/*! 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 17 An article in Annals of Mathemat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An article in Annals of Mathematical Statistics reports the following data on the number of borers in each of 120 groups of borers. Does the Poisson pmf provide a plausible model for the distribution of the number of borers in a group? [Hint: Add the frequencies for \(7,8, \ldots\), 12 to establish a single category " \(\geq 7 . "]\) $$ \begin{array}{l|llllllllllllll} \begin{array}{l} \text { Number } \\ \text { of Borers } \end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text { Frequency } & 24 & 16 & 16 & 18 & 15 & 9 & 6 & 5 & 3 & 4 & 3 & 0 & 1 \end{array} $$

Short Answer

Expert verified
Calculate combined frequencies, find mean \(\lambda=2.1\), get expected Poisson frequencies, and use Chi-square test to assess fit.

Step by step solution

01

Combine Frequencies for Values Greater than or Equal to Seven

Given that we need to combine the frequencies for the values 7 through 12, let's sum them up. The frequencies for each value are:- 7: 5- 8: 3- 9: 4- 10: 3- 11: 0- 12: 1Thus, the total frequency for the category '≥ 7' is:\[5 + 3 + 4 + 3 + 0 + 1 = 16\]
02

Calculate Expected Frequencies Using the Poisson Distribution

To obtain the expected frequencies for a Poisson distribution, we need to calculate its mean (λ). The mean can be computed by multiplying each value by its frequency, summing those products, and then dividing by the total number of samples (120). \[\lambda = \frac{(0 \times 24) + (1 \times 16) + (2 \times 16) + (3 \times 18) + (4 \times 15) + (5 \times 9) + (6 \times 6) + (7 \times 5) + (8 \times 3) + (9 \times 4) + (10 \times 3) + (11 \times 0) + (12 \times 1)}{120}\]Computing this gives:\[\lambda = \frac{252}{120} = 2.1\]
03

Calculate Poisson Probabilities

Using the mean \( \lambda = 2.1 \), compute the probability \( P(X = k) \) for each borer number \( k \) using the Poisson probability mass function:\[P(X = k) = \frac{e^{-\lambda} \times \lambda^k}{k!}\]Calculate for each number from 0 through 6, and cumulatively for '≥ 7'. This will involve calculating the Poisson probabilities and then multiplying by 120 to get expected frequencies.
04

Compare Observed to Expected Frequencies using Chi-square Test

Calculate the expected frequencies from Poisson probabilities and compare them to the observed frequencies. Use the Chi-square goodness-of-fit test to determine if the differences are statistically significant:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. Sum this formula for each category and compare with a Chi-square distribution with appropriate degrees of freedom to determine if the Poisson distribution is a good fit.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chi-square test
The Chi-square test is a statistical method used to assess whether observed frequencies differ significantly from expected frequencies. In this context, it's applied as a **goodness-of-fit** test. This means it checks how well the observed data from the borer groups fits a Poisson distribution model. To perform this test, you calculate the Chi-square statistic using the formula: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Here,
  • \(O_i\) represents the observed frequency for each category of borers.
  • \(E_i\) is the expected frequency under the Poisson model.
Working through this process allows us to quantify the difference between our observed counts and what we'd expect if the Poisson model accurately describes our data. After calculating the \(\chi^2\) value, it is compared to a critical value from the Chi-square distribution table with corresponding degrees of freedom. This comparison helps decide if any deviations from the expected values could be attributed simply to random chance.
Goodness-of-fit
When we talk about the **goodness-of-fit** for a distribution, we are essentially asking: "How well does our model fit the data?" In this exercise, a Poisson distribution is used as a model. It predicts the frequency of events (like the number of borers per group) by first determining the average rate at which they occur.

The goal is to see if the observed frequencies of borers matches what's expected according to the Poisson distribution. Here’s how it’s typically done:
  • Calculate the expected frequencies for each category based on the Poisson model.
  • Use these expected frequencies to perform a Chi-square test.
Once both observed and expected frequencies are compared, we gauge whether the differences are just random variation. A close match would indicate the Poisson distribution is a good fit; substantial differences would suggest our model might not be appropriate, signaling that a different distribution could describe the data better.
Expected frequencies
Expected frequencies are crucial when performing a Chi-square goodness-of-fit test. They represent the number of occurrences we'd anticipate for each category, assuming our model is correct. In the case of the Poisson distribution, these are computed using the known mean \((\lambda)\) of the data.First, calculate \(\lambda\), the average number of borers:
\[\lambda = \frac{252}{120} = 2.1\]With \(\lambda = 2.1\), the probability for each category \((k)\) is found using:
\[P(X = k) = \frac{e^{-2.1} \times 2.1^k}{k!}\]From these probabilities, multiply by the total sample size (120) to determine each expected frequency. For instance, if the probability \(P(X = 0)\) is calculated to be a certain value, multiplying by 120 gives the expected number of groups with zero borers.Finally, these computed expected frequencies are directly compared to the actual observed frequencies to assess the goodness-of-fit of the Poisson model. The comparison reveals whether the observed distribution aligns well with theoretical predictions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Many shoppers have expressed unhappiness because grocery stores have stopped putting prices on individual grocery items. The article "The Impact of Item Price Removal on Grocery Shopping Behavior" (J. Market., 1980: 73-93) reports on a study in which each shopper in a sample was classified by age and by whether he or she felt the need for item pricing. Based on the accompanying data, does the need for item pricing appear to be independent of age? $$ \begin{array}{lccccc} & \multicolumn{5}{c}{\text { Age }} \\ \cline { 2 - 6 } & <\mathbf{3 0} & \mathbf{3 0 - 3 9} & \mathbf{4 0 - 4 9} & \mathbf{5 0 - 5 9} & \geq \mathbf{6 0} \\ \hline \begin{array}{l} \text { Number } \\ \text { in Sample } \end{array} & 150 & 141 & 82 & 63 & 49 \\ \begin{array}{l} \text { Number } \\ \text { Who Want } \\ \text { Item Pricing } \end{array} & 127 & 118 & 77 & 61 & 41 \\ & & & & & \end{array} $$

The article "A Probabilistic Analysis of Dissolved Oxygen-Biochemical Oxygen Demand Relationship in Streams" \((J\). Water 91Ó°ÊÓ Control Fed., 1969: 73-90) reports data on the rate of oxygenation in streams at \(20^{\circ} \mathrm{C}\) in a certain region. The sample mean and standard deviation were computed as \(\bar{x}=.173\) and \(s=.066\), respectively. Based on the accompanying frequency distribution, can it be concluded that oxygenation rate is a normally distributed variable? Use the chisquared test with \(\alpha=.05\). $$ \begin{array}{lc} \hline \text { Rate (per day) } & \text { Frequency } \\ \hline \text { Below .100 } & 12 \\ .100 \text {-below .150 } & 20 \\ .150 \text {-below .200 } & 23 \\ .200 \text {-below .250 } & 15 \\ .250 \text { or more } & 13 \\ \hline \end{array} $$

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}

An information retrieval system has ten storage locations. Information has been stored with the expectation that the long-run proportion of requests for location \(i\) is given by the expression \(p_{i}=(5.5-|i-5.5|) / 30\). A sample of 200 retrieval requests gave the following frequencies for locations 1-10, respectively: \(4,15,23,25,38,31,32\), 14,10 , and 8 . Use a chi-squared test at significance level .10 to decide whether the data is consistent with the a priori proportions (use the \(P\)-value approach).

Criminologists have long debated whether there is a relationship between weather conditions and the incidence of violent crime. The author of the article "Is There a Season for Homicide?" (Criminology, 1988: 287-296) classified 1361 homicides according to season, resulting in the accompanying data. Test the null hypothesis of equal proportions using \(\alpha=.01\) by using the chi-squared table to say as much as possible about the \(P\)-value. \begin{tabular}{llll} Winter & Spring & Summer & Fall \\ \hline 328 & 334 & 372 & 327 \end{tabular}

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.