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Suppose I am measuring the time it takes to complete tasks, and I am looking across a year period, say 2019. At the end of the year, I will report on the median time taken across all tasks. Assume that all the tasks started on January 1st, and that I don't know the underlying distribution of task time $T_i$.

Suppose that thru June 50 tasks were completed, each with known time $T_i$ for $i \in \{1,...,50\}$ I know that there will be 70 more tasks completed this year, and that each has a known forecasted completion date, which provides estimates $T_{i,forecast}$ for $i \in \{51,...,120\}$.

Taking all 120 samples, I can get an estimate of the median for the year. I am wondering how I might provide a confidence/prediction interval around that estimate.

Note that I believe I can take the median of the first 50 samples and use the nonparametric confidence interval for the median for its confidence iterval, as mentioned here. Would that be valid?

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  • $\begingroup$ That would not be valid, because you are not estimating the median of independent samples from a distribution. The data used to make the forecasts and the method of forecasting are both important determinants of any estimate of the median, along with any models you might have for task completion time. $\endgroup$ – whuber Jun 6 at 16:58
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Well, any estimation of a confidence interval that you can make will rely on some assumption about the variance (or any other dispersion measure) of task durations.

Your approach will be OK (it could even be fine to assume normal distribution and work just with variance, although I have nothing against your preference for a non-parametric confidence interval) as long as there is no reason to suspect that the variance of the remaining tasks behaives in a wildly different way with respect to the previous observations.

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  • $\begingroup$ thanks, David. I think I'm clear on the Note part. Will likely use binomial model as described here. But I'm still wondering about my first question, whether I can provide a confidence interval around my estimate of the median across ALL 120 samples. Any thoughts? $\endgroup$ – Jason Jun 6 at 16:00

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