How to calculate overall confidence intervals from geometric means?

[Sal Watts]

Hello,
I have two geometric means (both of which are average antibody concentrations for a group of subjects calculated by multiplying all values and taking the nth root of this number, where n is the number of subjects with available data).

Group 1: Mean = 8985, confidence intervals = 5705–14150
67 people

Group 2: Mean = 7622, confidence intervals = 5486–10591
132 people

Overall mean: 8056.

Please could someone advise how I calculate the overall 95% confidence intervals?
Many thanks

[William Lisowski]

Welcome to Statalist.

The output of the Stata command help geometric points us to the ameans command.

. sysuse auto, clear
(1978 Automobile Data)

. ameans price

    Variable |    Type             Obs        Mean       [95% Conf. Interval]
-------------+---------------------------------------------------------------
       price | Arithmetic           74    6165.257        5481.914     6848.6
             |  Geometric           74    5656.907        5165.664   6194.865
             |   Harmonic           74    5296.672        4928.901    5723.75
-----------------------------------------------------------------------------

. by foreign, sort: ameans price

------------------------------------------------------------------------------------------------
-> foreign = Domestic

    Variable |    Type             Obs        Mean       [95% Conf. Interval]
-------------+---------------------------------------------------------------
       price | Arithmetic           52    6072.423        5210.184   6934.662 
             |  Geometric           52    5533.564        4945.587   6191.446 
             |   Harmonic           52    5171.918         4749.29   5677.111 
-----------------------------------------------------------------------------

------------------------------------------------------------------------------------------------
-> foreign = Foreign

    Variable |    Type             Obs        Mean       [95% Conf. Interval]
-------------+---------------------------------------------------------------
       price | Arithmetic           22    6384.682         5222.19   7547.174 
             |  Geometric           22    5959.483        5063.614   7013.853 
             |   Harmonic           22    5616.916        4904.397   6571.657 
-----------------------------------------------------------------------------

[Clyde Schechter]

-help means-. N.B. -means-, not -mean-.

[William Lisowski]

Note the common thread in posts #2 and #3: Google is not the only search engine available; Stata's built-in help and search commands do a good job of finding results that are relevant to Stata users without having to sort through answers for SPSS or SAS or R users.

[Sal Watts]

Hi thank you. Is this possible to do with the information I have provided?
I don’t have every individual score of the 67 participants in group one, or the individual scores of the 132 participants in group 2. I just have their mean scores from which I have calculated the overall mean. But I do not know the formula for calculating overall CIs.

i haven’t inputted my data into Stata I just inputted the above numbers as

di

to calculate the overall mean.

[William Lisowski]

Is this possible to do with the information I have provided?

The mean and variance of a set of numbers x_i can be expressed in terms of the sum of the x_i and the sum of the x_i². The usual way of calculating geometric means and their confidence intervals is to calculate z_i = ln(x_i), then calculate the arithmetic mean and confidence interval for the z_i, and then exponentiate each of these to get the geometric mean and confidence intervals.

So from what you show, it should be possible to

• start with the within group geometric means and confidence intervals for the x_i
• take logs to get the within group arithmetic means and confidence intervals for the z_i = log(x_i)
• derive from these arithmetic means and confidence intervals the within group sums of the z_i and the z_i²
• add the sums from the two groups to get the overall sums of the z_i and the z_i²
• use these sums to calculate the arithmetic means and confidence intervals of the z_i and the z_i²
• exponentiate the results to give the overall geometric mean and confidence interval for the x_i.

[Sal Watts]

Okay I got this to calculate the mean:

di  67*log(8985) + 132*log(7622)
777.318812122

di 777.318812122/199
3.90612468403

di exp(3.90612468403)
8056.09666003

Can you advise do I calculate the overall confidence intervals in exactly the same way?

i.e

67*log(5705)+132*log(5486)..... etc etc

(With group 1 confidence intervals = 5705–14150
67 people

Group 2: confidence intervals = 5486–10591
132 people)

[William Lisowski]

Can you advise do I calculate the overall confidence intervals in exactly the same way?

No, that's not what the results of the technique in post #6 are. In that post, I gave an abstract of the steps someone with a reasonable basic knowledge of mathematical statistics would go through to calculate the confidence intervals. I didn't know your background and didn't want to assume you wouldn't perhaps be saying "oh, yes, of course." But it appears what I wrote may be beyond your background in statistics.

I'm not a statistician so I generally spend my time on Statalist answering Stata programming questions. Your question is more about statistical methodology – the only connection to Stata is using the display command as a pocket calculator – so perhaps one of the more methodological members can help further with the math to give you the formulas you need.