The simplest statistical test: how to check for a difference between treatments
Open Access
- 25 May 2006
- Vol. 332 (7552) , 1256-8
- https://doi.org/10.1136/bmj.332.7552.1256
Abstract
The complexity of statistical methods for analysing clinical data can make interpreting clinical trial reports a daunting task for many readers. However, the key result of many trials could be presented and interpreted using quite basic statistical methods. The overall spirit of this article is to encourage all interested in understanding clinical trials to “feel the data” rather than get too absorbed in the technicalities (and occasional confusions) of advanced statistical techniques. For many trials the primary outcome is a disease event. This might be death or a composite outcome such as death, myocardial infarction, or stroke. The standard statistical methods—Cox proportional hazard models and log rank tests—take account of variation in patient follow-up times, but the consequent hazard ratios, confidence intervals, and P values seem a mysterious “black box” to some readers. Alternatively, if events relate to a fixed follow-up time then methods for comparing two proportions (for example, the χ2 test) may be used. This article describes a much easier method than these, which readers can use to assess quickly the strength of evidence for a treatment difference in an event outcome. It's surprising even how many statisticians don't know this simplest test: I first heard of it from a cardiologist. Consider a randomised clinical trial with two treatment groups of roughly equal size. Let the outcome of interest be a clinical event. The key data are the numbers of patients experiencing the event by treatment group. The figure shows how to perform a statistical test of significance based solely on these two numbers. Fig 1 How to perform the simplest statistical test of significance Calculate the difference in the two numbers of events and divide by the square root of their sum. Call the resulting number z. Under the null hypothesis that the two treatments have identical influence …Keywords
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