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It is known that 40% of a certain species of birds have characteristic B. Twelve birds of this species are captured in an unusual environment and 4 of them are found to have characteristic B.

Is it reasonable to assume that the birds in this environment have a smaller probability than that the species in general has?

I assumed that this is a binomial case with $p=0.4$ and $n=12.$ Then to figure out if the assumption of the birds having smaller probability than the species is correct, I tried to do it using hypothesis testing and finding the P-value, but I got confused. Any help will be appreciated!

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    Well, under the null hypothesis the probability that you'd see at most $4$ type B birds is $0.438178222$...the probability that you'd see exactly $4$ is $0.21284094$ so I don't see how you could reject the null hypothesis. – 2017-02-18
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    Not reasonable because your sample has 4 whereas the expected value is 3, so this is hardly evidence that this sample has fewer than expected. – 2017-02-18

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Null and Alternative Hypotheses. You want to test $H_0: p = .4$ against $H_a: p < .4.$

In $n = 12$ observations you observe $X = 4$ birds of Type B. If the null hypothesis is true $E(X) = np = 12(.4) = 4.8.$

While it is true that you observed fewer than the 'expected' number of birds of Type B, the question is whether 4 is enough smaller than 4.8 to reject $H_0,$ calling this a 'statistically significant' result.

Finding the P-value. The P-value is the probability (assuming $H_0$ to be true) of a result as extreme or more extreme (in the direction of the alternative) than the observed $X = 4.$

If $X \sim \mathsf{Binom}(n = 12,\, p = .4),$ then the P-value is $P(X \le 4) = 0.4382.$

This can be computed with a calculator using the PDF of $\mathsf{Binom}(12, .4),$ and evaluating $P(X \le 4) = P(X=0) + P(X=1) + \cdots + P(X=4),$ or by using software. The computation in R statistical software is as follows:

pbinom(4, 12, .4)
## 0.4381782

Conclusion. So the P-value of the test is about 0.44, which is not surprisingly small. Testing at the 5% level of significance, one would not reject $H_0$ unless the P-value is less than 0.05. Thus, seeing 4 birds of Type B is consistent with $H_0$ by the usual standards of statistical significance. (This is the same as the conclusion in the Comments of @lulu and @DavidQuinn, even if perhaps not for precisely the same reasons.)

P-value by Normal Approximation. Alternatively, an approximate value of this probability can be found by using the normal approximation to the binomial distribution (with continuity correction): $\mu = E(X) = 4.8,$ as above, and $\sigma = SD(X) = \sqrt{np(1-p)} = 1.6971.$ Then the 'best-fitting' normal distribution is $\mathsf{Norm}(\mu = 4.8, \sigma = 1.6971).$ The approximation is as follows:

$$P(X \le 4.5) = P\left(\frac{X-\mu}{\sigma} \le \frac{4.5-4.8}{1.6971} = -0.1768\right) \approx P(Z \le -0.18) = 0.4286,$$

where $Z$ has a standard normal distribution, so that the approximate probability can be found using printed normal tables. Slightly more accurately (without rounding to use tables), the normal approximation of the P-value can be found using software:

n = 12;  p = .4;  mu = n*p; sg = sqrt(n*p*(1-p))
pnorm(4.5, mu, sg)
## 0.4298419

Sketch of Null Binomial Distribution. Below is a plot of $\mathsf{Binom}(12, .4)$ (black bars) compared with the PDF of the 'best fitting' normal distribution (blue curve). The P-value is the sum of the heights of the bars to the left of the vertical red line.

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