# Showing the sample mean is a sufficient statistics from an exponential distribution

Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($$\theta$$), where $$\theta>0$$ is unknown. If we observe $$\overline x = 5.2$$ for a sample of $$20$$ light bulbs, record a representative likelihood function. Why is it that we only need to observe the sample average to obtain a representative likelihood?

The likelihood is pretty straightforward to find: $$L(\theta \mid x_1,\ldots,x_{20})=\prod_{i=1}^{20}\theta e^{-\theta \overline x} = \theta^{20}e^{-20 \theta \overline x}$$

I am having an issue showing this is a sufficient statistic however, as I cannot seem to factor it in the form of $$h(\overline x)g_\theta (T(\overline x))$$

You should write it as $$h(x_1,\ldots,x_{20}) g_\theta( (x_1 + \cdots + x_{20})/20),$$ or in other words as $$h(x_1,\ldots,x_{20}) g_\theta(T(x_1,\ldots, x_{20})).$$
What's going on is camouflaged by the fact that $$h(x_1,\ldots,x_{20}) = 1$$ regardless of the value of $$(x_1,\ldots,x_{20}).$$