If I have a complicated function of multivariables $f(x_1,x_2,x_3,\ldots,x_n)$, and I were to find the variance approximation through the delta method, say $\sigma^2_{approx}$, would the 95% confidence interval directly follow as:
$$
f(x_1,x_2,x_3,\ldots,x_n)\pm 1.96\cdot \sqrt{\sigma^2_{approx}}
$$
Or would there be something else I need to do on $f(x_1,x_2,x_3,\ldots,x_n)$?
Best Answer
you are missing a $\sqrt{n}$, ie
$$ f(x_1,\ldots,x_n) \pm 1.96 \sqrt{\sigma^2_{approx}/n} $$ Assuming by $\sigma^2_{approx}$ you mean $\sigma^2_{approx} = f'(x_1,\ldots,x_n)^2\sigma^2$.