Rejection region

Question

I need help with the following problem:
Let H0: p = 0.6

HA: p = 0.7

based on observing a binomial random variable with 10 trials. What is the rejection region for the most powerfil level
sigma = 0.046 test of H0 versus HA? Simplify your answer.
This is my approach,
since the distribution is binomial then,

$mean = np = (10)(.6) = 6$

$variance = np(1 - p) = (10)(.6)(1 - .6) = 2.4$

then,
$$Z = \frac{6 -.6}{2.4(3.1622)} = 7.11$$
the value 7.11 doesn't make sense.

Answer 1

We are choosing between $H_0: p = .6$ and $H_A: p = .7$. This is sometimes called a 'simple vs. simple' test because both hypotheses have only a single value.

For a test at level $\alpha = .046,$ we want $P(\text{Reject} H_0|p=.6) = .046.$

For a most-powerful test, we will reject for large numbers of $X$ of Successes. (If that is not obvious, please take a look at the Neyman-Pearson Fundamental Lemma.)

So we seek $c$ such that $P(X \ge c) = .046,$ where $X \sim Binom(n=10, p=.6).$

Because $n$ is small, I am not enthusiastic about using a normal approximation. From R, here is a PDF table of $Binom(10,.6).$ (Ignore row numbers in brackets.)

 n = 10;  p = .6;  x=0:10;  pdf = dbinom(x, n, p)
 cbind(x, pdf)
        x          pdf
  [1,]  0 0.0001048576
  [2,]  1 0.0015728640
  [3,]  2 0.0106168320
  [4,]  3 0.0424673280
  [5,]  4 0.1114767360
  [6,]  5 0.2006581248
  [7,]  6 0.2508226560
  [8,]  7 0.2149908480
  [9,]  8 0.1209323520
 [10,]  9 0.0403107840
 [11,] 10 0.0060466176

From the table it is clear that the critical value is $c = 9$ so that the critical (rejection) region is $\{X \ge 9\}.$ If you are familiar with the quantile function (inverse CDF) then you might confirm this in R as follows:

  qbinom(1-.046, 10, .6)
  ## 9

The power of this test not especially large, but it is as large as possible in the circumstances described:

$$P(\text{Reject} H_0|p = .7) = P(X \ge 9|p = .7) = 0.1493.$$

In R,

 sum(dbinom(9:10, 10, .7))
 ## 0.1493083