The data below are the intervals in hours between failures of the air conditioning equipment in 10 Boeing 707s.	The aim was to describe concisely the  variation within and between aircraft

plane no
1	2	3	4	5	6	7	8	9	10

413	90	74	55	23	97	50	359	487	102
14	10	57	320	261	51	44	9	18	209
58	60	48	56	87	11	102	12	100	14
37	186	29	104	7	4	72	270	7	57
100	61	502	220	120	141	22	603	98	54
65	49	12	239	14	18	39	3	5	32
9	14	70	47	62	142	3	104	85	67
169	24	21	246	47	68	15	2	91	59
447	56	29	176	225	77	197	438	43	134
184	20	386	182	71	80	188	 	230	152
36	79	59	33	264	1	79	 	3	27
201	84	27	15	21	16	88	 	130	14
118	44	153	104	42	106	46	 	 	230
34	59	26	35	20	206	5	 	 	66
31	29	326	 	5	82	5	 	 	61
18	118	 	 	12	54	36	 	 	34
18	25	 	 	120	31	22	 	 	 
67	156	 	 	11	216	139	 	 	 
57	310	 	 	3	46	210	 	 	 
62	76	 	 	14	111	97	 	 	 
7	26	 	 	71	39	30	 	 	 
22	44	 	 	11	63	23	 	 	 
34	23	 	 	14	18	13	 	 	 
 	62	 	 	11	191	14	 	 	 
 	130	 	 	16	18	 	 	 	 
 	208	 	 	90	163	 	 	 	 
 	70	 	 	1	24	 	 	 	 
 	101	 	 	16	 	 	 	 	 
 	208	 	 	52	 	 	 	 	 
 	 	 	 	95	 	 	 	 	 


Minitab output gives

		N	MEAN	MEDIAN	TRMEAN	STDEV	SEMEAN	
	plane1	23	95.7	57.0	83.2	119.3	24.9	plane2	29	83.5	61.0	77.9	70.8	13.1	plane3	15	121.3	57.0	100.4	154.3	39.8	plane4	14	130.9	104.0	124.7	98.2	26.2	plane5	30	59.6	22.0	49.1	71.9	13.1	plane6	27	76.8	63.0	74.3	63.7	12.3	plane7	24	64.1	41.5	60.3	62.7	12.8	plane8	9	200.0	104.0	200.0	225.9	75.3	plane9	12	108.1	88.0	80.7	136.2	39.3	plane10	16	82.0	60.0	76.3	66.3	16.6	


	MIN	MAX	Q1	Q3	
	plane1	7.0	447.0	22.0	118.0
	plane2	10.0	310.0	27.5	109.5	
	plane3	12.0	502.0	27.0	153.0	
	plane4	15.0	320.0	44.0	224.8
	plane5	1.0	261.0	11.8	87.7
	plane6	1.0	216.0	18.0	111.0	
	plane7	3.0	210.0	16.7	94.8
	plane8	2.0	603.0	6.0	398.5	
	plane9	3.0	487.0	9.8	122.5
	plane10	14.0	230.0	32.5	126

	After some reflection it seems that plane number 8 is rather different.  It may be worth trying to fit a probability distribution of the Gamma type, that is


The _ values are for each plane
	Plane	1	2	3	4	5	6	7	8	9	10
	_	____	____	____	____	____	____	____	____	____	____	

Once again plane eight stands out with a low value .  The question then is what distributions should one fit to the data
	
	(1) separate Gammas
	(2) common Gammas
	(3) Gammas with _=1 i.e. exponentials



	The exponential fit is quit good and gives a very simple and vivid summary of the data .  In an average sense the fit is not bad but there are some unexplained variations.  If we leave out the plane number 8 then a Chi-squared goodness of fit test would accept an exponential fit
	 To test the distribution we tried a probability plot of the pooled data (less no 8)



 MTB > plot c12 c11
 
         -
      6.0+
         -
 C12     -                                                  *
         -
         -                                              *
      4.0+                                        * *
         -                                 **
         -                           *    *
         -                        3*2
         -                      52
      2.0+                 **332
         -            24332*
         -         66+
         -      6++*
         -   +++5
      0.0+ 6+5
           +---------+---------+---------+---------+---------+------C11     
           0       100       200       300       400       500
The plot looks fairly linear.!


Example 2
Cycles to failure of worsted yarn
x1:	length of test specimen (250, 300, 350mm)
x2:	amplitude of loading cycle (8, 9 ,10 mm)
x3:	load (40, 45 ,50 grm)

		x1		x2		x3	obsns

	-1	-1	-1	674
	-1	-1	0	370
	-1	-1	1	292
	-1	0	-1	338
	-1	0	0	266
	-1	0	1	210
	-1	1	-1	170
	-1	1	0	118
	-1	1	1	90
	0	-1	-1	1414
	0	-1	0	1198
	0	-1	1	634
	0	0	-1	1022
	0	0	0	620
	0	0	1	438
	0	1	-1	442
	0	1	0	332
	0	1	1	220
	1	-1	-1	3636
	1	-1	0	3184
	1	-1	1	2000
	1	0	-1	1568
	1	0	0	1070
	1	0	1	556
	1	1	-1	1140
	1	1	0	884
	1	1	1	360



	There are some argument for taking log observations here, and for taking logs of the explanatory variables x1, x2, x3  so transforming back to our original data scales

  ROW     x1    x2    x3    cyc      lcyc
 
    1    250     8    40    674   6.51323
    2    250     8    45    370   5.91350
    3    250     8    50    292   5.67675
    4    250     9    40    338   5.82305
    5    250     9    45    266   5.58350
    6    250     9    50    210   5.34711
    7    250    10    40    170   5.13580
    8    250    10    45    118   4.77068
    9    250    10    50     90   4.49981
   10    300     8    40   1414   7.25418
   11    300     8    45   1198   7.08841
   12    300     8    50    634   6.45205
   13    300     9    40   1022   6.92952
   14    300     9    45    620   6.42972
   15    300     9    50    438   6.08222
   16    300    10    40    442   6.09131
   17    300    10    45    332   5.80513
   18    300    10    50    220   5.39363
   19    350     8    40   3636   8.19864
   20    350     8    45   3184   8.06589
   21    350     8    50   2000   7.60090
   22    350     9    40   1568   7.35756
   23    350     9    45   1070   6.97541
   24    350     9    50    556   6.32077
   25    350    10    40   1140   7.03878
   26    350    10    45    884   6.78446
   27    350    10    50    360   5.88610

we can take logs and try to fit the data.  I have not plotted all the diagrams you would have made up to this point, we should ;leave some trees standing.
	Using the regression in minitab we have
 
 The regression equation is
 lcyc = 3.93 + 4.94 lx1 - 5.65 lx2 - 3.51 lx3
 
 Predictor       Coef       Stdev    t-ratio        p
 Constant       3.930       2.254       1.74    0.095
 lx1           4.9447      0.2581      19.16    0.000
 lx2          -5.6540      0.3894     -14.52    0.000
 lx3          -3.5117      0.3894      -9.02    0.000
 
 s = 0.1844      R-sq = 96.6%     R-sq(adj) = 96.2%
 
 Analysis of Variance
 
 SOURCE       DF          SS          MS         F        p
 Regression    3     22.4217      7.4739    219.77    0.000
 Error        23      0.7822      0.0340
 Total        26     23.2039
 
 SOURCE       DF      SEQ SS
 lx1           1     12.4852
 lx2           1      7.1703
 lx3           1      2.7661
 
 Obs.     lx1      lcyc       Fit Stdev.Fit  Residual   St.Resid
   1     5.52    6.5132    6.5205    0.0847   -0.0073     -0.04  
   2     5.52    5.9135    6.1069    0.0722   -0.1934     -1.14  
   3     5.52    5.6768    5.7369    0.0838   -0.0602     -0.37  
   4     5.52    5.8230    5.8546    0.0722   -0.0315     -0.19  
   5     5.52    5.5835    5.4410    0.0571    0.1425      0.81  
   6     5.52    5.3471    5.0710    0.0712    0.2761      1.62  
   7     5.52    5.1358    5.2589    0.0838   -0.1231     -0.75  
   8     5.52    4.7707    4.8453    0.0712   -0.0746     -0.44  
   9     5.52    4.4998    4.4753    0.0830    0.0245      0.15  
  10     5.70    7.2542    7.4220    0.0720   -0.1679     -0.99  
  11     5.70    7.0884    7.0084    0.0568    0.0800      0.46  
  12     5.70    6.4520    6.6384    0.0710   -0.1864     -1.10  
  13     5.70    6.9295    6.7561    0.0568    0.1734      0.99  
  14     5.70    6.4297    6.3425    0.0356    0.0872      0.48  
  15     5.70    6.0822    5.9725    0.0556    0.1097      0.62  
  16     5.70    6.0913    6.1604    0.0710   -0.0691     -0.41  
  17     5.70    5.8051    5.7468    0.0556    0.0584      0.33  
  18     5.70    5.3936    5.3768    0.0700    0.0168      0.10  
  19     5.86    8.1986    8.1843    0.0834    0.0144      0.09  
  20     5.86    8.0659    7.7706    0.0707    0.2952      1.73  
  21     5.86    7.6009    7.4007    0.0826    0.2003      1.21  
  22     5.86    7.3576    7.5183    0.0707   -0.1608     -0.94  
  23     5.86    6.9754    7.1047    0.0552   -0.1293     -0.73  
  24     5.86    6.3208    6.7347    0.0697   -0.4139     -2.42R 
  25     5.86    7.0388    6.9226    0.0826    0.1162      0.70  
  26     5.86    6.7845    6.5090    0.0697    0.2755      1.61  
  27     5.86    5.8861    6.1390    0.0817   -0.2529     -1.53  
 
 R denotes an obs. with a large st. resid.
 
 MTB > histo c19
 
 Histogram of C19   N = 27
 
 Midpoint   Count
     -2.5       1  *
     -2.0       0
     -1.5       1  *
     -1.0       4  ****
     -0.5       5  *****
      0.0       5  *****
      0.5       5  *****
      1.0       3  ***
      1.5       3  ***
 
 MTB > plot c20 c5
 
         -
 C20     -
         -                                                        *
         -                                                      *
      7.5+                                           **   *
         -                                       *
         -                                      * 2
         -                             * **   *
         -                          *    *
      6.0+                      *2  *
         -                    * *
         -            *   *  *
         -               *
         -       *
      4.5+   *
         -
         -
           --------+---------+---------+---------+---------+--------lcyc    
                4.90      5.60      6.30      7.00      7.70
 
Thus we have cycle = x14.957x2-5.651x33.501  which can be written as a useful rule 
	cycle = ()-5x3-7/2
Example 3
	A brewer approached an AFRC research worker with a question about the types of yeast in his process.   The AFRC unit did a DNA analysis on the yeasts the result of which was that they could classify the 30 yeast cells that they examined into 5 types.  The results were

	Type	Frequency
	A	14
	B	8
	C	5
	D	2
	E	1
	
The question that then arose was as follows:
	We have found 5 classes in 30 how many classes are there in the population?  Natually we have probably not seen them all.
	To answer the question we make a few assumptions such as the classes are arranged in a sensible order.  We can then try and find a distribution that fits the data.  It is fairly easy to see that if we call A,B,C,D,E 1,..5 then S fx = 58  Sfx2 = 148  and so the mean estimate is 1.93333 while the estimate of the standard deviation is 1.19556.

	The distribution that might model the data is the Negative Binomial 
	
with mean and variance .  This is not quite right because we cannot observe classes with zero numbers in them, so although the Negative Binomial may be a good fit what we see is conditional on x ­0 t hat is  p[ X = x | x> 0] .  Since 
p[ X > 0 ] = 1 - pr and p[ X = x | x> 0] =we can work out the distribution.
Then all we have to do is estimate the parameters.  A simple we is to equate the mean and variance of the new distribution with that of the sample values and solve the equations
	so   = 1.93333   and  = 
Solving gives p= 0.63 and r = 2.1.  Now for the distribution with zero classes the mean number per class is  or 1.1597 so if the number of classes is N we have 1.1597N=30 or N=25.9 giving the number of unobserved classes as 20.

1