4
$\begingroup$

Let $X_{1}, \ldots, X_{n}$ be a sample from the Bernoulli distribution with parameter $p$ Consider testing $H_{0}: p=p_{0}$ versus $H_{1}: p=p_{1}$ where $p_{0}<p_{1}$ are known numbers.
(a) Using the Neyman-Pearson lemma, find the most powerful test (Non-Randomized) among tests with level at most $\alpha$. When can the level of the test be exactly $\alpha ?$
(b) Use the CLT to find a critical value such that the level of the test is approximately (asymptotically for large $n$ ) $\alpha$.

To use Neyman Peason lemma I calculate $r$ as : $$r= \frac{f_1}{f_0} = \frac{p_1^{\Sigma x_i}(1-p_1)^{n-\Sigma x_i}}{p_0^{\Sigma x_i}(1-p_0)^{n-\Sigma x_i}}$$ For $p_1>p_0$ we have non randomized test is $\phi(x) =$ Reject $H_0$ for $\Sigma x_i \geq k$.
For given at most level $\alpha$ : $$\alpha \geq E_{p_0} \phi(x) = P_{p_0}\{\Sigma x_i \geq k\} = \Sigma_{r=k+1}^n (n_{C_r}) p_{0}^r (1-p_{0})^{n-r}$$

(a) I don't know how to proceed after this and what to say about When can the level of the test be exactly $\alpha ?$

(b) How to use CLT to find the critical value which satisfies the given condition.

Please help me with this problem. Thankyou.

$\endgroup$
1
$\begingroup$

I will get you started with couple of particular numerical examples that illustrate the main issues involved in a general answer.

Example 1. Let $n = 10, p_0 = .5, p_1 = .7, \alpha = 0.5.$ Then NP says to reject when $T = \sum_{i=1}^{10}X_i \ge k.$

The issue is to find $k$ so that $\alpha \le 0.05.$ Under $H_0,$ $T \sim \mathsf{Binom}(n=10, p=.5).$ Then using $k = 8$ would give $\alpha = 0.0547 > 0.05,$ so use $k = 9$ to get $\alpha=0.0107 < 0.05.$ Computations in R,

qbinom(.95, 10,.5)
[1] 8
sum(dbinom(8:10, 10, .5))
[1] 0.0546875
sum(dbinom(9:10, 10, .5))
[1] 0.01074219

Thus, there is no nonrandomized test exactly at level $\alpha = 0.05.$

Under $H_0,$ we have $E(T) = np = 5,$ $SD(T) = \sqrt{np(1-p)} = \sqrt{2.5} = 1.581.$

So $T\stackrel{aprx}{\sim}\mathsf{\mu=5, \sigma=\sqrt{2.5}}.$ So an approximate value of $k = 7.6.$ This is not a good approximation for $n$ as small as $n=10,$ but this kind of approximation is useful for large $n.$

qnorm(.95, 5, sqrt(2.5))
[1] 7.600742

enter image description here

t = 0:10;  pdf = dbinom(t, 10, .5)
plot(t, pdf, type="h", lwd = 3, col="blue")
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = 8.5, lwd=2, lty="dotted")
 curve(dnorm(x, 5, sqrt(2.5)), add=T, col="red")

Example 2. Now suppose $n=100,$ keeping the other parameters the same. Then we have $T\sim\mathsf{Binom}(100, .5)$ under $H_0$ and $k = 59$ gives a test at level $\alpha = 0.044$ and the normal approximation says to use $k = 58.22.$

qbinom(.95, 100, .5)
[1] 58
sum(dbinom(58:100, 100, .5))
[1] 0.06660531
sum(dbinom(59:100, 100, .5))
[1] 0.04431304
qnorm(.95, 50, 5)
[1] 58.22427

enter image description here

t = 0:100;  pdf = dbinom(t, 100, .5)
plot(t, pdf, type="h", col="blue")
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = 58.5, lwd=2, lty="dotted")
 curve(dnorm(x, 50, 5), add=T, col="red")
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.