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I have a response variable that obtains different confidence intervals (CI) when calculated with different explanatory variables. I want to add up all values of the response variable and create a CI for the sum.

I would understand how to do if someone please helped me solve the following example from a triathlon where time (minutes) is the response variable, and distance (km) and discipline are the explanatory variables:

  • The 95% CI for swimming 1.5 km is 40 to 50 minutes
  • The 95% CI for cycling 40 km is 60 to 80 minutes
  • The 95% CI for running 10 km is 30 to 40 minutes

Q: Between how long does it take a person to complete the triathlon with a 95% CI?

[If it makes any difference, I assuming normal distribution and independence between disciplines]

Thank you

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  • $\begingroup$ Are these really confidence intervals? The concept does not usually apply to the results of a single individual: it pertains to estimates of a population. What, then, do these intervals actually mean and how have you computed them? This is a crucial point because it's likely the intervals are based on correlated data, which requires a different method of combination than if they were independent intervals. $\endgroup$
    – whuber ♦
    Commented Jul 20, 2016 at 16:03
  • $\begingroup$ @whuber Thank you for your comment, I have changed the question so it considers a population instead. Unfortunately i don't know how the CI were calculated. $\endgroup$
    – Williamstat
    Commented Jul 22, 2016 at 12:25
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    $\begingroup$ Thank you for the clarification. Since now it's likely the data are (highly) correlated--a good athlete typically has lower times and a poorer athlete has higher times--then the root-sum-of-squares method is sure to overstate the CI. $\endgroup$
    – whuber ♦
    Commented Jul 22, 2016 at 13:55

2 Answers 2

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In short:

  • Take as central point of your confidence interval the sum of central points of every confidence interval (45+70+35=150 minutes).
  • Take as radius of your interval the square root of the sum of the squares of the radius of every confidence interval $\sqrt{5^2+10^2+5^2}=12.25$

Therefore, a person does that triathlon in between 137.75 and 162.25 minutes with 95% probability. Anyway, beware of assumptions.

In long:

I assumed normal distribution and independence between disciplines, although first assumption may be reasonable as a rough approximation but second assumption is likely false, because I would expect that people performing well in one discipline are likely to perform well on the other ones (for example, I would expect myself to perform poorly in every discipline in a triathlon).

Assuming that times in every discipline is a normal variable, total time is just the sum of three random variables, and therefore normally distributed. Variance of the sum is also the sum of variances of the three variables, and since intervals radius is proportional to the square root of variance, you can just sum the squares of radius of every interval to get the square of the radius of the sum variable.

However, please notice that the (dubious) assumption that times for each discipline are independent narrows the resulting interval - I would say, unrealistically narrows it.

We could make the opposite assumption, that is that times for disciplines are absolutely correlated (that is, roughly, that the person who swam in 40 minutes is the same that cycled in 60 minutes and ran in 30 minutes). That assumption is probably as unrealistic as the assumption of independence was, but surely not a lot more unrealistic.

In this assumption, the radius of intervals just sum, and the triathlon is expected to be completed in between 130 and 170 minutes by 95% of athletes.

In the end, we should expect the real interval to be somewhere between [137.25,162.25] and [130,170] (both unrealistic extreme cases), but to give a more accurate result we would need to know (at least) what is the correlation between times in different discipline.

Edit after reviewing the answer a few years later: The assumption I made that results in different disciplines are likely positively correlated is reasonable if the sample includes people with different levels of fitness. However, if the sample only includes people with similar overall level in triathlon - for example, triathletes who took part in the 2020 Olympic Games - correlation between disciplines might be negative. Anyway, since assuming negative correlation yields smaller confidence intervals (or even zero length intervals), in case of lack of information about correlation I would take the conservative assumption that correlation is somewhere between 0 and 1. End of edit.

Edit about terminology

As Whuber points in his comments, it's not clear what is the meaning of the intervals given in the question. Although, the answer is valid interpreting the resulting intervals in the same way of the intervals in the question.

The two reasonable meanings of the intervals of the question are:

  • Intervals of confidence about the mean of each sport.
  • Or intervals containing the times of 95% of participants on each sport.

In spite of the wording of the question fitting better the second meaning (and hence the wording in my answer), the name "confidence interval" is usually not used with this meaning.

However, since individual times follow a normal distribution (according to assumption made in the question) and estimations of means also follow a normal distribution (if sample size is large enough or if we keep sticking to the assumption that individual times are normally distributed), the arithmetic of intervals is the same for both meanings and therefore the results given hold for both meanings.

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  • $\begingroup$ The final confidence interval could have zero length: it's plausible that the times are strongly negatively correlated and that this particular athlete might have consistent total times but variable split times during the race. Thus, we can't say much of anything until we understand just what these intervals actually are and how they were computed. $\endgroup$
    – whuber ♦
    Commented Jul 20, 2016 at 16:05
  • $\begingroup$ If we assumed correlation +1 between swimming and running and correlation -1 between other times, the interval would be zero length. Anyway, I think we can reasonably assume that correlation between times are somewhere 0 and 1, and I took those values as extreme values, but for some other variables in other problems, intervals of sum could be smaller than for each individual variable. $\endgroup$
    – Pere
    Commented Jul 20, 2016 at 16:34
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    $\begingroup$ I don't think we will need to assume anything. Because these are "response" variables, what we need to do is find out how the CIs were computed. (Likely they are based on a common set of predictor variables.) Then we will be in a better position to make reasonable recommendations about how to combine them, rather than having to speculate. $\endgroup$
    – whuber ♦
    Commented Jul 20, 2016 at 16:52
  • $\begingroup$ @Pere Thank you for a very comprehensive response. Unfortunately i don't know how the CIs were calculated. but I can assume that the correlation will be between 0 and +1, so I will use your example with the two extremes. $\endgroup$
    – Williamstat
    Commented Jul 22, 2016 at 12:50
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    $\begingroup$ I understand your point that since we do know what the intervals given in the question mean, we can't tell what the intervals in the result mean. But since, according to the question "A group of people swims 1.5 km in between 40 and 50 minutes with a 95% CI", whatever it means, the answer should be "The group does that triathlon in between 137.75 and 162.25 minutes with a 95% CI", with the same meaning. Anyway, whatever it means, I agree that it is a bizarre way to express such a meaning. $\endgroup$
    – Pere
    Commented Jul 22, 2016 at 19:09
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Pere has given a good answer. Adding up of variances is what happens when you add distributions. However in your case, he has assumed the radius as the standard deviation. but in fact, it is 1.96 times the standard deviation. So you will need to divide the radius (pere's terminology) by 1.96 (95% conf int) and then square, sum and take sq-root.

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  • $\begingroup$ Please beware that I'm not assuming the radius to be the standard deviation. Dividing by 1.96 (with additional assumptions) would be useful to find the standard deviation, but to find the radius of the CI for the sum, you need to square the radius, sum and take the square root. That is, the square root of the sum of squared standard deviations is the standard deviation of the sum, and doing the same operations on the radius yields the radius of the sum - which is what is the question is about. $\endgroup$
    – Pere
    Commented May 6, 2022 at 18:30
  • $\begingroup$ The original CI radius (i.e. 5, 10 and 5) are nothing but 1.96*std-deviation of the distribution they came from.... So, you will need to divide them by 1.96 and then square and sum before taking the square root again. i.e. SQ_ROOT((5/1.96)^2 + (10/1.96)^2 +(5/1.96)^2) will be the standard deviation of the added distributions..... And you need to multiply this by 1.96 again to get the radius of the final CI..... And that happens to be the same number you have arrived at.... :) So I just arrived this by a more purer way :) $\endgroup$
    – Jigidi Sarnath
    Commented May 17, 2022 at 15:12

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