Estimating the Deer Population on Campus

The university has noticed more deer wandering across campus, and Facilities Management is worried about overgrazing and deer–vehicle collisions. They have asked your wildlife statistics team to provide a reliable estimate of the deer population living on campus.

You decide to use a capture–recapture method combined with a fixed width sequential confidence interval approach so that you only collect as much data as needed to reach a target precision.

Preparation

Capture-recapture method?

Fixed Width sequential confidence interval

Phase I: Initial Marking

On the first night, you safely capture and tag \(M=20\) deer with bright tail markers. These deer are released unharmed back into the campus woods.

Phase 2: Recapture Sampling

On each subsequent night, you set up camera traps and spotlight surveys to observe a group of \(n\) deer. You record how many of them, \(m\), are already tagged.

If \(\frac{m}{n}\) of your sample is tagged, that should reflect the fraction \(____\) of the entire population that is tagged.

Estimate:

\[ \hat{N}=? \]

Phase 3: Estimation

Central Limit Theorem

By the Central Limit Theorem, when the sample size is sufficiently large, most estimators are approximately normally distributed. In particular, once we know (or can estimate) the standard deviation of an estimator, we can model its distribution using a normal curve and construct confidence intervals accordingly.

\[ \hat{\theta} \sim N(\theta, \text{SE}(\hat{\theta})^2) \]

where \(\text{SE}(\hat{\theta})\) is the standard deviation of \(\hat{\theta}\), \(\text{SE}(\hat{\theta})^2\) is the variance of \(\hat{\theta}\).

For example, we can use sample mean \(\bar{X}\) to estimate the population mean \(\mu\), (\(\hat{\mu} = \bar{X}\)) . Then \(\text{Var}(\hat{\mu})=\text{Var}(\bar{X})=\frac{\sigma^2}{n}\), which gives us

\[ \hat{\mu}=\bar{X}\sim N(\mu, \frac{\sigma^2}{n}) \]

If the value of \(\sigma\) is unknown, it can be replaced by its estimate, the sample standard deviation \(S\) . Thus

\[ \hat{\mu} = \bar{X} \sim N(\mu, \frac{S^2}{n})=N(\mu, \left(\frac{S}{\sqrt{n}}\right)^2) \]

That’s why we may use the following formula to construct the confidence interval for \(\mu\)

\[ \text{Estimate} \pm \text{Critical Value}\times \text{SE} = \bar{X} \pm 1.96 \frac{S}{\sqrt{n}} \]

Delta Method

What if our goal is to estimate a function of the parameter, such as \(g(\mu)\)? To get the variance of \(g(\hat{\mu})\), we will apply the delta method.

If \(g(\mu)\) is steep (large derivative), then small changes in \(\hat\mu\) lead to bigger swings in \(g(\hat{\mu})\).
If \(g(\mu)\) is flat, then \(g(\hat{\mu})\) hardly moves at all, so the variance shrinks.

Thus we will need to evaluate the slope of \(g(\mu)\), given by \(g'(\mu)\) .

\[ \text{Var}(g(\hat{\mu})) = (g'(\mu))^2 \text{Var}(\hat{\mu}) \]

For example, to find the variance of \(g(\mu)=\mu^2\), we first compute \(g'(\mu)=2\mu\), then

\[ \text{Var}(g(\hat{\mu})) = 4 {\mu}^2 \text{Var}(\hat{\mu}) \approx\frac{4 \mu^2 \sigma^2}{n} \]

Now plug in the estimate \(\hat{\mu}=\bar{X}\) and \(\hat{\sigma}=S\) ,

\[ \text{Var}(g(\hat{\mu})) = 4 {\mu}^2 \text{Var}(\hat{\mu}) \approx\frac{4 \bar{X}^2 S^2}{n} \]

Can you use the delta method to compute \(\text{SE}(\hat{N})\) , the standard deviation of \(\hat{N}\) in Phase 2? You may find it helpful to use some properties of the hypergeometric distribution.

Then the confidence interval is

\[ \hat{N} \pm 1.96 \times \text{SE}(\hat{N}) \]

Phase 4: Sequential Sampling

Check the width of this interval. We sample 20 deer on the first night. If the total width is less than or equal to the required precision (say, \(\pm\) 20 deer), stop the sampling. If not, continue to sample 5 more the next night.

Night \(k\)	Cumulative \(n_k\)	Cumulative Marked Recapture \(m_k\)
1	20	1
2	25	1
3	30	3
4	35	5
5	40	7
6	45	9
7	50	10
8	55	11
9	60	12
10	65	14
…	…	…