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Dear users,

The CI of a ratio of two samples is often difficult to calculate even so Fieller has published a method that applies mostly to large normal samples. I have write a small piece of code to calculate it.

'Two samples data'
  COPY (16 14 13 17 15) numerator
  COPY (32 28 29 30 31) denominator
'For information'
  MEAN numerator Mnumerator
  MEAN denominator Mdenominator
  DIVIDE Mnumerator Mdenominator Ratio
  PRINT Mnumerator Mdenominator Ratio
'Bootstrap'
  COPY 10000 rptCount
  REPEAT rptCount
    SAMPLE 5 numerator numBootstrap
    SAMPLE 5 denominator denBootstrap
    MEAN numBootstrap MnumBS
    MEAN denBootstrap MdenBS
    DIVIDE MnumBS MdenBS ratios
    SCORE ratios ratio
    SCORE MnumBS Mnum
    SCORE MdenBS Mden
  END
'results'
  MEAN Mnum MBSnum
  MEAN Mden MBSden
  MEAN ratio Mratio
  STDEV ratio SDratio
  PERCENTILE ratio (2.5 97.5) CIratio
  PRINT MBSnum MBSden Mratio SDratio CIratio
  HISTOGRAM ratio

You can easily adapt this code to dependent samples modifying the Sample syntax :
    SAMPLE 5 1,5 chooser
    TAKE numerator chooser numBootstrap
    TAKE denominator chooser denBootstrap

Take care to verify these codes. I am not a specialist.

I am also interested to complete these two examples with the CI of correlated samples... This is the case when a trend exists into each sample i.e. (17 16 15 14 13) / (32 31 30 29 28).
In such case, the trend adds variability to the samples, and increases artificially the CI of the ratio. Taking into accound the previous data, there is in fact none variability other than the linear trend.

Currently I am unable to build the code to directly calculate the CI with Statistics101. I suppose that a basic approach is to remove the trend before the calculation of the bootstrap ratio :
- determine the pivot point of the sample (the mean?)
- Calculate the parameters of the trend Y = ax + b
- rotate horizontally the regression line, so that the variability of the sample is reduced (practically, it means Y' = Y - (Ymean - Ypredicted).
- calculate the CI of the ratio as the ratio of two normal independant samples.

I will be pleased to received advices or commentaries from specialists.
Sincerely.

I'm not a statistician and I'm not sure exactly what you are trying to do, but if you haven't done so already, you might look into these chapters of Julian Simon's book, where he gives examples of confidence intervals and correlated data:
http://www.resample.com/content/text/24-Chap-20.pdf
http://www.resample.com/content/text/25-Chap-21.pdf
http://www.resample.com/content/text/27-Chap-23.pdf

Given your example of correlated data, it might be enough to simply compute the ratios and bootstrap from there.
For example:

'Since the two vectors are correlated, all we really have
'for samples is five ratios. They happen to be in order in the beginning,
'but when we SAMPLE them, the new samples will not be in order,
'and may have duplicates.
DIVIDE (17 16 15 14 13) (32 31 30 29 28) ratios
PRINT ratios

COPY 1000 rptCount
REPEAT rptCount
   SAMPLE 5 ratios sample
   MEAN sample meanSample
   SCORE meanSample meanSamples
END
PERCENTILE meanSamples (2.5 97.5) CIRatio
PRINT CIRatio
HISTOGRAM meanSamples