How to improve a regression model without making new measurements?

Question

Assuming I can't make more measurements, how can I improve a linear model regression?

I have the following data:

u    time(ms)
10    13.54
20    41.49
30    52.50
40    67.71
50   101.13
60   112.85

70   133.13
80   152.68
90   167.11
100   196.53

p-value: 3.718e-10
F-statistic: 1310 on 1 and 8 degrees of freedom
Multiple R-squared: 0.9939
Adjusted R-squared: 0.9932

Min:    -6.7411
1Q:     -2.2888
Median: -0.5126
Max:     7.0639

(Intercept): -3.9729 (estimate); 3.36107 (std. error); -1.182 (t value); 0.271 (Pr > |t|)
u:  1.96072 (estimate); 0.05417 (std. error); 36.197 (t value); 3.72e-10 (Pr > |t|)

Command in R-tool: > summary(lm(time ~ u))

I'm using linear model and R-tool gave me the equation -3,9729 + 1,96072x. Of course to x = 2 my model is not good because time cannot be negative. So how can I improve it?

If you give us your whole data set and tell us what the measurements are, we might be able to say something. $\qquad$ — Michael Hardy, Jul 06 '16 at 00:57

score 0 · Accepted Answer · edited Apr 13 '17 at 12:44

0

You're facing an issue because you're using an OLS regression when your outcome variable isn't continuous on the range of real numbers. In your case, this is because the DV must be non-negative; another example the DV is a probability and thus must be between 0 and 1. One solution for this is to take the natural log of time. You also might want to check out the answer for this question: Negative values in predictions for an always-positive response variable in linear regression

edited Apr 13 '17 at 12:44

Community

1

answered Jul 06 '16 at 03:30

robinson.es

56
3

Are you saying that a logarithmic transformation would be a good option? – eightShirt Jul 06 '16 at 03:34
Make a histogram of the log transform of time. If that looks normal, go ahead and do the log transformation – robinson.es Jul 07 '16 at 01:25

How to improve a regression model without making new measurements?

1 Answers1