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I have a problem that each customer gives us a revenue and a cost, one of the indicators of the company is the:

$\frac{\sum Costs}{\sum Revenue}$

To estimate the confidence interval for this parameter can I assume it's a mean and use the t or z distributions? (After checking for minimum sample size, normality, etc).

Roger Danilo Figlie
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    You can always just bootstrap it. Resample customers and calculate the statistic for each bootstrap sample, then use the quantiles of the bootstrap distribution as the confidence interval bounds. No need to think too deeply about how to compute the confidence interval for a ratio of two dependent means. – Noah Nov 13 '23 at 23:19
  • I think doing bootstrap as Noah suggeted is a good idea. You cannot assume to have a normal distribution for this ratio in general, e.g. suppose costs are fixed and revenue comes from a small number of customers only. But maybe if you can make some simplifications. E.g. Since you divide total costs by revenues it looks like customers pay a fixed or at least somewhat regular amount to the company and cause costs in a less regular/ frequent way. Maybe you can fix the total revenue and focus on the distribution of the costs if most of the variation comes from those. – ChrisL Nov 13 '23 at 23:49
  • If the revenue is fixed and costs per customer are iid with finite variance and mean then you can apply the central limit theorem to find that the ratio is approx normal. This might / should also be true if revenues are positive and vary little in comparison to costs. – ChrisL Nov 14 '23 at 00:04
  • Note that such a ratio is not necessarily normally distributed, see https://en.wikipedia.org/wiki/Ratio_distribution . If it is heavy tailed, bootstrap might still work, see https://stats.stackexchange.com/questions/425441/bootstrap-confidence-interval-on-heavy-tailed-distribution. If you want to compare means of two groups using bootstrap, see https://stats.stackexchange.com/questions/92542/how-to-perform-a-bootstrap-test-to-compare-the-means-of-two-samples . – ChrisL Nov 22 '23 at 16:44

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