Does the width of the confidence interval necessarily imply power? What does the width of the confidence interval imply about the statistical power of a given study? I specifically want to figure out if a narrow confidence interval reasonably excludes a lack of statistical power to detect a given effect size as an explanation for a large p value.
 A: If the CI is narrow in some meaningful way (e.g. the width would not be a meaningful effect size), then typically the analysis was well powered. Exceptions could include cases, where the assumptions of the used analysis are violated (e.g. using Poisson regression in the presence of substantial overdispersion, ignoring observed or unobserved factors influencing group assignment in an observational study, informative censoring in standard survival analysis etc.), the study design is flawed (i.e. at best provides a biased estimate of the estimand of interest) or the results are misinterpreted.
A: Short answer: For symmetrical CIs, the half confidence interval width corresponds to the effect (MDE) that has 50% power for the given sample size.
Details: Let's look at the simplest two-sided test confidence interval and power formula for a z-test. Assume that the treatment groups are of equal size ($N_C=N_T=N/2$) and a homogeneous treatment effect. The half confidence interval width (CIWH) for a given variance and $\alpha$ is given by
\begin{equation}
CIWH= z_{\alpha/2}\sqrt{\frac{\sigma^2}{n_C}+\frac{\sigma^2}{n_T}}= z_{\alpha/2}\sqrt{\frac{4\sigma^2}{N}}.
\end{equation}
By solving for N we get
\begin{align}
CIWH&=z_{\alpha/2}\sqrt{\frac{4\sigma^2}{N_{req}}}\\
CIWH^2&=z_{\alpha/2}^2\frac{4\sigma^2}{N_{req}}\\
N_{req}&=\frac{z_{\alpha/2}^2}{CIWH^2}\frac{4\sigma^2}{1}
\end{align}
The required sample size for a given variance, $\alpha$, MDE, and $\beta$ is given by
\begin{align}
N &= \frac{(z_{\alpha/2} +z_{\beta})^2}{MDE^2}\frac{4\sigma^2}{1}\\
MDE^2 &= (z_{\alpha/2} +z_{\beta})^2\frac{4\sigma^2}{N}\\
MDE &= (z_{\alpha/2} +z_{\beta})\sqrt{\frac{4\sigma^2}{N}}.
\end{align}
This implies that difference between MDE and CIWH is
\begin{align}
MDE - CIWH &=  (z_{\alpha/2} +z_{\beta})\sqrt{\frac{4\sigma^2}{N}} -z_{\alpha/2}\sqrt{\frac{4\sigma^2}{N}} \\
&=z_{\beta}\sqrt{\frac{4\sigma^2}{N}},
\end{align}
or equivalently $MDE =CIWH + z_{\beta}\sqrt{\frac{4\sigma^2}{N}}$.
Finally, it is clear that $MDE=CIWH$ if $z_{\beta}=0$ which is true iff $\beta=0.5$, or in other words when the power is 50%. This is logical, if the true effect is at MDE half of the point estimates will be below MDE which implies that the lower CI bound will cover zero.
