Unable to get the correct answer - what am I doing wrong!

[help12]

Hello,

I have two 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).

The first mean for group 1: 8985.
Group 1 is made up of 67 subjects.

The second mean for group 2: 7622
Group 2 is made up of 132 subjects.

I have been given the overall mean is 8056. However, I do not know how this number was arrived at. I have attempted the formula:

(67*8985)+(132*7622) = 1608099 / 67+132 = 8080.8995.

I don't understand where I am going wrong???

 

[lev888]

Hello,

I have two 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).

The first mean for group 1: 8985.
Group 1 is made up of 67 subjects.

The second mean for group 2: 7622
Group 2 is made up of 132 subjects.

I have been given the overall mean is 8056. However, I do not know how this number was arrived at. I have attempted the formula:

(67*8985)+(132*7622) = 1608099 / 67+132 = 8080.8995.

I don't understand where I am going wrong???

You are adding and dividing, while the first two means were obtained by "multiplying all values and taking the nth root".
Look up geometric mean vs arithmetic mean.

 

[JeffM]

I have no idea why the averages were calculated as geometric means rather than arithmetic means. To create a combined mean, you must use a consistent method. So your answer is

[MATH]y = (8985^{67} * 7622^{132})^{1/(67+132)}.[/MATH]
Unfortunately, my hand calculator cannot process that.

[MATH]\therefore y = 1000 * (8.985^{67} * 7.622^{132})^{1 \div 199}.[/MATH]
That still gives my calculator indigestion. Fortunately, there are logs on my calculator.

[MATH]log(y) = 3 + \dfrac{1}{199} * \{67 * log(8.985) + 132 * log(7.622)\} \implies y \approx 8029.[/MATH]
So I do not get 8056 either. Of course, I may have entered a number wrong.

 

[Steven G]

I have no idea why the averages were calculated as geometric means rather than arithmetic means. To create a combined mean, you must use a consistent method. So your answer is

[MATH]y = (8985^{67} * 7622^{132})^{1/(67+132)}.[/MATH]
Unfortunately, my hand calculator cannot process that.

[MATH]\therefore y = 1000 * (8.985^{67} * 7.622^{132})^{1 \div 199}.[/MATH]
That still gives my calculator indigestion. Fortunately, there are logs on my calculator.

[MATH]log(y) = 3 + \dfrac{1}{199} * \{67 * log(8.985) + 132 * log(7.622)\} \implies y \approx 8029.[/MATH]
So I do not get 8056 either. Of course, I may have entered a number wrong.

I just checked [MATH] (8985^{67} * 7622^{132})^{1/(67+132)}[/MATH] in WA and got

which is close enough to 8056

 

[Steven G]

I have no idea why the averages were calculated as geometric means rather than arithmetic means. To create a combined mean, you must use a consistent method. So your answer is

[MATH]y = (8985^{67} * 7622^{132})^{1/(67+132)}.[/MATH]
Unfortunately, my hand calculator cannot process that.

[MATH]\therefore y = 1000 * (8.985^{67} * 7.622^{132})^{1 \div 199}.[/MATH]
That still gives my calculator indigestion. Fortunately, there are logs on my calculator.

[MATH]log(y) = 3 + \dfrac{1}{199} * \{67 * log(8.985) + 132 * log(7.622)\} \implies y \approx 8029.[/MATH]
So I do not get 8056 either. Of course, I may have entered a number wrong.

I just checked (898567∗7622132)1/(67+132)(898567∗7622132)1/(67+132) in WA and got

which is close enough to 8056

 

[Harry_the_cat]

I calculated it as:
\(\displaystyle y= (8985^{67}*7622^{132})^{\frac{1}{199}}\)
\(\displaystyle y = 8985^{(\frac{67}{199})} * 7622^{(\frac{132}{199})}\)
\(\displaystyle y = 8056.1 \) approx
Calculator can handle that.

 

[help12]

Thanks for your help.

Can anyone advise how to calculate the overall 95% confidence intervals?

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

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

I do not have any individual data that made these means. Only the above data.

Overall mean: 8056.

Overall mean calculated as follows:

 67*log(8985)+132*log(7622)
1789.8427

1789.8427/199
8.9941844

exp(8.9941844)
8056.0964

.

I need to calculate the overall CIs.

Is it the same method?

So for example the lower overall CI calculated as:

Code:
67*log(5705)+132*log(5486)
1716.0036

1716.0036/199
8.6231337

exp(8.6231337)
5558.7787

Or is this incorrect?

Thanks

 

[JeffM]

I just checked (898567∗7622132)1/(67+132)(898567∗7622132)1/(67+132) in WA and got

which is close enough to 8056

Did it on my computer. Got 8056.0969.

Time to scrap that calculator.