EXAMPLE 1  -  CANCER MORTALITY RATES FOR 9 OREGON COUNTIES

			x - index of exposure (C1)
			y - cancer mortality rate per 100,000 people (C2)
a.  SCATTERPLOT
 MTB > plot c2 c1
 
          -
       210+                                       *               *
          -
  rate    -
          -
          -
       175+                             *
          -
          -                *
          -
          -         *
       140+     *
          -          *   *
          -
          -
          -   *
       105+
            ------+---------+---------+---------+---------+---------+index   
                2.0       4.0       6.0       8.0      10.0      12.0

	There is a fairly strong positive linear association between
	the two variables.  There are no apparent outliers.

b.   MTB > regress c2 on 1 in c1;
     SUBC > predict 4;
     SUBC > predict 15.
 
 The regression equation is
 rate = 115 + 9.23 index
 
 Predictor       Coef       Stdev    t-ratio        p
 Constant     114.716       8.046      14.26    0.000
 index          9.231       1.419       6.51    0.000
 
 s = 14.01       R-sq = 85.8%     R-sq(adj) = 83.8%
 
 Analysis of Variance
 
 SOURCE       DF          SS          MS         F        p
 Regression    1      8309.6      8309.6     42.34    0.000
 Error         7      1373.9       196.3
 Total         8      9683.5


	 Fit	Stdev.Fit	    95% C.I.	     95% P.I.

	151.64	   4.75		(140.40,162.88)   (116.65,186.63)

	253.19	  15.45		(216.64,289.74)	  (203.85,302.52)

	x denotes a row with x values away from the center
	xx denotes a row with very extreme x values


   Note that the slope is positive as expected from the scatterplot

c.	The slope = 9.23  This means that according to the regression
 equation for a one unit increase in ind4ex of exposure on average
 cancer mortality increases by 9.23 per 100,000.	

d.	According to the regression equation when index of exposure
  equals 4, the cancer mortality rate equals 115 + 9.23(4) = 151.92    
  (using the Minitab ouput  fit = 151.64)

e.      According to the regression equation when index of exposure
  equals 15, the cancer mortality rate equals 115 + 9.23(15) = 253.45  
  (using the Minitab ouput  fit = 253.19)  Note that there are two xx beside
  the output for  "predict 15". This indicates that we are extrapolating.
  The x value that we are predicting at is much larger than the largest value
  of x in the data set. The relationship between x and y described by the
  regression line may not hold for values of x as high as 15.

f.   MTB > regress c2 on 1 in c1 c3 c4;
     SUBC> resids c5.
 
 The regression equation is
 rate = 115 + 9.23 index
 
 Predictor       Coef       Stdev    t-ratio        p
 Constant     114.716       8.046      14.26    0.000
 index          9.231       1.419       6.51    0.000
 
 s = 14.01       R-sq = 85.8%     R-sq(adj) = 83.8%
 
 Analysis of Variance
 
 SOURCE       DF          SS          MS         F        p
 Regression    1      8309.6      8309.6     42.34    0.000
 Error         7      1373.9       196.3
 Total         8      9683.5
 
	c3 is a column of standardized residuals
	c4 is a column of fitted values
	c5 is a column of raw residuals (ei) 

	To confirm that c5 is a column of the raw residuals (ei)
		raw residual = y - fitted value
	c2 contains the y values and c4 the corresponding fitted values so
		raw residuals = c2 - c4

 MTB > let c6 = c2-c4  
 MTB > print c1 c2 c4 c6 c5
 
  ROW   index    rate        C4        C6        C5
 
    1    2.49   147.1   137.702    9.3980    9.3980
    2    2.57   130.1   138.440   -8.3405   -8.3405
    3    3.41   129.9   146.195  -16.2949  -16.2949
    4    1.25   113.5   126.255  -12.7550  -12.7550
    5    1.62   137.5   129.671    7.8294    7.8294
    6    3.83   162.3   150.072   12.2279   12.2279
    7   11.64   207.5   222.170  -14.6698  -14.6698
    8    6.41   177.9   173.889    4.0107    4.0107
    9    8.34   210.3   191.706   18.5940   18.5940

	Note that c5 and c6 are the same (as they should be)

	Remember that for the least squares line the sum of the residuals
	equals zero.  To confirm this

 MTB > sum c5
    SUM     =-0.000022888
	  (Because of rounding errors the calculated sum is not exactly zero.) 


	TO CALCULATE THE SUM OF SQUARES DUE TO ERROR
 MTB > let c7 = c5*c5
 MTB > sum c7
    SUM     =      1373.9 ( This corresponds to the value for SS Error
				 on the ANOVA table)
g.	R-squared is the coefficient of determination and it tells us what
   proportion of the variability in y is explained by x.
	In this case,  R-squared is fairly large - 85.8% of the variablity
    in y is explained by x.

h.  MTB >  CORR C1 C2  
		Correlation of index and rate = 0.926

            (CORR C2 C1 gives the same output.)


