Let $X_1,\ldots,X_n$ be exponentially distributed with parameter $\lambda$ This implies that $Y=\sum_{i=1}^nX_i$ has a gamma distribution with parameters $(\lambda,n)$
Can anyone help me show that $\left( \frac{a}{n\bar{x}}, \frac{b}{n\bar{x}} \right)$ Is an exact $95$% central confidence interval for $\lambda$ if
$\int_0^a \frac{y^{n-1}e^{-y}\;dy}{\Gamma(n)}= \int_b^\infty \frac{y^{n-1}e^{-y}\;dy}{\Gamma(n)}= 0.025$
Here is what I have so far. Basically i've been trying to construct anything to help use those given integrals
$P\left(\frac{a}{n\bar{x}}
$=\lambda^{-n}\int_0^\frac{a}{n\bar{x}} \frac{y^{n-1}e^{-y/\lambda}\;dy}{\Gamma(n)}+\lambda^{-n}\int_0^\frac{b}{n\bar{x}} \frac{y^{n-1}e^{-y/\lambda}\;dy}{\Gamma(n)}-1$
Which is where I get stuck as this doesn't really look salvagable. Any help here would be greatly appreciated!