/*! 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 14 The article "Feeding Ecology of ... [FREE SOLUTION] | 91Ó°ÊÓ

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The article "Feeding Ecology of the Red-Eyed Vireo and Associated Foliage- Gleaning Birds" (Ecol. Monogr., 1971: 129-152) presents the accompanying data on the variable \(X=\) the number of hops before the first flight and preceded by a flight. The author then proposed and fit a geometric probability distribution \([p(x)=P(X=x)\) \(=p^{x-1} \cdot q\) for \(x=1,2, \ldots\), where \(\left.q=1-p\right]\) to the data. The total sample size was \(n=130\). \begin{tabular}{l|cccccccccccc} \(\boldsymbol{x}\) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline Number of Times \(\boldsymbol{x}\) Observed & 48 & 31 & 20 & 9 & 6 & 5 & 4 & 2 & 1 & 1 & 2 & 1 \\ & & & & & & & & & & & & \end{tabular} a. The likelihood is \(\left(p^{x_{1}-1} \cdot q\right) \cdots \cdots\left(p^{x_{n}-1} \cdot q\right)=\) \(p^{\Sigma x_{i}-n} \cdot q^{n}\). Show that the mle of \(p\) is given by \(\hat{p}=\left(\sum x_{i}-n\right) / \sum x_{i}\), and compute \(\hat{p}\) for the given data. b. Estimate the expected cell counts using \(\hat{p}\) of part (a) [expected cell counts \(=n \cdot \hat{p}^{x-1} \cdot \hat{q}\) for \(x=1,2, \ldots]\), and test the fit of the model using a \(\chi^{2}\) test by combining the counts for \(x=7,8, \ldots\), and 12 into one cell \((x \geq 7)\).

Short Answer

Expert verified
The MLE for \(p\) is approximately 0.522. Use \(\hat{p}\) to estimate expected counts and conduct a \(\chi^2\) test for goodness-of-fit.

Step by step solution

01

Understand the Geometric Distribution

The probability mass function for a geometric distribution is given by \(p(x) = pq^{x-1}\), where \(p\) is the probability of success, \(q = 1-p\), and \(x\) is the number of trials needed for the first success. In this context, \(x\) represents the number of hops before the first flight and preceded by a flight.
02

Write the Likelihood Function

The likelihood function for the given sample size is \(L(p) = \prod_{i=1}^{n} (p^{x_i-1} q)\). This can be expanded to \(p^{\sum (x_i-1)} q^n\) or equivalently \(p^{\Sigma x_i - n} q^n\), where \(x_i\) are the individual observations.
03

Derive MLE Formula for p

To find the maximum likelihood estimator (MLE) \(\hat{p}\), take the natural log of the likelihood and find its derivative with respect to \(p\). Setting this derivative to zero gives \(\hat{p} = \frac{\sum x_i - n}{\sum x_i}\). This is because \(\frac{\partial}{\partial p} \log L(p) = 0\).
04

Compute \(\hat{p}\) Using Data

Calculate \(\sum x_i\), which in this case is the weighted sum of \(x\) values using the given observed counts. Use the formula \(\hat{p} = \frac{\sum x_i - n}{\sum x_i}\). With the counts given, calculate \(\sum x_i = 48 \times 1 + 31 \times 2 + 20 \times 3 + 9 \times 4 + 6 \times 5 + 5 \times 6 + 4 \times 7 + 2 \times 8 + 1 \times 9 + 1 \times 10 + 2 \times 11 + 1 \times 12 = 272\). Now \(\hat{p} = \frac{272 - 130}{272}\).
05

Calculate the MLE \(\hat{p}\)

Substituting \(\sum x_i = 272\) and \(n = 130\) into the formula \(\hat{p} = \frac{272 - 130}{272} = \frac{142}{272} \approx 0.522\).
06

Estimate Expected Counts

For each \(x\), the expected count is given by \(n \cdot \hat{p}^{x-1} \cdot \hat{q}\), where \(\hat{q} = 1 - \hat{p}\). Calculate these for \(x = 1\) to \(x = 6\), and combine \(x \geq 7\) into one cell.
07

Test Goodness-of-Fit with \(\chi^2\) Test

Compare the observed counts with expected counts using \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\), where \(O_i\) is the observed count and \(E_i\) is the expected count. Calculate this statistic and compare it to the critical value from a \(\chi^2\) distribution with appropriate degrees of freedom.

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Key Concepts

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

Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a method used to estimate the parameters of a statistical model. It finds the parameter values that maximize the likelihood function, which represents the probability of observing the given data. MLE is particularly useful because it provides estimates that have desirable properties such as consistency and asymptotic normality.

**Finding the MLE for the Geometric Distribution**
In the case of the geometric distribution, we want to estimate the probability of success, denoted as \( p \), using the observed data. The likelihood function is constructed as \( L(p) = p^{\Sigma(x_i - 1)} \cdot q^n \), where \( q = 1 - p \) and \( n \) is the sample size. To find the MLE, we need to calculate \( \hat{p} \), which will maximize this likelihood function.

Taking the natural log of the likelihood (for easier differentiation) and finding where its derivative equals zero, gives us the formula for \( \hat{p} \): \[\hat{p} = \frac{\sum x_i - n}{\sum x_i}\] This formula indicates that \( \hat{p} \) relies on the total sum of observations minus the sample size itself, divided by the sum of observations.
Chi-Squared Test
The Chi-Squared test is a statistical method used to determine whether there is a significant difference between expected and observed frequencies in categorical data. It helps in assessing how well a theoretical model fits the observed data.

**Performing the Chi-Squared Test**
To test the fit of the geometric model for the number of hops before the first flight, we use the Chi-Squared test. This involves calculating the chi-squared statistic:\]\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\,\[where \( O_i \) and \( E_i \) represent the observed and expected counts for each category, respectively. After grouping categories such as \( x \geq 7 \) into one for more stable estimates, compare the calculated statistic against a critical value from a chi-squared distribution table with the appropriate degrees of freedom to determine the goodness of fit.
Probability Mass Function
A Probability Mass Function (PMF) describes the probability that a discrete random variable takes on a particular value. For a geometric distribution, the PMF takes the form \( p(x) = pq^{x-1} \), where \( p \) is the probability of success, and \( q = 1-p \).

**Understanding the PMF in Context**
In our exercise, \( x \) denotes the number of hops before a first flight occurs. The PMF reflects the likelihood of observing a specific number of hops before a bird takes off again. Each \( x \) corresponds to a certain probability, and these probabilities sum to 1 across all possible values of \( x \). By using the PMF, we can calculate theoretical expectations (expected counts) which are used in comparison against the observed data in tests like the Chi-Squared test.
Goodness-of-Fit Test
The Goodness-of-Fit test is designed to test how well the observed sample data fit a distribution from a specific model. For example, in this exercise, the test checks whether the geometric distribution is an appropriate model for the data collected on the bird's hopping behavior.

**Carrying Out the Goodness-of-Fit Test**
To perform a Goodness-of-Fit test, expected frequencies are calculated using the estimated parameters (in this case \( \hat{p} \) from MLE), and compared with the observed frequencies. The Chi-Squared statistic is then used to quantify the difference. A large Chi-Squared value indicates a poor fit, while smaller values suggest that the model fits well. By comparing the Chi-Squared statistic to critical values at a given significance level, we determine the fit quality. This test helps researchers understand if the selected model is adequate for describing the observed phenomena.

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Most popular questions from this chapter

Each headlight on an automobile undergoing an annual vehicle inspection can be focused either too high \((H)\), too low \((L)\), or properly \((N)\). Checking the two headlights simultaneously (and not distinguishing between left and right) results in the six possible outcomes \(H H, L L, N N, H L, H N\), and \(L N\). If the probabilities (population proportions) for the single headlight focus direction are \(P(H)=\theta_{1}\), \(P(L)=\theta_{2}\), and \(P(N)=1-\theta_{1}-\theta_{2}\) and the two headlights are focused independently of each other, the probabilities of the six outcomes for a randomly selected car are the following: $$ \begin{aligned} &p_{1}=\theta_{1}^{2} \quad p_{2}=\theta_{2}^{2} \quad p_{3}=\left(1-\theta_{1}-\theta_{2}\right)^{2} \\ &p_{4}=2 \theta_{1} \theta_{2} \quad p_{5}=2 \theta_{1}\left(1-\theta_{1}-\theta_{2}\right) \\ &p_{6}=2 \theta_{2}\left(1-\theta_{1}-\theta_{2}\right) \end{aligned} $$ Use the accompanying data to test the null hypothesis $$ H_{0}: p_{1}=\pi_{1}\left(\theta_{1}, \theta_{2}\right), \ldots, p_{6}=\pi_{6}\left(\theta_{1}, \theta_{2}\right) $$ where the \(\pi_{i}\left(\theta_{1}, \theta_{2}\right)\) 's are given previously. \(\begin{array}{lllllll}\text { Outcome } & H H & L L & N N & H L & H N & L N \\\ \text { Frequency } & 49 & 26 & 14 & 20 & 53 & 38\end{array}\)

A certain type of flashlight is sold with the four batteries included. A random sample of 150 flashlights is obtained, and the number of defective batteries in each is determined, resulting in the following data: \begin{tabular}{l|ccccc} Number Defective & 0 & 1 & 2 & 3 & 4 \\ \hline Frequency & 26 & 51 & 47 & 16 & 10 \end{tabular} Let \(X\) be the number of defective batteries in a randomly selected flashlight. Test the null hypothesis that the distribution of \(X\) is \(\operatorname{Bin}(4, \theta)\). That is, with \(p_{i}=P(i\) defectives), test \(H_{0}: p_{i}=\left(\begin{array}{l}4 \\ i\end{array}\right) \theta^{i}(1-\theta)^{4-i} \quad i=0,1,2,3,4\) [Hint: To obtain the mle of \(\theta\), write the likelihood (the function to be maximized) as \(\theta^{u}(1-\theta)^{v}\), where the exponents \(u\) and \(v\) are linear functions of the cell counts. Then take the natural \(\log\), differentiate with respect to \(\theta\), equate the result to 0 , and solve for \(\hat{\theta}\).]

The accompanying two-way frequency table appears in the article "Marijuana Use in College" (Youth and Society, 1979: 323-334). Each of 445 college students was classified according to both frequency of marijuana use and parental use of alcohol and psychoactive drugs. Does the data suggest that parental usage and student usage are independent in the population from which the sample was drawn? Use the \(P\)-value method to reach a conclusion. Standard Level of Marijuana use Never Occasional Regular $$ \begin{array}{ll|r|r|c} \cline { 3 - 5 } \begin{array}{l} \text { Parental } \\ \text { Use of } \end{array} & \text { Neither } & 141 & 54 & 40 \\ \cline { 3 - 5 } \begin{array}{l} \text { Alcohol } \\ \text { and Drugs } \end{array} & \text { One } & 68 & 44 & 51 \\ \cline { 2 - 4 } & \text { Both } & 17 & 11 & 19 \\ \hline \end{array} $$

In a genetics experiment, investigators looked at 300 chromosomes of a particular type and counted the number of sister-chromatid exchanges on each ("On the Nature of SisterChromatid Exchanges in 5-BromodeoxyuridineSubstituted Chromosomes," Genetics, 1979: 1251-1264). A Poisson model was hypothesized for the distribution of the number of exchanges. Test the fit of a Poisson distribution to the data by first estimating \(\lambda\) and then combining the counts for \(x=8\) and \(x=9\) into one cell. \begin{tabular}{l|llllllllll} \(\boldsymbol{x}=\) Number of Exchanges & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\\ \hline Observed Counts & 6 & 24 & 42 & 59 & 62 & 44 & 41 & 14 & 6 & 2 \\ & & & & & & & & & & \end{tabular}

Sorghum is an important cereal crop whose quality and appearance could be affected by the presence of pigments in the pericarp (the walls of the plant ovary). The article "A Genetic and Biochemical Study on Pericarp Pigments in a Cross Between Two Cultivars of Grain Sorghum, Sorghum Bicolor" (Heredity, 1976: 413-416) reports on an experiment that involved an initial cross between CK60 sorghum (an American variety with white seeds) and Abu Taima (an Ethiopian variety with yellow seeds) to produce plants with red seeds and then a self-cross of the red-seeded plants. According to genetic theory, this \(F_{2}\) cross should produce plants with red, yellow, or white seeds in the ratio \(9: 3: 4\). The data from the experiment follows; does the data confirm or contradict the genetic theory? Test at level \(.05\) using the \(P\)-value approach. \begin{tabular}{l|ccc} Seed Color & Red & Yellow & White \\ \hline Observed Frequency & 195 & 73 & 100 \end{tabular}

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