 Solution

Put $T={\sum }_i{X}_i$. For interior parameters $0<{\theta }_0<{\theta }_1<1$, the alternative-to-null likelihood ratio isThe second base is greater than one, so this is strictly increasing in $T$. The Neyman-Pearson lemma therefore gives an upper-tail rejection rule. Under the null hypothesis, $T\sim \mathrm{Bin}(n,{\theta }_0)$, and an ordinary conservative threshold can be chosen asThere is an important discrete qualification: an integer cutoff need not have test size exactly $\alpha$. To obtain exact size, choose an integer $b$ with ${\mathrm{Pr}}_0(T>b)\le \alpha \le {\mathrm{Pr}}_0(T\ge b)$, reject certainly when $T>b$, and at $T=b$ reject with probabilityThis randomized statistical test is a most powerful test at exact size $\alpha$; when a binomial upper tail equals $\alpha$, no randomization is needed and the stated region is literal. Otherwise the unrandomized upper-tail region has size at most $\alpha$, and the exact-size result includes the boundary randomization. Endpoint parameter cases retain the same upper-tail ordering by taking limits, with ties resolved on the null-supported outcomes.