How do I construct a symmetric, two sided confidence interval for a given statistic?

Question

Let $X_1,...,X_n$ be a random sample of iid random variables, $X_i\sim Exp(\lambda),\lambda>0$. Consider the statistic $T_1(X_1,...,X_n,\lambda)=2\lambda n\bar{X_n}$. The task is to construct a symmetric, two-sided confidence interval for the parameter $\lambda$ with confidence level $1-\alpha$ with the help of $T_1$.

How do I proceed here ? I know that $T_1\sim\Gamma(\frac{1}{2},n)$, since $n\bar{X_n}\sim\Gamma(\lambda,n)$. So I guess $P(\lambda\in[T_1(X_1,...,X_n,\lambda)-\varepsilon,T_1(X_1,...,X_n,\lambda)+\varepsilon])\geq 1-\alpha $ for a $\varepsilon>0$ must hold. But how do I proceed here ? If I compute the inverse $T_1^{-1}$ with regards to $\lambda$ I get $T_1^{-1}(X_1,...,X_n,\lambda)=\frac{\lambda}{2n\bar{X_n}}$. Furthermore the symmetric confidence interval is always given by $I=[T_1^{-1}(F^{-1}(\frac{\alpha}{2})),T_1^{-1}(F^{-1}(1-\frac{\alpha}{2}))]$, where F is a knows distribution. My guess is that $F(x)=1-e^{-\lambda x},x\geq0$, since $X_i\sim Exp(\lambda)$. Then $F^{-1}(x)= -\frac{ln(1-x)}{\lambda},x\in[0,1)$. But now if we were to compute $I$, we would get $I=[-\frac{ln(1-\frac{\alpha}{2})}{\lambda 2n\bar{X_n}},-\frac{ln(\frac{\alpha}{2})}{\lambda 2n\bar{X_n}}],$, which does not really make sense.