Question
Every July, as part of a research project, Rita collects data about sightings of a particular kind of bird.
Each day in July she notes whether she sees this kind of bird or not, and she records the number X of days on which she sees it. She models the distribution of X by B(31, p), where p is the probability of seeing this kind of bird on a randomly chosen day in July.
Data from previous years suggests that p = 0.3, but in 2022 Rita suspected that the value of p had been reduced. She decided to carry out a hypothesis test.
In July 2022, she saw this kind of bird on 4 days.
(a) Use the binomial distribution to test at the 5% significance level whether Rita’s suspicion is justified.
In July 2023, she noted the value of X and carried out another test at the 5% significance level using the same hypotheses.
(b) Calculate the probability of a Type I error.
Rita models the number of sightings, Y, per year of a different, very rare, kind of bird by the distribution B(365, 0.01).
(c) (i) Use a suitable approximating distribution to find P(Y=4).
(ii) Justify your approximating distribution in this context.
▶️Answer/Explanation
Solution: –
7(a) $H_0: p=0.3$
$H_1: p<0.3$
$P(X\le4) =$
\[0.7^{31}+31\times0.7^{30}\times0.3+{}^{31}C_{2}\times0.7^{29}\times0.3^{2}+{}^{31}C_{3}\times0.7^{28}\times0.3^{3}+^{31}C_{4}\times 0.7^{24}\times 0.3^{4}\]
\[= 0.00001577 + 0.0002096 + 0.0013475 + 0.0055826 + 0.016748\]
$$= 0.0239 (3sf)$$
‘There is sufficient evidence (at 5% level) to support Rita’s
suspicion’, or ‘There is sufficient evidence to suggest the probability of seeing
this type of bird has decreased’
7(b) $P(X\le5) = [0.0239 + ^{31}C_{5}\times0.7^{26}\times0.3^{5}] = 0.0627$ [which is > 0.05]
7(c) (i) $[\lambda=] 3.65$
$e^{-3.65} \times \frac{3.65^{4}}{4!}$
= 0.192 (3sf)
7(c) (ii) $n = 365 > 50$
$np = 3.65 < 5$ or $p = 0.01 < 0.1$