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I am a health economist. I often look at information on costs, which is generally assumed to come from a Gamma distribution because costs are constrained to be zero or positive. Typically this information is taken from the literature, where a point estimate is reported with a range around this estimate (a max and a min, say 50% larger, and 50% less than the point estimate).

I would like to calculate the standard error (se) value when I come across the mean and this range, as typically reported. The following post: https://math.stackexchange.com/questions/2873763/is-it-possible-to-determine-shape-and-scale-for-a-gamma-distribution-from-a-mean

Suggests:

se <- ((Maximum) - (Mean)) / 2
se           

This post didn't provide the intuition for this approach, but I assume that because the central limit theorem states that the distribution of a sample mean will approach a normal distribution regardless of the population distribution, I can assume a normal distribution of the sample mean m such that the confidence interval is CI = m ± t*SE. Thus, this formula can be solved for the SE: CIupper = m + t*SE ---> SE = (CIupper-m)/t. While the t-quantile can be looked up for the level of confidence when the total sample size (n)/the number of coefficients in the model (k) are known, for large n (and 95% CI) the quantile approaches 2.0, which is why I assume the post suggests a division by 2.

(Above formula and rearrangement comes from a post by Jochen Wilhelm here: https://www.researchgate.net/post/Formula_for_calculate_Standard_errorSE_from_Confidence_IntervalCI)

To investigate this approach for calculating the standard error I use:

Goldstein, D. A., Chen, Q., Ayer, T., Howard, D. H., Lipscomb, J., Harvey, R. D., ... & Flowers, C. R. (2014). Cost effectiveness analysis of pharmacokinetically-guided 5-fluorouracil in FOLFOX chemotherapy for metastatic colorectal cancer. Clinical colorectal cancer, 13(4), 219-225. https://www.sciencedirect.com/science/article/abs/pii/S1533002814000978

In the attached Table they report the max, min and average cost values. Crucially, they report the gamma distribution shape and scale values for these cost parameters. This lets me test the standard error I create when following the above approach, by entering this into the methods of moments, along with the mean supplied in the paper, and determining if I get the same shape and scale values.

I can replicate almost all of their results as follows using the R programming language:

Maximum <- SOMEVALUEHERE
Mean <- SOMEVALUEHERE

se <- ((Maximum) - (Mean)) / 2
se                                  

mean <-Mean
mean

mn.cIntervention <- mean ## mean cost of intervention
se.cIntervention <- se ## standard error of cost of intervention

a.cIntervention <- (mn.cIntervention/se.cIntervention)^2 ## alpha value for cost of intervention (shape)
b.cIntervention <- (se.cIntervention^2)/mn.cIntervention ## beta value for cost of intervention (scale)

a.cIntervention
b.cIntervention

Bar the first row, when I generate the standard error as per the above my resulting shape and scale values are identical to those reported by the author down to the decimal point, so I conclude my standard error is calculated appropriately.

For the first row they report:

Administration cost MEAN: 284.77 MIN: 177.70 MAX: 375.44 gamma(28.295, 10.064), whereas with the above method I get: gamma(39.45674, 7.217271).

Knowing that CI = m ± t*SE, I assume that the formula can be solved for the SE: as CIupper = m + t*SE ---> SE = (CIupper-m)/t, but also as CIlower = m - t*SE ---> SE = (m-CIlower)/t.

Thus, I also assume that where the difference between the upper interval and the mean, and the lower interval and the mean wasn't symmetric, the authors checked which of what they call the max or the min (per attached Table) was further away, and chose to calculate the standard error using the lower interval (minimum) as SE = (m-CIlower)/t, because this was further away and thus incorporated more of the variability in the SE.

My questions is thus, is my approach to calculating a standard error for a gamma distribution correct where only the mean and a min/max range about this mean is reported?

Attached Table 4 as image, per comments:

Table 4 from referenced publication

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    $\begingroup$ Maybe I'm not reading carefully enough but are you considering Cllower the same as min and CLupper the same as max ? And if you're given the confidence interval for the mean (as opposed to a prediction interval for a single observation), do you know the sample size used to calculate the interval? $\endgroup$
    – JimB
    Commented Oct 29, 2021 at 22:59
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    $\begingroup$ Hi @JimB good question, yes I am considering the Cllower the same as min and CLupper the same as max. When given the confidence interval for the mean (as opposed to a prediction interval for a single observation), the sample size used to calculate the interval is rarely reported. I attach an image of the Table from the study I quote which reflects the information that is typically supplied in the literature (albeit the shape and scale are generally not included, only the mean, min and max). $\endgroup$ Commented Oct 31, 2021 at 10:40

1 Answer 1

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The minimum and maximum values in that table relate to the values:

$$\mu \pm 2 \sigma $$

where $$\begin{array}{}\mu = k \theta \\ \sigma = \sqrt{k\theta^2}\end{array}$$ and $k$ and $\theta$ are the parameters for the $\gamma$ distribution.

there is a little typo in the table, with the values $k=28.295$ and $\theta = 10.064$ you should get $391.83$ as the maximum. But the formula matches the rest, so I guess that this is a typing error.


In the article they write:

Utilities were varied over their 95% confidence intervals.

It is not very clear how they did that exactly. They used 10 000 samples in the sensitivity analysis and should have roughly 500 samples outside the min and max values. Possibly they used some sort of truncated distribution.


Your approach is correct. And using the $t$ value instead of the factor $2$ (which was a simplification) is even better.

However,

I would like to note that, while those values in the table happen to correspond with $\pm 2\sigma$, the minimum and maximum values do not generally follow such simple formula with mean plus-minus some standard deviation.

In this case, the minimum and maximum values only correspond to the interval $\mu \pm 2 \sigma$ because the distribution seems to have been truncated at those values.

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    $\begingroup$ I've figured out a simple test for whether the max/min reported is ±2σ: se <- ((Max) - (Mean)) / 2 MaxMatch <- Mean + 2*se MinMatch <- Mean - 2*se I can then check if the max/min reported in a Table match the above, and if so I know that the max/min reported is ±2σ. But, in situations where studies report just a point estimate, how could I construct a max/min ±2σ? My R thought process is something like: Mean<- SOMEVALUEHERE Max <- Mean + 0.95*Mean Min <- Mean - 0.95*Mean Because approx 95% of data fall within two standard deviations of the mean. Thank you @Sextus Empiricus. $\endgroup$ Commented Nov 4, 2021 at 13:20
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    $\begingroup$ Ok, I've done some trial and error and when I do the following: Max <- Mean + 0.20*Mean and Min <- Mean - 0.20*Mean I can replicate the max and the min values just from the mean reported the Table, but I'm not sure why this works? Is this a standard formula? If you have the mean, you can multiply it by this percentage and add or subtract this to get a min and a max that is ±2σ? $\endgroup$ Commented Nov 7, 2021 at 11:42
  • $\begingroup$ I took the @SextusEmpiricus point that the min/max values correspond to the interval μ±2σ as indicating that a confidence interval must be symmetric about the mean to generate the se using the above method. However, as I have shown in the question for Row 1 (administration costs) although the 95% CI wasn't symmetric about the mean, when calculating the se using the interval furthest from the mean I got the same shape/scale as reported in the Table. Does this mean that I can also apply the above method for a non-symmetric CI, provided I use the furthest interval to calculate the standard error? $\endgroup$ Commented Nov 10, 2021 at 18:03

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