Finding the p-value for a one-tailed test

The whale-scaring hypothesis test involved a random sample of size n from a population with probability π of success (whale leaving immediately). The data collected were x successes, and we tested the hypotheses...

H0 :   π  =  π0
HA :   π  >  π0

where π0 was the constant of interest (e.g. 0.4 in this example). The following steps were followed to obtain the p-value for the test.

  1. The sample proportion of successes, p, was identified as the most informative summary statistic about π.
  2. The number of successes, x = np has a standard binomial distribution with no unknown parameters when H0 holds, so it is a better test statistic.
  3. The p-value is a sum of tail probabilities for this binomial distribution.

The diagram below illustrates these steps

The rat-learning example was similar, but the alternative hypothesis involved high values of π and the p-value was found by counting upper tail probabilities.

Finding the p-value for a two-tailed test

The appropriate tail probability to use depends on the alternative hypothesis. If the alternative hypothesis allows either high or low values of x, the test is called a two-tailed test,

H0 :   π  =  π0
HA :   π  ≠  π0

The p-value is then double the smaller tail probability since values of x in both tails of the binomial distribution would provide evidence for HA.

Somali blood groups

In a study of sab bondsmen, a population sub-group in Northern Somalia, blood tests were conducted on a sample of 54, in order to investigate whether they differed genetically from the main population of 'free-born noble Somali'.

It is known that a proportion 0.574 of free-born noble Somali have blood group O. (Actually 574 had blood group O in a sample of 1000, but this sample size was large enough to provide a reasonably accurate estimate.) Is there any evidence that the sample proportion with blood group O in the sab bondsmen, 26 out of 54, does not come from a population with π  = 0.574. This can be expressed as the hypotheses

H0 :   π  =  0.574
HA :   π  ≠  0.574

We would expect (0.574 x 54) = 31 of the sab bondsmen to have blood group O. A sample count that is either much greater than 31 or much less than 31 would suggest a genetic difference between the sab bondsmen and the free-born noble Somali. Use the slider below to obtain the p-value.

The probability of getting as few as 26 is 0.1084. Since this is a 2-tailed test, we must also take account of the probability of getting a count that is as unusually high, so the p-value is twice this, 0.2169. Getting 26 sab bondsmen with blood group O is therefore not unlikely, so we conclude that there is no evidence from these data of a genetic difference between sab bondsmen and the free-born Somali.