Poisson binomial distribution hypothesis test Let $X_i$, $i=1,  \dots, n$, be independent non-identically distributed random variables with Bernoulli distributions with unknown probability of successes $p_i$, $i=1,  \dotsc, n$. Then $Y:=\sum_{i=1}^n X_i$ has Poisson binomial distribution. 
Let $p_i^0 \in (0,1)$, $i=1,  \dots, n$, be some known real numbers and let
$$p:=\frac{1}{n}\sum_{i=1}^n p_i$$, 
$$p^0:=\frac{1}{n}\sum_{i=1}^n p_i^0$$.
How do I test a hypothesis:
$\text{H}_0: p=p^0$ vs. $\text{H}_1: p \neq p^0$?
Similarly, how to test one sided hypothesises? 
 A: Do you have a context for this problem, any additional information? As there is $n$ (unrelated) parameters $p_1, p_2, \dotsc, p_n$ with $n$ independent observations $X_1, X_2, \dotsc,X_n$, there is not much to go on ... but since you have defined the focus (or interest) parameter $p=\frac1{n}\sum p_i$, maybe there are some possibiities ... in an applied setting, I would go for any scrap there must be of prior information, build a prior distribution for the $p_i$, and go for bayes. But without that: We can estimate $p$ with $\bar{X}_n=\frac1n\sum_i X_i$, which is unbiased for  $p$, and is also the maximum likelihood etimator. We can even bound its variance with
$$ \DeclareMathOperator{\V}{\mathbb{V}}
\V \bar{X}_n = (\frac1n)^2 \sum_i p_i(1-p_i) \leq \frac1{4n}
$$
which tends to zero when $n$ grows without bound, so this estimator is consistent. 
Your hypothesis testing problem is more difficult, but you could use, with large $n$, the above variance bound to construct a very conservative confidence interval, and invert it to get a test. I'm unsure if we can do much better than that, without getting more information. 

But can we do better than this? Let us see if we can define a profile likelihood function. The likelihood function (which in itself will not be very useful here) is
$$ L(p_1, \dotsc,p_n)=\prod_i p_i^{X_i} (1-p_i)^{1-X_i} $$
which is saturated and the maximized value is always 1. The profile likelihood for $p=\frac1n \sum_i p_i$ is
$$
 L_{\text{Prof}}(p) = \sup_{\sum_i p_i=p} L(p_1, \dotsc,p_n)
$$ and this will have value 1 only if $\sum_i p_i =\sum_i X_i$, else it will be less. Now write $Y=\sum_i X_i$, we can show that
$$
L_{\text{Prof}}(p) = (1-\underline{q})^{n-Y}\bar{q}^Y
$$
where 
$$
\underline{q}=\begin{cases} \frac{p-Y/n}{1-Y/n} &\text{if $Y/n \leq p$}\\
                              0         & \text{otherwise}
               \end{cases} \\
\bar{q}=\begin{cases} 1  & \text{If $Y/n \leq p$}\\
                      np/Y & \text{otherwise}
         \end{cases}
$$
Let us look at this for a case with $n=100, Y=35$.  Then $\sqrt{\frac1{4 n}}=0.05$, so our conservative about 95% interval is $(0.35-0.1,0.35+0.1)$.
 
which does not relly lok like a loglikelihood as we know them ... not being differentiable at the maximum, not looking parabolic at least close to the maximum, and contemplating a bit we see that this properties will survive irrespective $n$.  So we cannot expect to be able to use this like a usual loglikelihood function. For the record, R code for the plot:
lproflik <- function(n, Y) {
    Vectorize(function(p) {
        qu <- if (Y/n <= p) 1.0 else p/(Y/n)
        ql <- if (Y/n <= p) (p-Y/n)/(1-Y/n) else 0.0
        (n-Y)*log(1-ql) + Y*log(qu)
        } )
}

plot( lproflik(100, 35), from=0.25, to=0.45, col="blue",
     main="Profile loglikelihood for p")
abline(h=-qchisq(0.95, 1)/2, col="red")

