/*! 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 3 Explain why goodness-of-fit test... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Explain why goodness-of-fit tests are always right-tailed tests.

Short Answer

Expert verified
Goodness-of-fit tests are right-tailed because significant deviations between observed and expected data increase the test statistic, indicating a right-side extreme in the distribution.

Step by step solution

01

Understanding Goodness-of-Fit Tests

Goodness-of-fit tests are statistical tests used to determine how well a set of observed data matches a particular distribution, such as a normal or chi-square distribution. The goal is to assess whether the observed data significantly differ from what is expected under the assumed distribution.
02

Hypotheses in Goodness-of-Fit Tests

In a goodness-of-fit test, the null hypothesis ( H_0 ) is that the observed data follows the specified distribution. The alternative hypothesis ( H_a ) is that the data does not follow the distribution. The test checks whether there is a significant discrepancy between the observed and expected frequencies.
03

Calculating the Test Statistic

Typically, the test statistic is calculated as follows: \[ X^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]where O_i represents the observed frequencies and E_i represents the expected frequencies based on the distribution. Larger values of X^2 indicate greater differences between observed and expected values.
04

Interpreting the Test Statistic

In goodness-of-fit tests, large test statistic values imply that observed frequencies deviate significantly from expected frequencies. Since we're interested in whether the observed data is significantly different from the expected data under H_0 , we focus on the right tail of the distribution where large discrepancies lie.
05

Concluding the Test

The critical region for rejecting the null hypothesis is in the right tail because only large values of the test statistic suggest a significant difference from the expected distribution. Thus, if the calculated X^2 exceeds the critical value from the chi-square distribution table at a given significance level, H_0 is rejected.

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.

Statistical Hypothesis Testing
Statistical hypothesis testing is a method used to make inferences about populations using sample data. It involves formulating two opposing hypotheses—the null hypothesis, denoted as \( H_0 \), and the alternative hypothesis, \( H_a \). The null hypothesis assumes that there is no effect or no difference, while the alternative hypothesis suggests that there is an effect or a difference.

When conducting a hypothesis test, the aim is to determine whether there is enough evidence in the sample data to reject the null hypothesis in favor of the alternative. To do this, a test statistic is calculated from the sample data. This statistic helps us decide whether the observed data is consistent with the null hypothesis or if it supports the alternative hypothesis.

  • The test statistic is compared to a critical value derived from a theoretical distribution, which could be normal, t, chi-square, etc., depending on the test type.
  • If the test statistic is extreme in the direction specified by the alternative hypothesis, \( H_0 \) is rejected.
  • This process involves determining a significance level (commonly 0.05), below which the null hypothesis is rejected.


In goodness-of-fit tests, the goal is to assess how well the sample data adheres to a specified distribution by evaluating the discrepancy between observed and expected frequencies.
Chi-Square Distribution
The chi-square distribution is a key component in goodness-of-fit tests and other statistical tests like the test for independence. It is defined only for non-negative values and is skewed to the right, especially when the degrees of freedom are low. This type of distribution is very useful in evaluating how observed frequencies deviate from expected frequencies.

The chi-square distribution is characterized by its degrees of freedom, which affect its shape. The degrees of freedom in a goodness-of-fit test are usually calculated as the number of categories minus one. More degrees of freedom make the distribution more symmetrical.

  • In goodness-of-fit tests, the test statistic follows a chi-square distribution under the null hypothesis.
  • As the sample size increases, the chi-square distribution approaches a normal distribution.


Because we assess the likelihood of observing a value as extreme or more extreme than the test statistic, the right tail of the chi-square distribution is of particular interest. Large values in this tail indicate that our observed data significantly deviate from what was expected, thereby suggesting that the null hypothesis might not be valid.
Test Statistic Interpretation
In the context of goodness-of-fit tests, interpreting the test statistic involves determining how well the observed data fits the theoretical distribution. This is primarily determined using the chi-square test statistic, computed as \[X^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency for each category.

The resulting test statistic \(X^2\) is compared against a critical value from the chi-square distribution table. The interpretation is straightforward:
  • If \(X^2\) is larger than the critical value at a given significance level, it suggests a significant discrepancy between observed and expected data, prompting us to reject \( H_0 \).
  • If \(X^2\) is less than the critical value, there is insufficient evidence to reject \( H_0 \), indicating that the observed data fits the expected distribution well.


The emphasis on the right tail for this test means we're looking for large values of \( X^2 \) to determine if the observed frequencies differ significantly from what's expected under the null hypothesis. This inherent focus on larger discrepancies helps explain why goodness-of-fit tests are right-tailed.

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

A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a \(0.05\) level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries. $$ \begin{array}{rrcr} \text { Country I } & \text { Country II } & \text { Country III } & \text { Country IV } \\ 12.7 & 8.3 & 20.3 & 17.2 \\ 9.2 & 17.2 & 16.6 & 8.8 \\ 10.9 & 19.1 & 22.7 & 14.7 \\ 8.9 & 10.3 & 25.2 & 21.3 \\ 16.4 & & 19.9 & 19.8 \end{array} $$

For a chi-square goodness-of-fit test, how are the degrees of freedom computed?

When using the \(F\) distribution to test two variances, is it essential that each of the two populations be normally distributed? Would it be all right if the populations had distributions that were mound-shaped and more or less symmetrical?

For the study regarding mean cadence (see Problem 1), two-way ANOVA was used. Recall that the two factors were walking device (none, standard walker, rolling walker) and dual task (being required to respond vocally to a signal or no dual task required). Results of two-way ANOVA showed that there was no evidence of interaction between the factors. However, according to the article, "The ANOVA conducted on the cadence data revealed a main effect of walking device." When the hypothesis regarding no difference in mean cadence according to which, if any, walking device was used, the sample \(F\) was \(30.94\), with \(d . f \cdot \mathrm{N}=2\) and \(d . f \cdot D=18\). Further, the \(P\) -value for the result was reported to be less than \(0.01\). From this information, what is the conclusion regarding any difference in mean cadence according to the factor "walking device used"?

The following problem is based on information from Archaeological Surveys of Chaco Canyon, New Mexico, by A. Hayes, D. Brugge, and W. Judge, University of New Mexico Press. A transect is an archaeological study area that is \(1 / 5\) mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let \(x\) represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that \(x\) has a population variance \(\sigma^{2}=42.3 .\) In a different section of Chaco Canyon, a random sample of 23 transects gave a sample variance \(s^{2}=46.1\) for the number of sites per transect. Use a \(5 \%\) level of significance to test the claim that the variance in the new section is greater than \(42.3 .\) Find a \(95 \%\) confidence interval for the population variance.

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.