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For Weibull distribution,
S(t) = $e^{-(\lambda * e^(x * \beta)*t)^\rho}$

"$^{(1/rho)}$" will be only for log(v)

so, I modified like this

Tlat <- - log(v)^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.

For Weibull distribution,
S(t) = $e^{-(\lambda * e^(x * \beta)*t)^\rho}$

"$^{(1/rho)}$" will be only for log(v)

so, I modified like this

Tlat <- (- log(v))^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.

For Weibull distribution,
S(t) = exp(-(lambda * exp(x * beta)*t)^rho)

"^(1/rho)" will be only for log(v)

so, I modified like this
Tlat <- - log(v)^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.

For Weibull distribution,
S(t) = $e^{-(\lambda * e^(x * \beta)*t)^\rho}$

"$^{(1/rho)}$" will be only for log(v)

so, I modified like this

Tlat <- - log(v)^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.

For Weibull distribution,
S(t) = exp(-(lambdaexp(xbeta)*t)^rho)

"^(1/rho)" will be only for log(v)

so, I modified like this
Tlat <- - log(v)^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.

For Weibull distribution,
S(t) = exp(-(lambda * exp(x * beta)*t)^rho)

"^(1/rho)" will be only for log(v)

so, I modified like this
Tlat <- - log(v)^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.