Air pressure as a function of elevation, p. 5

     Since air pressure is never negative, we can take its
     logarithm (to base e). I did a simple linear regression
     of that on elevation in feet and got the following output:

     - - - - - - - -
     Call:
     lm(formula = I(log(mmHg)) ~ feet, data = airtable)
 
     Residuals:
          Min        1Q    Median        3Q       Max 
     -0.089660 -0.037796 -0.001926  0.034928  0.097122 

     Coefficients:
                  Estimate  Std. Error  t value  Pr(>|t|)    
     (Intercept)  6.657e+00  9.828e-03   677.4   <2e-16 ***
     feet        -4.444e-05  2.896e-07  -153.4   <2e-16 ***
     ---
     Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

     Residual standard error: 0.05165 on 38 degrees of freedom
     Multiple R-Squared: 0.9984,	Adjusted R-squared: 0.9983 
     F-statistic: 2.354e+04 on 1 and 38 DF,  p-value: < 2.2e-16 

     - - - -

     Here the multiple R-squared of 0.9984 or 0.9983 is the largest
     of the three we've seen, quite close to 1, and the model has the
     advantage that it has just two parameters, as opposed to 3 for
     the quadratic regression. The size of the residuals isn't
     directly comparable to the previous ones, we'd have to take
     the exponential of log p  to get  p  and then see how large
     the errors in mmHg would be. Anyhow, let's look on the
     next page at the residuals from this regression for
     a pattern.