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As a check on the relative abundance of certain species of fish in two lakes, \(n=50\) observations are taken on results of net trapping in each lake. For each observation, the experimenter merely records whether the desired species was present in the trap. Past experience has shown that this species appears in lake A traps approximately \(10 \%\) of the time and in lake \(\mathrm{B}\) traps approximately \(20 \%\) of the time. Use these results to approximate the probability that the difference between the sample proportions will be within. 1 of the difference between the true proportions.

Short Answer

Expert verified
The probability that the sample difference is within 0.1 of the true difference is the area between the Z-scores for the values -0.2 and 0.

Step by step solution

01

Identify given values

We are given that in Lake A, the probability of catching the desired species (\(p_A\)) is 0.1, and in Lake B, it is 0.2 (\(p_B\)). The sample size \(n\) for each lake is 50. We need to find the probability that the difference between the sample proportions is within 0.1 of the difference between the true proportions \(p_A - p_B\).
02

Determine true difference in proportions

Calculate the true difference in proportions: \(p_A - p_B = 0.1 - 0.2 = -0.1\). We want the probability that the sample difference \((\hat{p}_A - \hat{p}_B)\) is within 0.1 of this true difference \(-0.1\).
03

Calculate standard deviation of sample proportions

Calculate the standard deviation for each lake's sample proportion using the formula for a proportion's standard deviation: \[\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\]For Lake A, \(\sigma_{\hat{p}_A} = \sqrt{\frac{0.1(0.9)}{50}}\) and for Lake B, \(\sigma_{\hat{p}_B} = \sqrt{\frac{0.2(0.8)}{50}}\).
04

Calculate combined standard deviation

The standard deviation of the difference \((\hat{p}_A - \hat{p}_B)\) is given by: \[\sigma_{diff} = \sqrt{\sigma_{\hat{p}_A}^2 + \sigma_{\hat{p}_B}^2}\]Substitute the values from Step 3 to find \(\sigma_{diff}\).
05

Use Normal Approximation

Assume the distribution of \(\hat{p}_A - \hat{p}_B\) is approximately normal. Use the Z-score to find the probability that the difference is within 0.1 of \(-0.1\):\[P(-0.2 < \hat{p}_A - \hat{p}_B < 0)\] Calculate the Z-scores for \(-0.2\) and \(0\), then find the probability using standard normal distribution tables or a calculator.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
In the context of this problem, the sample proportion is a vital concept. When collecting data, researchers often measure the frequency of a desired outcome within a population. In our case, we are observing whether a specific species of fish is captured in lake traps.
  • For Lake A, the sample proportion (\(\hat{p}_A\)) represents 10% of successful captures in many tries.
  • In Lake B, the sample proportion (\(\hat{p}_B\)) represents a higher capture rate of 20%.
Breaking this down, the sample proportion is simply the ratio of successful events to the total number of trials. It is useful because it serves as a point estimate of the true proportion in the entire population. However, given different sample sizes or rates in various environments (like different lakes), these proportions may vary. Therefore, knowing the limitations and possibilities for variation in these proportions helps us assess differences accurately and judge if our samples provide a good representation of reality. To gain a better understanding, making repeated trials and observing sample proportions becomes essential.
Standard Deviation
Standard deviation is a measure of variation or dispersion within a set of values. In statistical inference, the standard deviation of sample proportions indicates how much the sample proportions are expected to vary from the true proportion. When dealing with proportions, we use a specific formula:\[\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\]Here, \(p\) is the true proportion, and \(n\) is the sample size.
  • For Lake A, we substitute \(p = 0.1\) and \(n = 50\) into the formula to get the standard deviation of lake A's proportion.
  • Similarly, for Lake B, \(p = 0.2\) yields a different standard deviation due to the distinct catch rate.
The general purpose of doing this calculation is to understand the likely variation around the average catch rate in each lake over different samples. If you only draw a few samples, the proportion might fluctuate due to random chance, which is why considering the standard deviation is important. It helps settle the expectation about how spread-out our observed results could be. This makes it an indispensable part of assessing risk and variability in statistical reports.
Normal Approximation
The Normal Approximation is a principle used to simplify the problem of probability estimation in large-sample scenarios. When we deal with probabilities of sample proportions, the sample distribution resembles a bell-shaped curve as sample size increases, which ties to the Central Limit Theorem.In our exercise, the approximation allows us to approximate the distribution of the difference between sample proportions as normal. Given the number of observations, our sample size fulfills the requirements to use Normal Approximation:
  • The distribution of the difference between sample proportions (\(\hat{p}_A - \hat{p}_B\)) can be assumed normal.
  • Once we calculate the combined standard deviation, we can use a standard normal curve (Z-score) to see where these proportions might fall.
The Z-score tells us how many standard deviations an element is from the mean. By identifying areas in the normal distribution with the Z-score, we can predict probabilities of different outcomes. Here, it helps us find how likely the difference between trap success rates in lakes A and B will fall within a specific range, thereby providing insights into the statistical likelihood of observed results relative to known proportions.

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Most popular questions from this chapter

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