title: How is 95% CI calculated using confint in R?
body: <p>I use the example provided in <code>confint</code> help page </p>
  
  <pre><code>&gt; fit &lt;- lm(100/mpg ~ disp + hp + wt + am, data=mtcars)
  &gt; summary(fit)
  
  Call:
  lm(formula = 100/mpg ~ disp + hp + wt + am, data = mtcars)
  
  Residuals:
      Min      1Q  Median      3Q     Max 
  -1.6923 -0.3901  0.0579  0.3649  1.2608 
  
  Coefficients:
              Estimate Std. Error t value Pr(&gt;|t|)   
  (Intercept) 0.740648   0.738594   1.003  0.32487   
  disp        0.002703   0.002715   0.996  0.32832   
  hp          0.005275   0.003253   1.621  0.11657   
  wt          1.001303   0.302761   3.307  0.00267 **
  am          0.155815   0.375515   0.415  0.68147   
  ---
  Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
  
  Residual standard error: 0.6754 on 27 degrees of freedom
  Multiple R-squared: 0.8527,     Adjusted R-squared: 0.8309 
  F-statistic: 39.08 on 4 and 27 DF,  p-value: 7.369e-11 
  
  &gt; confint(fit)
                     2.5 %      97.5 %
  (Intercept) -0.774822875 2.256118188
  disp        -0.002867999 0.008273849
  hp          -0.001400580 0.011949674
  wt           0.380088737 1.622517536
  am          -0.614677730 0.926307310
  &gt; confint(fit, "wt")
         2.5 %   97.5 %
  &gt; wt 0.3800887 1.622518
  
  &gt;confint.default(fit,"wt")
         2.5 %   97.5 %
  wt 0.4079023 1.594704
  
  &gt; 1.001303 + 1.96*0.302761
  [1] 1.594715
  &gt; 1.001303 - 1.96*0.302761
  [1] 0.4078914
  </code></pre>
  
  <p>So the 95% CI obtained from <code>confint.default</code> is based on asymptotic normality. What about for <code>confint</code>?</p>
  
  <p>Thanks</p>