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It is proposed to fit the Poisson distribution to the following data: \begin{tabular}{c|ccccc} \(x\) & 0 & 1 & 2 & 3 & \(3

Short Answer

Expert verified
The chi-square test gives us the chi-square statistic and the degrees of freedom. The comparison of the computed p-value with the significance level allows the rejection or the acceptance of the Poisson model.

Step by step solution

01

Calculate the Mean

Firstly, to fit the Poisson model, we need to find its parameter, lambda (λ), which is the mean of the distribution. We can calculate the mean using the formula: \(\lambda = \frac{\sum x \cdot f}{\sum f}\), where x represents the given x values and f being their corresponding frequencies. For \(3<x\), treat \(x=4\). Therefore, \(\lambda = \frac{(0\cdot20)+(1\cdot40)+(2\cdot16)+(3\cdot18)+(4\cdot6)}{20+40+16+18+6}\).
02

Compute the Expected Frequencies

Next, use the computed mean from step 1 to calculate the expected frequency for each x using the Poisson model formula: \(E= n \cdot p\), where \(n\) is the total number of observations and \(p\) is the probability of each x which can be calculated using the Poisson model formula: \(p = \frac{e^{-\lambda} \cdot \lambda^x}{x!}\).
03

Compute the Chi-Square Statistic

With the expected and observed frequencies at hand, we can now calculate the chi-square statistic. The formula for this is \(\chi^2 = \sum \frac{(o - e)^2}{e}\) where \(o\) stands for observed frequency and \(e\) stands for expected frequency.
04

Calculate Degrees of Freedom

Next, we need to compute the degrees of freedom for the chi-square test. The formula used is: \(df = k - 1 - p\), where k represents the number of possible outcomes (in this case, 5) and p represents the number of estimated parameters (in this case, 1 since we've only computed λ). Thus, \(df = 5 - 1 - 1\).
05

Interpretation

To understand whether the observed frequencies significantly deviate from the expected frequencies as per Poisson model we use the calculated Chi-square value and the calculated degrees of freedom to find the p-value from Chi-square distribution tables. If the p-value is less than the significance level (which is 0.05 in this problem), then the deviation is significant and we will reject the Poisson model. If it is greater than 0.05, there is no significant deviation so we will not reject the Poisson model.

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Key Concepts

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

Poisson distribution
The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. It is characterized by the mean rate of occurrence, \( \lambda \), known as lambda, which is the expected number of occurrences in the given interval.

For example, it can be used to model the number of emails one receives in an hour or the number of stars in a given volume of space. In the exercise provided, the Poisson distribution is used to analyze the frequency of occurrences for different values of \( x \) and see if the distribution model fits the observed data.
Chi-square statistic
The chi-square statistic is a measure used in statistics to quantify how a set of observed values compares to the set of expected values if a certain hypothesis is true. It's essential in tests of significance and can indicate whether the observed distribution differs from the expected distribution according to a specific statistical model.

To calculate the chi-square statistic, one would use the formula \( \chi^2 = \sum \frac{(o - e)^2}{e} \), where \( o \) is the observed frequency, and \( e \) is the expected frequency under the theoretical model. A high chi-square value suggests a significant difference between the observed and expected values, which implies that the model may not fit the data well.
Degrees of freedom
Degrees of freedom (df) in statistics refers to the number of independent values or quantities which can be assigned to a statistical distribution. In the context of the chi-square test, the degrees of freedom are calculated as the number of observed categories reduced by the number of parameters estimated from the data and by one (since total frequency is fixed).

The formula used is \( df = k - 1 - p \), where \( k \) is the number of categories and \( p \) is the number of parameters estimated. Degrees of freedom is crucial because it determines the shape of the chi-square distribution, which in turn dictates the critical value needed to assess the significance of the chi-square statistic.
Goodness-of-fit test
A goodness-of-fit test is a type of statistical test used to determine how well a set of observed values fit a specific distribution. It is a way to assess if a model is appropriate for the data.

In this case, the chi-square goodness-of-fit test is employed to evaluate whether the observed frequency distribution of events fits the Poisson distribution. The test compares the observed frequencies to the expected frequencies calculated under the assumption that the Poisson model holds true and computes a chi-square statistic. Depending on this statistic and corresponding p-value, one can conclude whether or not to reject the hypothesized distribution as a valid model for the observed data. If the p-value is less than the chosen significance level (e.g., \( \alpha=0.05 \) in the exercise), one rejects the model, suggesting that the data do not fit the proposed distribution well.

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Most popular questions from this chapter

Let the observed value of the mean \(\bar{X}\) and of the sample variance of a random sample of size 20 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) be \(81.2\) and \(26.5\), respectively. Find respectively \(90 \%, 95 \%\) and \(99 \%\) confidence intervals for \(\mu .\) Note how the lengths of the confidence intervals increase as the confidence increases.

Recall For the baseball data (bb.rda), 15 out of 59 ballplayers are lefthanded. Let \(p\) be the probability that a professional baseball player is left-handed. Determine an exact \(90 \%\) confidence interval for \(p .\) Show first that the equations to be solved are: $$ \sum_{j=0}^{14}\left(\begin{array}{c} n \\ j \end{array}\right) \underline{\theta}^{j}(1-\underline{\theta})^{n-j}=0.95 \text { and } \sum_{j=0}^{15}\left(\begin{array}{c} n \\ j \end{array}\right) \bar{\theta}^{j}(1-\bar{\theta})^{n-j}=0.05 $$ Then do the following steps to obtain the confidence interval. (a) Show that \(0.10\) and \(0.17\) bracket the solution to the first equation. (b) Show that \(0.34\) and \(0.38\) bracket the solution to the second equation. (c) Then use the \(\mathrm{R}\) function binomci. \(\mathrm{r}\) to solve the equations.

Let \(X\) and \(Y\) denote independent random variables with respective probability density functions \(f(x)=2 x, 0

Let the result of a random experiment be classified as one of the mutually exclusive and exhaustive ways \(A_{1}, A_{2}, A_{3}\) and also as one of the mutually exhaustive ways \(B_{1}, B_{2}, B_{3}, B_{4}\). Say that 180 independent trials of the experiment result in the following frequencies: \begin{tabular}{|c|c|c|c|c|} \hline & \(B_{1}\) & \(B_{2}\) & \(B_{3}\) & \(B_{4}\) \\ \hline\(A_{1}\) & \(15-3 k\) & \(15-k\) & \(15+k\) & \(15+3 k\) \\ \hline\(A_{2}\) & 15 & 15 & 15 & 15 \\ \hline\(A_{3}\) & \(15+3 k\) & \(15+k\) & \(15-k\) & \(15-3 k\) \\ \hline \end{tabular} where \(k\) is one of the integers \(0,1,2,3,4,5\). What is the smallest value of \(k\) that leads to the rejection of the independence of the \(A\) attribute and the \(B\) attribute at the \(\alpha=0.05\) significance level?

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be the values of a random sample. A bootstrap sample, \(\mathbf{x}^{* \prime}=\left(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\right)\), is a random sample of \(x_{1}, x_{2}, \ldots, x_{n}\) drawn with replacement. (a) Show that \(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\) are iid with common cdf \(\widehat{F}_{n}\), the empirical cdf of \(x_{1}, x_{2}, \ldots, x_{n}\) (b) Show that \(E\left(x_{i}^{*}\right)=\bar{x}\) (c) If \(n\) is odd, show that median \(\left\\{x_{i}^{*}\right\\}=x_{((n+1) / 2)}\). (d) Show that \(V\left(x_{i}^{*}\right)=n^{-1} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\).

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