Theil`s Method for Estimation of Slope and testing
  H_0:beta=beta_0 for a simple regression
See Sections 9.1-9.3 in text
 
Data: Amount of pre-packed engine lubricant and engine survival times:
(Source of data unknown.)
 
Lubricant  Survival          Lubricant Survival
   1.  29     23           11.   84    429
   2.  21     36           12.   37    489
   3.  30     67           13.   94    504
   4.  45    104           14.   56    925
   5.  48    147           15.   67    980
   6.  92    164           16.   79   1254
   7.  74    247           17.   70   2961
   8.  61    304           18.   93   5351
   9.  99    355           19.   76   9915
  10.  24    408           20.   72  11301
 
Let X=Lubricant and Y=Survival (in days)

Sorting data into one 20x3 matrix

Within-sample tie groups:  X: 0   Y: 0
NO TIES in Xs or Ys
 
Theil`s test of H_0:beta=0  (same as Kendall`s test)
(NOT assuming that X(i) are increasing, so different from text)
  C=72   Z=2.3360   P=0.0195 (Assuming no ties)

Theil`s test of H_0:beta=5  (similar to Kendall`s test)
  C=21   Z=0.6813   P=0.4957 (Assuming no ties)
 
Normal-theory test of H_0:beta=0  (Student-t test)
  beta_hat=37.9242   T=1.2883   P=0.2140  (df=18)
  (For normal-theory test, note huge outliers in Y_i.)
 
SIMPLE REGRESSION OF Y on X  (Y = beta*X + mu)
 
Theil`s estimate of beta is essentially the Hodges-Lehmann
  estimator of beta based on the Kendall (or Kendall tau) test.
That is, that value of beta such that the Kendall K statistic
  for  Y_i - beta X_i  vs  X_i  is 0 (or 1 or -1)
EXERCISE: If X_i<X_j for i<j, prove that this estimate of beta
  is the median of the ratios  (Y_j-Y_i)/(X_j-X_i) (for X_i < X_j).
  This estimator is also called Theil`s estimator of the slope beta.

The nd=190 difference ratios (Y_j-Y_i)/(X_j-X_i)
  range between -5527 (minimum) and 5187 (maximum)
The median is beta_hat=7.1050, which is Theil`s estimate
  of the slope
 
Given beta, mu=mu_hat is the median of the Walsh averages
  of Y_i - beta X_i.
The median of nw=210 Walsh averages is 179.2804
Thus Theil`s regression line in this case is
    Y = 7.1050 + 179.280
 
Theil also derived a nonparametric confidence interval
  for beta based on quantiles of the nd=190 difference ratios
Using a large-sample estimate of the quantiles,
  the nonparametric 95% Conf.Int. for beta bracketing beta_hat
  and the corresponding offsets in the slopes array are
    (1.371(65)   7.105(95.5)   28.313(126))
 
Regression line errors:
  E1 is mean of absolute values of residuals
  E2 is root-mean-square of residuals
  Theil:          Y =  7.1050 X +  179.280   E1=1614.72   E2=3354.75
  Least-Squares:  Y = 37.9242 X + -573.961   E1=2010.12   E2=3050.92
 
LET`S LOOK AT LOG(Y) on X instead of Y on X:
SIMPLE REGRESSION OF log(Y) on X  (log(Y) = beta*X + mu)
 
Regression line and errors:
  Theil:          log(Y) =  0.0340 X + 3.984   E1=1.230   E2=1.400
  Least-Squares:  log(Y) =  0.0373 X + 3.826   E1=1.231   E2=1.397
 
For log(Y_i) and X_i:
Theil`s test of H_0:beta=0  (same as Kendall`s test)
(NOT assuming that X(i) are increasing, so different from text)
  C=72   Z=2.3360   P=0.0195 (Assuming no ties)
 
Normal-theory test of H_0:beta=0  (Student-t test)
  beta_hat=0.0373   T=2.7692   P=0.0126  (df=18)
 
For log(Y_i) and X_i:
Nonparametric 95% Conf.Int. for beta bracketing beta_hat and
  the corresponding offsets in the slopes array:
  (0.0053(65)   0.0340(95.5)   0.0695(126))