A coin was flipped 10 times and landed on head 9 times. We want to test $$H_0: p = \frac{1}{2} \ vs \ H_1: p > \frac{1}{2}, p = \mathbb{P}(head)$$ with a confidence level of $\alpha = 0.05$. The question specifically ask for:

  • The test statistic
  • Distribution under $H_0$
  • The conclusion of the test and the $p$-value.

My attempt:

I don't know if the test statistic is given by $\hat{p} = 9/10 = 0.9$ or $$\mathbb{P}(X = 9) = \binom{10}{9} 0.5^9 \cdot 0.5^1 \approx 0.00977.$$

The distribution under the null hypothesis is $X \sim binomial(10, 1/2)$ and with a 0.05 signficance level, our conclusion would be to reject the null hypothesis. I think that the $p$-value would be $$p = \binom{10}{10} \frac{1}{2^{11}}$$ but i'm really confused about this question.


1 Answer 1


The test statistic and the sample mean are not the same thing. Recall the test statistic is what we compare to a known distribution to obtain inference - parts 1 and 2 of the question are essentially the same. The derivation is simple, or you can outline it if this a graduate class, or else (elementary non-calculus stats) you probably have covered tests of proportion in which case you just need to reference the correct formula.

Question 3 is about applying what you know of hypothesis tests. Having found the distribution under the null, you can find the area in the tail of that distribution (again using calculus or a lookup table) that's indexed by your test statistic.

  • $\begingroup$ The distribution is the one I pointed out, right? We have 10 flips and $p=1/2$. In this case, I think that what I wrote is correct. $\endgroup$ Commented Dec 2, 2022 at 3:00
  • $\begingroup$ Not knowing what level you are answering this at makes it difficult to answer precisely. But no, $\hat{p}$ does not follow a binomial distribution, but $X$ does - neither of which is a "test statistic". If you do a large sample test, you need to refer to or derive the test statistic. Lastly your calculation of the $p$-value is wrong, and without showing steps it's not clear what you mean. $\endgroup$
    – AdamO
    Commented Dec 2, 2022 at 16:47

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