From the Online Stat Book, Chapter 11.5:

The null hypothesis for the two-tailed test is π = 0.5. By contrast, the null hypothesis for the one-tailed test is π ≤ 0.5.

Why is that so? The null hypothesis for a binomial distribution is that a person has a 50% chance of guessing whether the martini was shaken or stirred. Why would the use of the one-tailed test make it less than 50%?

  • $\begingroup$ Two-tailed one tests hypothesis that true population probability is equal to 0.5, while one-tailed states null-hypothesis differently: true population probability is less or equal to 0.5. $\endgroup$ Commented Mar 6, 2017 at 7:20
  • $\begingroup$ I will try reading a different explanation of one/two tailed tests. I don't understand this one at all. There must be a more thorough explanation somewhere.. $\endgroup$ Commented Mar 6, 2017 at 7:24
  • $\begingroup$ @AndreyKolyadin - I don't understand why exactly it makes it less than 50%. $\endgroup$ Commented Mar 6, 2017 at 7:40
  • $\begingroup$ It doesn't make it less, we say that we care only if our succcess rate is greater than 50%. Two paragraphs under Figure 2 explains it quite well (At least i think so). $\endgroup$ Commented Mar 6, 2017 at 8:20
  • $\begingroup$ I'm too dense.. I've just re-read the whose chapter several times. When I come to the sentence "By contrast, the null hypothesis for the one-tailed test is π ≤ 0.5.", it all breaks down. I cannot understand what it means. $\endgroup$ Commented Mar 6, 2017 at 11:28

3 Answers 3


You wrote

The null hypothesis for a binomial distribution is that

Distributions do not have hypotheses. You have hypotheses. You make them up. The text is awkwardly worded because, while correct, it confused you. The test doesn't make the hypothesis, the hypothesis makes the test.

That is, if the null hypothesis that you wish to test is that $\pi < 0.5$ then you choose a one tailed test. If the null you wish to test is that $\pi \ne 0.5$ then you choose a two tailed test.


I think I know where you are stuck. The linked text states

On the other hand, if our question is whether Mr. Bond is better than chance at determining whether a martini is shaken or stirred, we would use a one-tailed probability

Now, the statement "better than chance" suggests we should test with $H_0: \theta = \frac{1}{2}$, and you are right. So we could test

$$ H_0 : \theta = \frac{1}{2}, \text{ against } H_1: \theta > \frac{1}{2}$$

So why author wrote $H_0: \theta \le \frac{1}{2}$? Maybe he wanted to emphasize that we do not check for $\theta < \frac{1}{2}$ in alternative $H_1$. It may also be important to note that in this case the complex hypothesis

$$H_0 : \theta \le \frac{1}{2}, \text{ against } H_1: \theta > \frac{1}{2}$$

Has the same uniformly most powerful test as the above simple one, and the author just assumed it is known. That is, we use the same test statistic to determine p-value of the test.


I believe that the answer to the original question is that the union of the two hypotheses must be the whole θ space, in this example all θ's between 0 and 1. The complement of θ = 1/2 is θ ≠ 1/2 , and the complement of θ>1/2 is θ≤1/2. In the "Bond" example, if we cannot prove that θ>1/2, it does not necessarily mean that θ=1/2, it only means that θ≤1/2 (he may have a chance of less than half for all we know).


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