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Supposed I have only this data point available:

Concentration = (1.1 - 2.0 g/L).

Is it acceptable to conclude :

(Theoretical) average = ((1.1 + 2.0) * 0.5) = 1.55 g/L?

Or would that considered to be statistically incorrect by a reviewer of a scientific thesis (in a medical field)?

EDIT 3: Complete rephrase: If I have only 2 data points out of a range, and nothing else, could I declare this as the only available data and therefore, calculate an arithmetic mean out of it?
Like: "I have a range which represents two samples, the lowest one and the highest one. I don't know how many samples were in between. Therefore, I assume there were just those two samples, and calculate the average of those two. And add a note that this average value has been calculated out of a range, and should be considered as less reliable than others with more samples".

EDIT 4: whubers comment on the question is what I was looking for: Mid-range. I don't have enough points to upvote the other answers, sorry. If whuber will write an answer instead of a comment I will mark it as correct.

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    $\begingroup$ Acceptable to whom? Applying what criteria? What counts as "statistically incorrect" for this question? $\endgroup$ – Glen_b Jul 23 '15 at 11:39
  • $\begingroup$ Glen_b, I edited my question. I hope I could explain better what I am getting at. $\endgroup$ – poshtad Jul 23 '15 at 12:20
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    $\begingroup$ The average you are interested in would lie between the average of the lower bounds of the intervals and the average of the higher bounds of the intervals. That information seems more complete to communicate what you actually know. $\endgroup$ – kasterma Jul 23 '15 at 14:28
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    $\begingroup$ Edit 3 can be made legitimate simply by replacing the word "average" by "midrange." You might report it as "The midrange of the samples is 1.55 g/L." $\endgroup$ – whuber Jul 23 '15 at 15:06
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    $\begingroup$ THANKS!! I found the wikipedia entry, this is exactly what I was looking for. Please add this as an answer, I will mark it. $\endgroup$ – poshtad Jul 23 '15 at 17:50
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The average calculation will be correct if the underlying distribution is symmetric and the endpoints of your interval were chosen by the same criterion.

For example, the above calculation is correct if the distribution of your concentration is normal and the interval refers to the interquartile range.

The calculation will not in general be true if the distribution is asymmetric such as a log-normal distribution.

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  • $\begingroup$ I am not sure wether the distribution is symmetric or not, I just have this range from a different source. Would it be acceptable to show an average, if I mark it as "calculated by this formula:.."? By I acceptable I mean: "I could not find an average and only have this range, so I show you at least a calculated average since no other data is available." $\endgroup$ – poshtad Jul 23 '15 at 12:19
  • $\begingroup$ Given that you have access to the range, you could just show the range. Alternatively, can you contact the source and ask how they define their range? $\endgroup$ – Till Hoffmann Jul 23 '15 at 12:40
  • $\begingroup$ Unfortunately no. I have a large number of those ranges, they come from different sources and are sometimes a few years old. Sometimes I have an average, sometimes just a range. If I could create a hypothetical average from those ranges, I could average all data points together. The result doesn't have to be exact, it's supposed to give you an idea where an average result could be. But if the whole idea of calculating an average from a range is fundamentally wrong, I'd rather exclude those datapoints. I'm sorry if this description is confusing,I guess it's because I am :) $\endgroup$ – poshtad Jul 23 '15 at 13:26
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In the ordinary setting where you actually observe the values in your sample the standard approach would be to report a confidence interval on the mean of the distribution. Since we don't know the actual values here and only have upper and lower bounds, a possible alternative might be to construct an interval which "covers" the confidence interval you would have constructed had you known the true values of the observations. If we're willing to assume we have an adequate sample to appeal to the central limit theorem this might look like $$ (\bar{x}_L - z \cdot s / \sqrt{n}, \bar{x}_U + z \cdot s / \sqrt{n}) $$ where $\bar{x}_L$ and $\bar{x}_U$ are the averages of the lower and upper bounds, $n$ is the number of pairs of bounds we have, $s$ the sample standard deviation of the sample comprised of all the endpoints, and $z$ a quantile from the standard normal distribution. It should be possible to show that this interval will contain the "true" interval based on the actual data points, and so have proper coverage probability if the standard interval does.

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  • $\begingroup$ Thank you very much. I am trying to apply this to the data I submitted...but I think I would need at least an additional standard deviation, right? $\endgroup$ – poshtad Jul 23 '15 at 14:31
  • $\begingroup$ You would only calculate a single standard deviation based on the sample of all the endpoints. This can be thought of as an upper bound on the standard deviation that would have been calculated from the original data, but I can't think of a proof of this off the top of my head. $\endgroup$ – dsaxton Jul 23 '15 at 14:41
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Having only information on range of your data makes it hard to make conclusions about mean, but it is still possible. As a matter of fact, it is even possible to make some educated guesses about population mean given a single data point. In this kind of cases, you have to make some assumptions about distribution of your data. Let's say that you could assume that your data comes from Normal distribution, with unknown parameters $\mu$ and $\sigma$. Using Bayesian approach you could choose some prior distribution for those parameters, take samples from those distributions, assess the likelihood of your data given those parameters, and so, infer about the parameters. In this case, you could use Approximate Bayesian Computation. For example, you could look for such parameters of Normal distribution that make 95% of values fit the interval of your interest. Looking for exact match seems to be overtly strict in here, so let's assume some margin of error, say $\pm$2%. Below I post an R code that illustrates the case.

x <- c(1.1, 2.0)        # data
crit <- c(0.025, 0.975) # 95% coverage criteria

# function to simulate a single value
simf <- function(crit) {
  mu <- rnorm(1, 1.5, 0.5)                # sampling mu
  sigma <- runif(1, 0, 2)                 # sampling sigma
  p <- pnorm(x, mu, sigma)                # checking coverage
  c(accept = all(abs(p - crit) <= 0.02),  # acceptance
    mu = mu,
    sigma = sigma)
}

sim <- t(replicate(n = 1e6, simf(crit))) # simulate
sim_accepted <- sim[sim[,1] == 1, -1]    # take only accepted values

t(apply(sim_accepted, 2, function(v) c(mean = mean(v),
                                       sd = sd(v),
                                       quantile(v, c(0.025, 0.975)))))
##            mean         sd      2.5%     97.5%
## mu    1.5500938 0.03816268 1.4775592 1.6229234
## sigma 0.2173743 0.01868164 0.1843378 0.2537986

mean(x)
## [1] 1.55

library(ggplot2)

ggplot(as.data.frame(sim_accepted), aes(x=mu, y=sigma)) +
  geom_point(color = "lightgray") +
  geom_density2d() +
  geom_vline(xintercept = x, color = "red") +
  theme_bw()

Simulation result

Using this approach you could also try fitting different distributions (e.g. non-symmetric ones) or make different prior assumptions. As you can see, using symmetric distribution such as Normal, leads to pretty the same result as with taking an arithmetic mean of the points, but it will not be the case with non-symmetric distributions as Till Hoffmann already noted in his answer. This is basically similar way of thinking as in dsaxton's answer, but using only the two data points for the interval and with using a simulation.

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  • $\begingroup$ Thank you very much for your effort. I feel so sorry that "mid-range" is what I was looking for but weren't able to articulate. $\endgroup$ – poshtad Jul 27 '15 at 12:03

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