AP Statistics 1.11 Random Sampling- Exam Style Questions - FRQs - New Syllabus

(a)

Step 1: State hypotheses.
Let \(p_A\) = true proportion of banded birds on Island A, and \(p_B\) = true proportion of banded birds on Island B.
\(H_0: p_A – p_B = 0 \qquad H_a: p_A – p_B \neq 0\)
Step 2: Identify the test and check assumptions.
We use a two-sample \(z\)-test for a difference in proportions. The test statistic is:
\(z = \dfrac{\hat{p}_A – \hat{p}_B}{\sqrt{\hat{p}(1-\hat{p})\left(\dfrac{1}{n_1} + \dfrac{1}{n_2}\right)}}\)
The problem states the samples are random. Since the two islands are separate, the samples are independent. We check the large sample condition using the pooled estimate:
\(\hat{p} = \dfrac{n_A\hat{p}_A + n_B\hat{p}_B}{n_A + n_B} = \dfrac{12 + 35}{180 + 220} = \dfrac{47}{400} = 0.1175\)
Expected counts: \(n_A\hat{p} = 21.15,\quad n_A(1-\hat{p}) = 158.85,\quad n_B\hat{p} = 25.85,\quad n_B(1-\hat{p}) = 194.15\)
All expected counts are well above 5, so the large sample condition is satisfied.
Step 3: Compute the test statistic and p-value.
\(\hat{p}_A = \dfrac{12}{180} = 0.067 \qquad \hat{p}_B = \dfrac{35}{220} = 0.159\)
\(z = \dfrac{0.067 – 0.159}{\sqrt{\dfrac{(0.1175)(0.8825)}{180} + \dfrac{(0.1175)(0.8825)}{220}}} = \dfrac{-0.092}{\sqrt{0.00105}} = \dfrac{-0.092}{0.032} = -2.875\)
\(\text{p-value} = 2 \times P(Z < -2.875) \approx 0.00429\)
Step 4: State conclusion in context.
Since the p-value of \(0.00429\) is less than \(\alpha = 0.05\), we reject the null hypothesis. There is convincing statistical evidence that the proportions of banded birds on the two islands are different — Island B has a notably higher proportion of banded birds than Island A.

(b)

We use the capture-recapture logic: the proportion of banded birds in the subsequent sample estimates the proportion of banded birds in the whole population.
For Island A, the number of birds banded in the initial sample is \(n_I = 200\), and the proportion of banded birds observed in the subsequent sample is:
\(\hat{p}_S = \dfrac{12}{180} \approx 0.06667\)
Setting this equal to the fraction of banded birds in the population:
\(\hat{p}_S \approx \dfrac{n_I}{\text{population size}}\)
Solving for the estimated population size:
\(\text{Estimated population size} = \dfrac{n_I}{\hat{p}_S} = \dfrac{200}{12/180} = \dfrac{200 \times 180}{12} = \dfrac{36{,}000}{12} = \boxed{3{,}000 \text{ birds}}\)

(c)

Two concerns that should be addressed before assuming the captures can be treated as random samples are:
Concern 1 — Differential catchability: Some birds may be more likely to be captured than others — for example, slower, older, or less wary birds might be caught at a higher rate than the general population. If the same birds that were easy to capture in the initial sample are also more likely to appear in the subsequent sample, then banded birds would be overrepresented in the subsequent sample, leading us to underestimate the true population size.
Concern 2 — Behavioural change after banding: Birds that were captured and banded in the initial sample may become more trap-shy (avoiding capture in the future) or, conversely, may be more conspicuous to predators due to the bands, altering their survival or behaviour. If banded birds are less likely to be recaptured, we would overestimate the population size. In either case, if banding changes the birds’ behaviour or survival, the subsequent sample can no longer be treated as a true random sample of the population.