Meta-Analysis: Converting Hazard Ratio to Effect Size while neither SE nor CI are reported

[Mahdie Raj]

Hi Everyone,
I am conducting a meta-analysis using metan function. All of the studies have reported completed information (HR, CI). As a result, I can convert them to effect size and estimate the corresponding weights as well. Unfortunately, one study reported neither SE nor CI associated with HR. However, the study has reported HR, sample sizes, and the number of deaths in both groups. Does stata have a function to calculate the SE or CI in this case?
Thanks,
Nader

[Marcos Almeida]

Welcome to the Stata Forum / Statalist,

The help file of the user-written - metan - presents an example.

[David Fisher]

Hi Nader,

If you have the number of deaths in each group, a good approximation to the variance of the log HR is: (1/d₁) + (1/d₂). From there, you can obtain the SE and CI.

A good reference describing methods for obtaining HRs or SEs when you have other data is: Tierney et al. Practical methods for incorporating summary time-to-event data into meta-analysis. Trials 2007; 8: 16.

Best wishes,

David.

[Mahdie Raj]

Hi David,
Thanks for your advice.
My HR comes from a population-based longitudinal study. I do not believe that your formula ((1/d₁) + (1/d₂)) gives a good proximation of the variance. According to Tierney and colleagues' paper, I think the formula is appropriate for a randomised 1:1 trial design.
I was wondering if someone could give me some hints to estimate the variance of HR in the case I described in my first post. Unfortunately, the study has not reported any survival curve.
Best,
Nader

[Matt Warkentin]

Does the study provide a P-value for testing the statistical significance of the HR?

[Mahdie Raj]

No, it does not. Exact p values were not reported. "P<0.05*" was reported!! I just know that HRs are/are not statistically significant at certain alpha level.

[Matt Warkentin]

Well, it isn't an ideal option but you could take a conservative approach and use P=0.049 and the point estimate to back calculate the SE/CI. I'm sure someone else here will have a more effective solution.

[David Fisher]

Hi Nader,

I believe that Tierney et al are too restrictive in their statement of the application of the formula Var(logHR) = (1/d₁) + (1/d₂) [or, equivalently, the form given in Tierney is "logrank V" = d₁d₂/(d₁+d₂), where "logrank V" = 1/Var(logHR)]. The only assumptions made are proportional-hazards and a constant baseline hazard (i.e. hazard follows an exponential distribution). So the formula should give a reasonable result in the majority of cases.

An alternative formula, also given by Tierney et al, is "logrank V" = (d₁+d₂)*n₁*n₂/((n₁+n₂)²); or, equivalently, Var(logHR) = ((n₁+n₂)²)/((d₁+d₂)*n₁*n₂). You could try calculating both, and see how much they differ by.

I don't know of any other suitable methods. In particular, I don't know of any methods specific to longitudinal data. I suspect you will have to make do with methods developed for randomised data. The only other suggestion I can make is to contact the study authors, if possible, and ask to be provided with the correct variance for their study.

Best wishes,

David.