Backing out the standard deviation from information on baseline mean/s.d., and coefficient mean/s.d

Question

I am trying to run a power calculation for a randomized control trial. For this I need a mean and standard deviation for our 'baseline'. There are papers out there which would have a mean and standard deviation, but what we would want would be to use their final outcomes as our baseline. In other words, these papers applied treatment $A$. I would like to apply treatment $B$ on top of treatment $A$, so treatment $A$ outcomes will be my new baseline.

In looking at their results tables, I can find the mean and SD of their baseline, plus the coefficient of the effect and that coefficient's SD. To find the new mean of my baseline, I would add the coefficient to the mean. However, is there a way to back out the SD of the new mean from the information I have from these papers?

To give an example: one paper has the mean in control as 5847 (SD=16784). The change of the mean under treatment $A$ is 267 (SD=527). So my new baseline would be 5847 + 267 = 6114. But would there be any way of getting this SD without the data? (Almost all of it is not available publicly.)

Thanks so much!

Answer 1

When conducting a power calculation for a randomised control trial, estimating the standard deviation (SD) for a new baseline from existing literature can be quite challenging, especially without access to raw data. However, you can make educated assumptions based on the information available. Let's give that a go !

Given:

Mean in control: $\mu_C = 5847$ with $SD_C = 16784$ Change in mean under treatment A: $\Delta \mu_A = 267$ with $SD_{\Delta A} = 527$ Your new baseline mean is $\mu_{new} = \mu_C + \Delta \mu_A$.

Estimating the New SD

The challenge is in estimating the standard deviation of the new baseline. The total variance in the treatment group is a combination of the variance in the control group and the variance of the treatment effect.

Assuming Independence: If we assume that the baseline measurements and the treatment effects are independent, we can combine their variances:

$\text{Var}_{new} = \text{Var}_C + \text{Var}_{\Delta A}$

Since $\text{SD} = \sqrt{\text{Var}}$, we obtain:

$\text{SD}_{new} = \sqrt{SD_C^2 + SD_{\Delta A}^2}$

Calculation: Using the provided numbers:

$\text{SD}_{new} = \sqrt{16784^2 + 527^2}$

This is a simplification and assumes that the variability of the treatment effect is independent of the baseline. This assumption might not always be valid. If the variability of the treatment effect is related to the baseline (for instance, if the effect of the treatment varies more in subjects with higher baseline values), then this calculation might not be accurate.

Considerations and Limitations

Data Heterogeneity: The method assumes that the populations and conditions of the original study and your study are similar.

Independence Assumption: This assumption is crucial in the calculation. If the treatment effect is not independent of the baseline, the estimation could be inaccurate. In that case you would need to incorporate the covariance between them, and you may not know that, in which case it would be a good idea to perform some sensitivity analysis with various assumptions about their covariance. The calculation then becomes:

$$ \text{SD}_{new} = \sqrt{\text{Var}_C + \text{Var}_{\Delta A} + 2 \times \text{Cov}(C, \Delta A)} $$

Access to Individual Participant Data (IPD): Access to IPD from original studies, ideally, would allow for more precise estimations, including more complex statistical modelling.

Consult with a Statistician: Given the assumptions and simplifications involved, consulting with a statistician, particularly one experienced in clinical trial design and power analysis, is advisable.

This method produces an approximation of the new baseline SD, which may be used in your power calculation while keeping the assumptions and potential mistakes in mind.