You have to be very careful with just how the Weibull distribution is parameterized within a function, as there are alternate choices. You will note that the value and the variance for the
Intercept in the first model are the same as those for
scale in the second, and the value for
log(scale) in the first model is the negative of the
shape in the second, with the same variances, indicating that the parameterizations differ between the models. The help page for
There are multiple ways to parameterize a Weibull distribution. The
survreg function embeds it in a general location-scale family, which is a different parameterization than the
rweibull function, and often leads to confusion.
survreg's scale = 1/(
survreg's intercept = log(
The manual page for
The Weibull parameterisation is different from that in survreg, instead it is consistent with dweibull. The "scale" reported by survreg is equivalent to 1/shape as defined by dweibull and hence flexsurvreg. The first coefficient (Intercept) reported by survreg is equivalent to log(scale) in dweibull and flexsurvreg.
The manual page for
dweibull describes its parameterization as:
The Weibull distribution with shape parameter a and scale parameter b has density given by
f(x) = (a/b) (x/b)^(a-1) exp(- (x/b)^a), for x > 0.
That parameterization by
dweibull agrees with that referred to by the reliability measure you cite, with the
a corresponding to shape $\beta$ and the
b corresponding to scale $\eta$.
Be ruthlessly consistent in the parameterization. It would seem to be safest here to use the
flexsurvreg parameterization, as that agrees with your intended downstream use. The confidence intervals based on the covariance matrix just use the coefficient estimates and their (co)variances directly, so there's no need to transform. Even then, look at plots, check carefully, and make sure that things make sense whenever you come upon a potentially confusing parameterization like you face with the Weibull distribution.