/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 The Environmental Protection Age... [FREE SOLUTION] | 91Ó°ÊÓ

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The Environmental Protection Agency is concerned with the problem of setting criteria for the amounts of certain toxic chemicals to be allowed in freshwater lakes and rivers. A common measure of toxicity for any pollutant is the concentration of the pollutant that will kill half of the test species in a given amount of time (usually 96 hours for fish species). This measure is called LC50 (lethal concentration killing 50\% of the test species). In many studies, the values contained in the natural logarithm of LC50 measurements are normally distributed, and, hence, the analysis is based on \(\ln (\mathrm{LC} 50)\) data. Studies of the effects of copper on a certain species of fish (say, species A) show the variance of In(LC50) measurements to be around. 4 with concentration measurements in milligrams per liter. If \(n=10\) studies on \(L C 50\) for copper are to be completed, find the probability that the sample mean of \(\ln (\text { LC50 })\) will differ from the true population mean by no more than .5.

Short Answer

Expert verified
The probability that the sample mean will differ by no more than 0.5 is approximately 98.76%.

Step by step solution

01

Identify the Population Variance

The problem states that the variance of \( \ln(\text{LC50}) \) measurements is around 0.4. This value represents the population variance \( \sigma^2 \). The population standard deviation, \( \sigma \), is the square root of the variance, thus \( \sigma = \sqrt{0.4} \approx 0.632 \).
02

Calculate the Standard Error

The standard error (SE) of the sample mean for a sample size \( n \) is given by \( SE = \frac{\sigma}{\sqrt{n}} \). Here, \( n = 10 \), so the standard error is \( SE = \frac{0.632}{\sqrt{10}} \approx 0.200 \).
03

Determine the Z-score for the Given Error Margin

We need to find the Z-score that corresponds to a difference of 0.5 in the sample mean from the population mean. This is calculated using \( Z = \frac{0.5}{SE} \). Substituting the values, we get \( Z = \frac{0.5}{0.200} = 2.5 \).
04

Calculate the Probability

We seek the probability that the difference between the sample mean and the population mean is no more than 0.5, which means we are looking for \( P(-2.5 < Z < 2.5) \). Using the standard normal distribution table, we find that \( P(Z < 2.5) \approx 0.9938 \) and \( P(Z < -2.5) \approx 0.0062 \). Thus, \( P(-2.5 < Z < 2.5) = 0.9938 - 0.0062 = 0.9876 \).
05

Present the Result

The probability that the sample mean of \( \ln(\text{LC50}) \) will differ from the true population mean by no more than 0.5 is approximately 0.9876, or 98.76%.

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Key Concepts

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

LC50 (Lethal Concentration)
The LC50, or Lethal Concentration 50, is a critical measure used in toxicology to assess the potency of a chemical substance. Specifically, it denotes the concentration of a pollutant required to kill 50% of a specified test species within a given time frame. This measure provides a standardized way to compare the toxic impacts of different substances on various organisms.
In environmental studies, LC50 values are extremely important. They help in setting safe chemical exposure limits to protect fish and other aquatic life. For instance, when scientists investigate copper's effects on a fish species, they determine the concentration at which 50% of the fish population does not survive under controlled conditions, commonly set over 96 hours.
The data from such studies is often transformed using the natural logarithm. This transformation can simplify data analysis, especially when such LC50 values tend to follow a log-normal distribution. This is crucial because it allows researchers to apply various statistical methods to make inferences about the toxicity levels effectively.
Standard Error
Standard error is a statistic that measures the accuracy with which a sample mean represents the population mean. It is essential when working with sample data, as it provides insight into how much sample means might deviate from the actual population mean.
In the context of measuring toxic substances, standard error helps assess the reliability of the calculated sample LC50. For example, if scientists conduct multiple studies (let's say 10), and find an average LC50, the standard error gives an approximation of how much this average might vary from the true population LC50.
The formula for standard error is given by: \[\text{SE} = \frac{\sigma}{\sqrt{n}}\]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. In our fish study, with a variance of 0.4, the standard deviation (\( \sigma \)) is about 0.632. Therefore, with ten measurements, the standard error computes to approximately 0.200. This means any sample LC50 mean we calculate is likely to be within 0.200 milligrams per liter of the true population mean.
Normal Distribution
A normal distribution, often referred to as a bell curve, is a fundamental concept in statistics. It represents a continuous probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean.
This distribution plays a pivotal role in statistical analysis because many natural phenomena approximate this pattern, including measurements in biological studies like LC50.
When we analyze LC50 data, assuming that the natural logarithm of LC50 follows a normal distribution simplifies the work. This assumption allows researchers to use well-established statistical tools, such as standard deviation and Z-scores, to make predictions. For instance, determining how likely it is for the sample LC50 mean to be within a specific range of the population mean requires using the normal distribution properties.
In our example, we calculated the probability that the sample mean of \(\ln(\text{LC50})\) will vary no more than 0.5 from the population mean by using the Z-score. From the normal distribution table, this probability was found to be approximately 98.76%, highlighting the robustness of normal distribution in practical analyses.

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Most popular questions from this chapter

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.

The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2. a. What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds \(14.5 ?\) b. Find an interval that includes, with probability \(0.95,\) the average fracture strength of 100 randomly selected pieces of this glass.

An airline finds that \(5 \%\) of the persons who make reservations on a certain flight do not show up for the flight. If the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

a. Find \(t_{05}\) for a \(t\) -distributed random variable with 5 df. b. Refer to part (a). What is \(P\left(T^{2}>t_{.05}^{2}\right) ?\) c. Find \(F_{10}\) for an \(F\) -distributed random variable with 1 numerator degree of freedom and 5 denominator degrees of freedom. d. Compare the value of \(F_{10}\) found in part \((c)\) with the value of \(t_{.05}^{2}\) from parts \((a)\) and \((b)\). e. In Exercise 7.33 , you will show that if \(T\) has a \(t\) distribution with \(\nu\) df, then \(U=T^{2}\) has an \(F\) distribution with 1 numerator degree of freedom and \(\nu\) denominator degrees of freedom. How does this explain the relationship between the values of \(F_{.10}\) (1 num. df, 5 denom df) and \(t_{05}^{2}(5 \mathrm{df})\) that you observed in part \((d)?\)

An experimenter is comparing two methods for removing bacteria colonies from processed luncheon meats. After treating some samples by method A and other identical samples by method B, the experimenter selects a 2 -cubic-centimeter subsample from each sample and makes bacteria colony counts on these subsamples. Let \(X\) denote the total count for the subsamples treated by method A and let \(Y\) denote the total count for the subsamples treated by method B. Assume that \(X\) and \(Y\) are independent Poisson random variables with means \(\lambda_{1}\) and \(\lambda_{2}\), respectively. If \(X\) exceeds Y by more than \(10,\) method \(\mathrm{B}\) will be judged superior to A. Suppose that, in fact, \(\lambda_{1}=\lambda_{2}=50 .\) Find the approximate probability that method B will be judged superior to method A.

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