + 271 WeightThat means that if you put nothing on the scale, it predicts -16.7 ft of tubing! The constant should be zero if you are going to have any chance of correctly predicting lengths. Look at the three measurements that were farthest off. You cut 50 ft, but it predicted 34, which is off by that -16,7 ft! When you cut 130 ft, it predicted 116, again off by about -16.7. I suspect the tape might be part of the problem. Why use tape at all? The weight of the tape itself will throw off your estimates of length. Each foot of tubing must be around 5 grams, so even a bit of tape could be a problem.Tim F0December 9, 2004 at 5:43 pm #112128I thought about the tape too. Thanls!

0December 9, 2004 at 7:03 pm #112129

BeenThereDoneThatParticipant@BeenThereDoneThat**Include @BeenThereDoneThat in your post and this person will**

be notified via email.If you have a long section of tubing (say 100 ft) and THEN cut it into 100 1 foot sections, the distribution of lengths will be non-normal. The lengths are not independent: for every section that is a bit too long, you have generated one that is a bit too small. These distributions will be skewed, but still does not explain the non-normal appearance of your residuals.I’m not sure how you can have the residuals look log normal or exponential, those both require data to be positive. If your residuals have been made positive by taking the absolute value, then your residuals will follow a half-normal plot. This will appear log-normal.

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