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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.

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  • $\begingroup$ Its odd that $H_o \cup H_a \neq \mathbb{R}$ $\endgroup$
    – user237392
    Mar 11, 2016 at 5:24
  • $\begingroup$ @Bey I do see that some sources stat that the union of the hypothesis is the entire parameter space (not $\mathbb{R}$ though), but also many sources will teach this simple vs simple hypoethsis testing. To OP: Do you mean the test has a type-I error of $\alpha = 0.046$? And you are using the proportion test with normal approximation? I think the test itself may encouraging you to compute the binomial pmf/cdf for exact test, due to the small sample size. $\endgroup$
    – BGM
    Mar 11, 2016 at 6:23
  • $\begingroup$ @BGM yea, the parameter space is $[0,1]$, but for brevity, most people would write $H_0: p=0.6,H_a: p\neq 0.6$ or $H_0: p\leq 0.6, H_a:p>0.6$ etc. The problem is that there is no p-value for this test. I think a likelihood ratio test is what you're looking for here. $\endgroup$
    – user237392
    Mar 11, 2016 at 12:45
  • $\begingroup$ @Bey: I believe this is a 'simple vs. simple' situation and that the significance level is $\alpha = .046.$. In my experience, it is a common typo for those not familiar with the Greek alphabet to confuse $\alpha$ and $\sigma.$. Anyhow, this is the only reasonable interpretation I can see. $\endgroup$
    – BruceET
    Mar 13, 2016 at 6:26

1 Answer 1

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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

enter image description here

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  • $\begingroup$ How did you decide that the rejection region should be one-sided? $\endgroup$
    – user237392
    Mar 13, 2016 at 15:10
  • $\begingroup$ Also, how does the $H_a$ play into the conclusion? Normally, you are trying to decide whether $H_0$ should be rejected. If you reject it, then generally you'd expect the $H_A$ to be better supported. However, if $H_a$ were $p=0.001$, and $X=10$ then we'd reject but the usual interpretation of hypothesis tests would be odd in this situation, as $H_0$ is far more likely than $H_A$. This is why its generally good to have your hypotheses partition the parameter space. $\endgroup$
    – user237392
    Mar 13, 2016 at 15:12
  • $\begingroup$ Not intended to be a practical problem. Parameter space has two points. You can look at Neyman-Pearson Fundamental Lemma for the framework. $H_0$ and $H_A$ have essentially equal 'status'. $\endgroup$
    – BruceET
    Mar 13, 2016 at 15:21
  • $\begingroup$ I know the NP lemma and why it applies. But you aren't comparing $H_0$ and $H_A$, you're just evaluating $H_0$ under and exact probability model. The NP lemma says that the most powerful test is a likelihood ratio test for some cutoff. $\endgroup$
    – user237392
    Mar 13, 2016 at 15:28
  • $\begingroup$ @Bey: I took OP's main question to be about computing probabilities, and tried to answ that. I left it to OP to determine from N-P that rej reg is for large X. Mainly in terms of the notation in Wikipedia on 'Neyman-Pearson', $\Lambda(x) = P(X=x|.6)/P(X=x|.7) \leq \eta = 0.3329796$ describes the critical region, where $P(\Lambda(X) \leq \eta|.6) = P(X \ge c = 9) = \alpha = .046.$ If you care to post an Answ you believe is more helpful, I will read it with interest. $\endgroup$
    – BruceET
    Mar 13, 2016 at 21:03

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