Confidence interval for geometric mean of fractions

Question

Is there any opportunity to create such interval where a variable ($\{\ln(X_i)\}^n_{i=1}$) is the fraction of prices for two periods?

$$ X_i = \frac{price.new_i}{price.old_i} $$

Please, look at my attempt below. Is everything correct?

Answer 1

Here is my attempt

Geometric mean: $$ ( \prod_{i=1}^{n} X_i ) ^{1 / n} = Const. $$

I use the excellent tip of Dmitrij Celov at Confidence interval for geometric mean, who stated that the geometric mean can be transformed to an arithmetic mean of logarithms (if strictly positive), we have

$$ \ln{(\prod_{i=1}^{n}X_i)^{1 / n}} = \ln{(Const.)} \\ \overline{\ln(X)} =\frac{ \sum_{i=1}^{n}\ln{X_i} }{n} \\ $$

where $\overline{\ln(X)}$ arithmetic mean of $\{\ln(X_i)\}^n_{i=1}$.

Let's construct confidence interval $$ P [\ln(X_i) \in (\overline{\ln(X)} \pm Z_{\alpha/2} \cdot \frac{sd}{ sqrt (n) })] = 95 \% $$ where

$$ sd = sqrt{\frac{\sum_{i=1}^{n}(\ln{X_i} - \overline{\ln(X)})^2}{n-1}} $$

Then for the "real" values we exponent it

$$ P [\exp(\ln(X_i) \in \exp(\overline{\ln(X)} \pm Z_{\alpha/2} \cdot \frac{sd}{sqrt (n)})] = 95 \% \Leftrightarrow \\ P [X_i \in \exp({\overline{\ln(X)}} \pm {Z_{\alpha/2} \cdot \frac{sd}{sqrt (n)}})] = 95 \% $$

Or, without mess, $$ P (\ln{X_i} \in [L; U]) = 95 \% \Leftrightarrow \\ P (X_i \in [\exp(L); \exp(U)]) = 95 \%. $$