91Ó°ÊÓ

The concept of a point estimate is fundamental in statistics when dealing with sample data. It refers to the use of sample data to estimate an unknown population parameter. In this exercise, we are tasked with finding a point estimate for the mortality rate, represented by the symbol \( p \). The point estimate for \( p \) is calculated using the sample proportion. This is done by dividing the number of observed events— in this case, the fish that died (26) — by the total sample size (855).
The formula for the sample proportion is:\[ \hat{p} = \frac{x}{n}, \]where

• \( \hat{p} \) is the point estimate for the population proportion,
• \( x \) is the number of successes (fish that died),
• and \( n \) is the total number of observations (total fish caught and released).

For this specific instance:\[ \hat{p} = \frac{26}{855} \approx 0.0304. \]This means that the estimated proportion of all pike and trout that die upon release is approximately 3.04%.

A confidence interval provides a range of values that is likely to contain the population parameter, here the true mortality rate \( p \). A 99% confidence interval means that we can be 99% confident that the population proportion lies within this interval. The confidence interval is computed using the formula:\[ \hat{p} \pm z \cdot SE, \] where

• \( z \) is the critical value for the standard normal distribution corresponding to the desired confidence level,
• and \( SE \) is the standard error of \( \hat{p} \).

For a 99% confidence interval, the critical value \( z \) is approximately 2.576.Given our previously calculated standard error of 0.0059, the interval becomes:\[ 0.0304 \pm 2.576 \times 0.0059. \]Breaking this down, you calculate:\[ 0.0304 \pm 0.0152, \]which results in the interval:\[ (0.0152, 0.0456). \]This interval suggests that, with 99% confidence, the true mortality rate ranges between 1.52% and 4.56%.

The normal approximation is a method used to approximate the binomial distribution with a normal distribution. This is particularly helpful for calculating probabilities and confidence intervals for large sample sizes. However, for this approximation to be valid, certain conditions must be met. The rule of thumb is that both \( np \) and \( n(1-p) \) should be greater than 5.In our problem:

• \( np = 855 \times 0.0304 \approx 26 \)
• \( n(1-p) = 855 \times (1-0.0304) \approx 829 \)

Both values are far greater than 5, indicating that the normal approximation to the binomial distribution is justified. This ensures that our use of the normal approximation in constructing the confidence interval is appropriate.

The standard error (SE) is a measure of the variation or spread of a sampling distribution of a statistic, often the sample proportion. It provides insight into how far the sample estimate (point estimate) is likely to be from the real population parameter. The formula for computing the standard error of a sample proportion is:\[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \]where

• \( \hat{p} \) is the sample proportion,
• and \( n \) is the sample size.

Plugging in the values from our exercise:\[ SE = \sqrt{\frac{0.0304 \times (1 - 0.0304)}{855}} \approx 0.0059. \]A smaller standard error indicates a more precise estimate of the population proportion, and in this case, our standard error tells us that the point estimate of 3.04% would not vary much.

The binomial distribution is a discrete probability distribution that arises from a sequence of independent experiments or trials, each with two possible outcomes. In this exercise, we're examining the number of fish dying, which fits a binomial context.A random variable follows a binomial distribution when:

• There are a fixed number of trials (in this case, 855 fish).
• Each trial has only two possible outcomes—"success" or "failure." (Fish survives or dies).
• The probability of success (or failure) is the same for each trial (probability \( p \)).
• The trials are independent.

In our scenario, catching and releasing the fish using barbless hooks, with a certain number dying, fits these conditions well, justifying our use of a binomial model to determine the mortality rate. The usefulness of binomial distribution lies in its simplicity and ability to model binary outcomes effectively.