/*! 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 6 Suppose that the assumptions ass... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 6

Suppose that the assumptions associated with a multinomial experiment are all satisfied. Then (see Section 5.9 ) each of the \(n_{i}\) 's, \(i=1,2, \ldots, k\), have a binomial distribution with parameters \(n\) and \(p_{i} .\) Further, \(\operatorname{Cov}\left(n_{i}, n_{j}\right)=-n p_{i} p_{j}\) if \(i \neq j\). a. What is \(E\left(n_{i}-n_{j}\right) ?\) b. Refer to part (a). Give an unbiased estimator for \(p_{i}-p_{j}\). c. Show that \(V\left(n_{i}-n_{j}\right)=n\left[p_{i}\left(1-p_{i}\right)+p_{j}\left(1-p_{j}\right)+2 p_{i} p_{j}\right]\). d. Refer to part (c). What is the variance of the unbiased estimator that you gave in part (b)? e. Give a consistent estimator for \(n^{-1} V\left(n_{i}-n_{j}\right)\). f. If \(n\) is large, the estimator that you gave in part (b) is approximately normally distributed with mean \(p_{i}-p_{j}\) and variance \(n^{-2} V\left(n_{i}-n_{j}\right) .\) If \(\hat{p}_{i}=n_{i} / n\) and \(\hat{p}_{j}=n_{j} / n,\) show that a large sample \((1-\alpha) 100 \%\) confidence interval for \(p_{i}-p_{j}\) is given by $$\hat{p}_{i}-\hat{p}_{j} \pm z_{\alpha / 2} \sqrt{\frac{\hat{p}_{i}\left(1-\hat{p}_{i}\right)+\hat{p}_{j}\left(1-\hat{p}_{j}\right)+2 \hat{p}_{i} \hat{p}_{j}}{n}}$$.

Short Answer

Expert verified
a) \(n(p_i - p_j)\); b) \(\hat{p}_i - \hat{p}_j\); c) Derived variance expression; d) \(\frac{1}{n}\) times it; e) Use \(\hat{p}_i, \hat{p}_j\); f) CI using formula.

Step by step solution

01

Calculate Expectation

To find \( E(n_i - n_j) \), use the linearity property of expectation: \( E(n_i - n_j) = E(n_i) - E(n_j) \). Since \( n_i \) and \( n_j \) are binomial distributed, \( E(n_i) = np_i \) and \( E(n_j) = np_j \). Therefore, \( E(n_i - n_j) = np_i - np_j = n(p_i - p_j) \).
02

Unbiased Estimator for Difference in Proportions

To provide an unbiased estimator for \( p_i - p_j \), consider the estimator \( \hat{d} = \hat{p}_i - \hat{p}_j \), where \( \hat{p}_i = \frac{n_i}{n} \) and \( \hat{p}_j = \frac{n_j}{n} \). This is an unbiased estimator because its expected value aligns with the true parameter: \( E(\hat{p}_i - \hat{p}_j) = p_i - p_j \).
03

Calculate Variance of Difference

Use the variance formula for binomial differences: \( V(n_i - n_j) = V(n_i) + V(n_j) - 2 \cdot \text{Cov}(n_i, n_j) \). Calculate each term: \( V(n_i) = np_i(1-p_i) \), \( V(n_j) = np_j(1-p_j) \), and \( \text{Cov}(n_i, n_j) = -np_ip_j \). Thus: \[ V(n_i - n_j) = np_i(1-p_i) + np_j(1-p_j) + 2np_ip_j \].
04

Variance of Unbiased Estimator

The variance of the unbiased estimator \( \hat{d} = \hat{p}_i - \hat{p}_j \) is found by dividing the variance of \( n_i - n_j \) by \( n^2 \). So, the variance of \( \hat{p}_i - \hat{p}_j \) is: \[ V(\hat{d}) = \frac{1}{n^2} \cdot V(n_i - n_j) = \frac{1}{n} \left(p_i(1-p_i) + p_j(1-p_j) + 2p_ip_j\right) \].
05

Consistent Estimator for Variance

To estimate \( n^{-1}V(n_i-n_j) \), use sample proportions: \( \hat{p}_i = n_i/n \) and \( \hat{p}_j = n_j/n \), giving the estimator \[ \frac{1}{n} (\hat{p}_i(1-\hat{p}_i) + \hat{p}_j(1-\hat{p}_j) + 2\hat{p}_i\hat{p}_j) \].
06

Confidence Interval for Proportion Difference

For large \( n \), \( \hat{d} \) is approx normal. A \((1-\alpha)\%\) CI is: \[ \hat{p}_i - \hat{p}_j \pm z_{\alpha/2} \sqrt{\frac{\hat{p}_i(1-\hat{p}_i) + \hat{p}_j(1-\hat{p}_j) + 2\hat{p}_i\hat{p}_j}{n}} \]. This interval is derived using the standard error of the estimator.

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 91影视!

Key Concepts

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

Binomial Distribution
The binomial distribution is a specific probability distribution that describes the number of successes in a fixed number of independent and identically distributed Bernoulli trials. Each trial has two possible outcomes: success with probability \( p \) and failure with probability \( 1-p \). The binomial distribution is defined by two parameters: \( n \), the number of trials, and \( p \), the probability of success in a single trial.
In a multinomial experiment, each category \( n_i \) is binomially distributed when considering the fixed number of trials \( n \) and probability \( p_i \). This links each category's count to the common underlying process characterized by \( n \) and their respective probabilities \( p_i \).
Key features of the binomial distribution include:
  • Mean: \( E(n_i) = np_i \)
  • Variance: \( V(n_i) = np_i(1-p_i) \)
These properties are instrumental in determining characteristics of multinomial outcomes and further analyses, such as confidence intervals and covariance relationships between different categories.
Unbiased Estimator
An unbiased estimator is a statistical technique used to estimate parameters wherein the expected value of the estimator is equal to the true parameter value. This means that, on average, the estimator neither overestimates nor underestimates the parameter.
For estimating the difference in proportions \( p_i - p_j \) within the context of our multinomial setup, the unbiased estimator is \( \hat{d} = \hat{p}_i - \hat{p}_j \). This estimator is calculated as the difference between the estimated proportions \( \hat{p}_i = \frac{n_i}{n} \) and \( \hat{p}_j = \frac{n_j}{n} \).
The effectiveness of this estimator lies in its property that \( E(\hat{d}) = p_i - p_j \), ensuring that it provides a true representation of the parameter differences across repeated samples.
Confidence Interval
A confidence interval gives a range of plausible values for an unknown parameter and is constructed to capture this parameter a certain percentage of the time in repeated sampling. The most common confidence interval is constructed for proportion differences in statistics.
In the multinomial experiment context, when the sample size \( n \) is large, the difference of proportions estimator \( \hat{d} \) can be approximated by a normal distribution. As a result, a \((1-\alpha)100\%\) confidence interval for \( p_i - p_j \) can be expressed as:
\[ \hat{p}_i - \hat{p}_j \pm z_{\alpha / 2} \sqrt{\frac{\hat{p}_i(1-\hat{p}_i) + \hat{p}_j(1-\hat{p}_j) + 2\hat{p}_i \hat{p}_j}{n}} \]
The term \( z_{\alpha / 2} \) is the critical value from the standard normal distribution, and the square root part expresses the standard error for the difference of proportions. This setup allows for a robust assessment of where the true difference \( p_i - p_j \) lies, with a specified level of certainty.
Covariance
Covariance is a measure indicating the extent to which two random variables change in tandem. In the context of a multinomial experiment, covariance helps describe how different outcomes might co-vary or change together, reflecting inter-relationships between categories.
For the binomials \( n_i \) and \( n_j \), their covariance is given by \( \text{Cov}(n_i, n_j) = -np_ip_j \), which is negative. This negative covariance highlights that increases in one category likely result in decreases in another. This makes sense in a multinomial context where the total trials \( n \) are fixed, thus one category's gains imply another's losses.
Understanding covariance is crucial for forming more complex estimates and calculations of variance, such as those required when dealing with differences in category proportions across the multinomial outcomes.

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

On the 40th anniversary of President John F. Kennedy's assassination, a FOX news poll showed that most Americans disagree with the government's conclusions about the killing. The Warren Commission found that Lee Harvey Oswald acted alone when he shot Kennedy, but many Americans are not so sure about this conclusion. Do you think that we know all of the relevant facts associated with Kennedy's assassination, or do you think that some information has been withheld? The following table contains the results of a nationwide poll of 900 registered voters. \(^{\star}\) $$\begin{array}{lccc} \hline & \begin{array}{c} \text { We Know All } \\ \text { Relevant } \\ \text { Facts } \end{array} & \begin{array}{c} \text { Some Relevant } \\ \text { Facts Withheld } \end{array} & \text { Not Sure } \\ \hline \text { Democrat } & 42 & 309 & 31 \\ \text { Republican } & 64 & 246 & 46 \\ \text { Other } & 20 & 115 & 27 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate a dependence between party affiliation and opinion about a possible cover-up? Test using \(\alpha=.05\) b. Give bounds for the associated \(p\) -value and interpret the result. c. Use the \(\chi^{2}\) applet to obtain the approximate \(p\) -value. d. Why is the value you obtained in part (c) "approximate"?

Counts on the number of items per cluster (or colony or group) must necessarily be greater than or equal to \(1 .\) Thus, the Poisson distribution generally does not fit these kinds of counts. For modeling counts on phenomena such as number of bacteria per colony, number of people per household, and number of animals per litter, the logarithmic series distribution often proves useful. This discrete distribution has probability function given by $$p(y | \theta)=-\frac{1}{\ln (1-\theta)} \frac{\theta^{y}}{y}, \quad y=1,2,3, \ldots, 0<\theta<1$$, where \(\theta\) is an unknown parameter. a. Show that the MLE \(\hat{\theta}\) of \(\theta\) satisfies the equation $$\bar{Y}=\frac{\hat{\theta}}{-(1-\hat{\theta}) \ln (1-\hat{\theta})}, \quad \text { where } \quad \bar{Y}=\frac{1}{n} \sum_{i=1}^{n} Y_{i}$$.

Suppose that \(\left(Y_{1}, Y_{2}, \ldots, Y_{k}\right)\) has a multinomial distribution with parameters \(n, p_{1}, p_{2}, \ldots, p_{k},\) and \(\left(X_{1}, X_{2}, \ldots, X_{k}\right)\) has a multinomial distribution with parameters \(m, p_{1}^{*}, p_{2}^{*}, \ldots, p_{k}^{*} .\) Construct a test of the null hypothesis that the two multinomial distributions are identical; that is, test \(H_{0}\) : \(p_{1}=p_{1}^{*}, p_{2}=p_{2}^{*}, \ldots, p_{k}=p_{k}^{*}\).

An interesting and practical use of the \(\chi^{2}\) test comes about in testing for segregation of species of plants or animals. Suppose that two species of plants, \(\mathrm{A}\) and \(\mathrm{B}\), are growing on a test plot. To assess whether the species tend to segregate, a researcher randomly samples \(n\) plants from the plot; the species of each sampled plant, and the species of its nearest neighbor are recorded. The data are then arranged in a table, as shown here. If \(a\) and \(d\) are large relative to \(b\) and \(c,\) we would be inclined to say that the species tend to segregate. (Most of A's neighbors are of type \(A\), and most of \(B\) 's neighbors are of type \(B\).) If \(b\) and \(c\) are large compared to \(a\) and \(d\), we would say that the species tend to be overly mixed. In either of these cases (segregation or over mixing), a \(\chi^{2}\) test should yield a large value, and the hypothesis of random mixing would be rejected. For each of the following cases, test the hypothesis of random mixing (or, equivalently, the hypothesis that the species of a sample plant is independent of the species of its nearest neighbor). Use \(\alpha=.05\) in each case. a. \(a=20, b=4, c=8, d=18\) b. \(a=4, b=20, c=18, d=8\) c. \(a=20, b=4, c=18, d=8\)

Suppose that the entries in a contingency table that appear in row \(i\) and column \(j\) are denoted \(n_{i j},\) for \(i=1,2, \ldots, r\) and \(j=1,2, \ldots, c ;\) that the row and column totals are denoted \(r_{i},\) for \(i=1,2, \ldots, r,\) and \(c_{j},\) for \(j=1,2, \ldots, c ;\) and that the total sample size is \(n\) a. Show that $$ X^{2}=\sum_{j=1}^{c} \sum_{i=1}^{r} \frac{\left[n_{i j}-E \widehat{\left(n_{i j}\right)}\right]^{2}}{E\left(n_{i j}\right)}=n\left(\sum_{j=1}^{c} \sum_{i=1}^{r} \frac{n_{i j}^{2}}{r_{i} c_{j}}-1\right) $$ Notice that this formula provides a computationally more efficient way to compute the value of \(X^{2}\) b. Using the preceding formula, what happens to the value of \(X^{2}\) if every entry in the contingency table is multiplied by the same integer constant \(k>0 ?\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
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.