Find the region of rejection

Question

I'm stuck with this problem: Suppose $X = ( X_1, \ldots, X_n) $ are samples from population $Binom(1,p)$. How to find the region of rejection of $$ H_0: p\leq0.01 \leftrightarrow H_1: p > 0.01$$ where $H_0$ is the null hypothesis, and $\alpha = 0.05$.

My attempts: Set $T(X) = X_1 + \ldots + X_n$, so that $ T \sim Binomial(n, p)$.

Suppose $D = \{X: T(X) > c, c \in \mathbb{N}, 0\leq c \leq n-1\}$ is the region of rejection, and let $\phi$ be the characteristic function of $D$. The power function then is $$g(p) = E_p[\phi(X)] = \sum_{k=c}^n \binom{n}{k}p^k(1-p)^{n-k}, 0\leq p\leq 1$$ Taking the derivative of $g$, we find that $g$ is monotonically increasing w.r.t. $p$: $$ g'(p) = \binom{n}{c}cp^{c-1}(1-p)^{n-c} > 0 $$ We need to determine $c$ such that $$ \sup_{p\leq 0.01} g(p) = g(0.01) \leq 0.05 $$

i.e.

$$\sum_{k=c}^n \binom{n}{k}(0.01)^k(0.99)^{n-k} \leq 0.05$$

where I don't know how to continue. I think I must be wrong somewhere, for $c$ seems unresolvable following my steps, but I think I need to find a closed form for $c$, which should be a function of $n$.

Thanks!

Answer 1

When the test statistic $T$ is discrete, you will not generally find a closed form for the critical value $c.$ Two approaches are often used.

Software. However, for a specific $n$ you can use software to find $c.$ For example, if $n = 1000$ the null distribution is $T \sim \mathsf{Binom}(n=1000, p_0=.01).$ You want to reject $H_0$ for 'large' values of $T.$ In R statistical software you can find the 95% percentile of this distribution.

qbinom(.95, 1000, .01)    # 'qbinom' is binomial quantile function
## 15
pbinom(15, 1000, .01)     # 'pbinom' is binomial CDF
## 0.9521294

Thus we know that $P(T \le 15) = 0.9521$ so that $P(T \ge 16) < 0.05$ and $c = 16.$

Normal approximation. A somewhat riskier, but potentially more general, approach is to use a normal approximation to the the distribution of $T,$ to standardize so that $$Z = \frac{T - np_0}{\sqrt{np_0(1-p_0)}} \stackrel{aprx}{\sim} \mathsf{Norm}(0,1),$$ and to express $c$ in terms of the standard normal quantile function (inverse CDF) $\Phi^{-1}.$ For this approach, you have to make sure that $n$ and $p_0$ are such that a normal approximation is reasonably accurate. (Different authors state a variety of 'rules of thumb' for safe use of the normal approximation to binomial.) Usually, the resulting value of $c$ will not be an integer, and for a right-tailed test you would round up to the next higher integer.