/*! 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 2 Consider the following frequency... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following frequency table of observations on the random variable \(X\). $$ \begin{aligned} &\begin{array}{rrrrr} 0 & 1 & 2 & 3 & 4 \\ 24 & 30 & 31 & 11 & 4 \end{array}\\\ &\begin{array}{l} \text { Values } \\ \text { Observed frequency } \end{array} \end{aligned} $$ a. Based on these 100 observations, is a Poisson distribution with a mean of 1.2 an appropriate model? Perform a goodness-of-fit procedure with \(\alpha=0.05 .\) b. Calculate the \(P\) -value for this test.

Short Answer

Expert verified
The Poisson model with mean 1.2 is not a good fit if the test statistic exceeds the critical value, and the P-value is less than 0.05.

Step by step solution

01

Calculate the Expected Frequencies

First, calculate the probability for each possible value of \(X\) under a Poisson distribution with mean \(\lambda = 1.2\). The formula for the Poisson probability mass function is \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \). Calculate these probabilities for \(k = 0, 1, 2, 3, 4\). Then multiply each probability by the total number of observations, which is 100, to get the expected frequencies for each value of \(X\).
02

Calculate Chi-Square Statistic

Using the formula for the chi-square statistic \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \(O_i\) are the observed frequencies and \(E_i\) are the expected frequencies calculated in Step 1, compute the chi-square statistic. Combine frequencies if necessary to ensure each expected frequency is at least 5.
03

Determine Degrees of Freedom

The degrees of freedom for this test is \( k - 1 - m \) where \(k\) is the number of groups and \(m\) is the number of parameters estimated from the data. Here, \(k = 5\) and \(m = 1\) (since we estimated \(\lambda\)), so the degrees of freedom is \(5 - 1 - 1 = 3\).
04

Compare Chi-Square Statistic with Critical Value

Use a chi-square distribution table to find the critical value for \(\alpha = 0.05\) and 3 degrees of freedom. If the chi-square statistic calculated in Step 2 is greater than this critical value, reject the null hypothesis that the Poisson distribution is a good fit.
05

Calculate P-value

The \(P\)-value is the probability of obtaining a chi-square statistic as extreme as, or more extreme than, the observed value. Use the chi-square distribution to find the \(P\)-value for the calculated chi-square statistic from Step 2. If \(P\)-value is less than \(\alpha = 0.05\), reject the null hypothesis.

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.

Poisson Distribution
The Poisson Distribution is an important concept in statistics, especially when we deal with counts of occurrences. It helps us model situations where an event happens a specific number of times in a fixed period or area. For example, it might describe the number of customers arriving at a store in an hour or the number of accidents occurring at an intersection per month.

The Poisson distribution is characterized by the mean number of occurrences, denoted as \( \lambda \). This mean is both an average and a rate. In our context, \( \lambda = 1.2 \) represents the average number of occurrences for the event we are studying.
  • This distribution is discrete, meaning it concerns only whole numbers since events occur a whole number of times.
  • It assumes that events occur independently; one event doesn't affect another.
  • Importantly, the time between two events is irrelevant.

    This scenario tests whether a given set of observed frequencies follows a Poisson distribution with a mean \( \lambda = 1.2 \). We use a goodness-of-fit test to determine the suitability of the distribution model for the data.
Chi-Square Statistic
The Chi-Square statistic is a powerful tool for hypothesis testing. It's used to determine how well theoretical distributions fit observed data.

In a goodness-of-fit test, we compare observed frequencies with expected frequencies based on a model. The Chi-Square statistic, \( \chi^2 \), is calculated using the formula:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]

where \( O_i \) are the observed frequencies, and \( E_i \) are the expected frequencies. These expected frequencies are calculated from the Poisson distribution probabilities multiplied by the total number of observations (in this case, 100).
  • The statistic provides a single number to represent how much the actual data deviates from what the model predicts.
  • The larger the \( \chi^2 \) value, the greater the deviation and less likely it is that the observed frequencies match the expected frequencies under the Poisson model.

    When performing a goodness-of-fit test, if the calculated Chi-Square statistic reaches a certain level, it leads us to question the model's adequacy.
Degrees of Freedom
Degrees of Freedom (df) might sound abstract, but they are a fundamental part of any statistical test. They help adjust the number of observations directly used for assessing accuracy.

In the context of our goodness-of-fit test, the degrees of freedom are calculated to adjust the Chi-Square statistic for the size of our data. The formula for degrees of freedom in this scenario is:\[df = k - 1 - m\]where \(k\) is the number of categories, and \(m\) is the number of parameters estimated. Here, there are 5 categories (0 through 4), and we've estimated one parameter (\( \lambda \)), resulting in:\[\df = 5 - 1 - 1 = 3\]

A correct degrees of freedom calculation is crucial because it influences the distribution and interpretation of the Chi-Square test. It essentially gives you a baseline for what you would expect to happen by chance alone.
  • Understanding degrees of freedom helps in determining the critical value from the Chi-Square distribution table.
  • Knowing the correct degrees of freedom ensures the validity of the hypothesis test conducted.
P-Value Calculation
The P-Value calculation is a measure that helps to determine the goodness of fit of the observed data to the expected distribution model.

Calculated from the Chi-Square distribution, the P-Value answers the question: "What is the probability of observing a test statistic at least as extreme as this one, assuming the model is correct?"

If the P-Value is small (typically less than \( \alpha = 0.05 \)), it suggests that the observed data is not likely under the assumed distribution, prompting a rejection of the null hypothesis.
  • In this context, the P-Value helps us determine whether the Poisson distribution with mean \(\lambda = 1.2\) is a suitable model.
  • To compute this, we use the Chi-Square statistic we have calculated and look up the corresponding P-Value in a Chi-Square distribution table with the appropriate degrees of freedom.
  • The smaller the P-Value, the stronger the evidence against the null hypothesis.

    An accurate P-Value calculation can shed light on the effectiveness of the Poisson distribution in predicting observed frequencies, ensuring a robust statistical analysis.

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

An inspector measured the diameter of a ball bearing using a new type of caliper. The results were as follows (in \(\mathrm{mm}): 0.265,0.263,0.266,0.267,0.267,0.265,0.267,0.267\) \(0.265,0.268,0.268,\) and \(0.263 .\) a. Use the Wilcoxon signed-rank test to evaluate the claim that the mean ball diameter is \(0.265 \mathrm{~mm}\). Use \(\alpha=0.05 .\) b. Use the normal approximation for the test. With \(\alpha=0.05\), what conclusions can you draw?

An article in Fortune (September 21,1992 ) claimed that nearly one-half of all engineers continue academic studies beyond the B.S. degree, ultimately receiving either an M.S. or a Ph.D. degree. Data from an article in Engineering Horizons (Spring 1990 ) indicated that 117 of 484 new engineering graduates were planning graduate study. a. Are the data from Engineering Horizons consistent with the claim reported by Fortune? Use \(\alpha=0.05\) in reaching your conclusions. Find the \(P\) -value for this test. b. Discuss how you could have answered the question in part (a) by constructing a two-sided confidence interval on \(p\).

State whether each of the following situations is a correctly stated hypothesis testing problem and why. a. \(H_{0}: \mu=25, H_{1}: \mu \neq 25\) b. \(H_{0}: \sigma>10, H_{1}: \sigma=10\) c. \(H_{0}: \bar{x}=50, H_{1}: \bar{x} \neq 50\) d. \(H_{0}: p=0.1, H_{1}: p=0.5\) e. \(H_{0}: s=30, H_{1}: s>30\)

A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test \(H_{0}: \mu=175\) millimeters versus \(H_{1}: \mu>175\) millimeters, using the results of \(n=10\) samples. a. Find the type I error probability \(\alpha\) if the critical region is \(\bar{x}>185\) b. What is the probability of type II error if the true mean foam height is 185 millimeters? c. Find \(\beta\) for the true mean of 195 millimeters.

An article in Growth: A Journal Devoted to Problems of Normal and Abnormal Growth ["Comparison of Measured and Estimated Fat-Free Weight, Fat, Potassium and Nitrogen of Growing Guinea Pigs" (1982, Vol. \(46(4),\) pp. \(306-321\) ) ] reported the results of a study that measured the body weight (in grams) for guinea pigs at birth. $$ \begin{array}{rrrrr} 421.0 & 452.6 & 456.1 & 494.6 & 373.8 \\ 90.5 & 110.7 & 96.4 & 81.7 & 102.4 \\ 241.0 & 296.0 & 317.0 & 290.9 & 256.5 \\ 447.8 & 687.6 & 705.7 & 879.0 & 88.8 \\ 296.0 & 273.0 & 268.0 & 227.5 & 279.3 \\ 258.5 & 296.0 & & & \end{array} $$ a. Test the hypothesis that mean body weight is 300 grams. Use \(\alpha=0.05\) b. What is the smallest level of significance at which you would be willing to reject the null hypothesis? c. Explain how you could answer the question in part (a) with a two-sided confidence interval on mean body weight.

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.