/*! 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 79 Consider the test of \(H_{0}: \s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) What are the critical values for the test statistic \(\chi_{0}^{2}\) for the following significance levels and sample sizes? (a) \(\alpha=0.01\) and \(n=20\) (b) \(\alpha=0.05\) and \(n=12\) (c) \(\alpha=0.10\) and \(n=15\)

Short Answer

Expert verified
(a) 7.63, (b) 4.57, (c) 6.57.

Step by step solution

01

Understanding the Chi-Square Test

We are conducting a chi-square test for variance where the null hypothesis is \( H_0: \sigma^2 = 5 \) and an alternative hypothesis \( H_1: \sigma^2 < 5 \). We will use the chi-square distribution with \( n-1 \) degrees of freedom to find critical values based on the given significance levels.
02

Determine the Degrees of Freedom

For a chi-square test statistic, the degrees of freedom \( u \) are given by \( u = n - 1 \), where \( n \) is the sample size. Calculate them for each part:- (a) \( n = 20 \), so \( u = 19 \).- (b) \( n = 12 \), so \( u = 11 \).- (c) \( n = 15 \), so \( u = 14 \).
03

Find the Critical Value for \( \alpha = 0.01 \) and \( n = 20 \)

Using a chi-square distribution table, find the critical value for \( \alpha = 0.01 \) and \( u = 19 \). For lower tail test, refer to the cumulative probability that corresponds to \( 1 - \alpha = 0.99 \). The critical value is \( \chi^2_{1-0.01, 19} = 7.63 \).
04

Find the Critical Value for \( \alpha = 0.05 \) and \( n = 12 \)

Look up the chi-square table for \( \alpha = 0.05 \) with \( u = 11 \). Find the critical value, which corresponds to a cumulative probability of \( 0.95 \). The critical value is \( \chi^2_{1-0.05, 11} = 4.57 \).
05

Find the Critical Value for \( \alpha = 0.10 \) and \( n = 15 \)

Using the chi-square distribution table, locate the critical value for \( \alpha = 0.10 \) and \( u = 14 \). The cumulative probability required is \( 0.90 \). The critical value is \( \chi^2_{1-0.10, 14} = 6.57 \).

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.

Significance Level
The significance level, often denoted by \( \alpha \), is a crucial concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is true, which is known as a Type I error. A lower significance level means a stricter criterion for rejecting the null hypothesis.

In a chi-square test like the one discussed in the exercise, choosing a significance level dictates how extreme our test statistic must be before we decide against the null hypothesis. Common significance levels are 0.01, 0.05, and 0.10.

Understanding the significance level is essential because it influences the critical value. The exercise gives different \( \alpha \) values:

  • For \( \alpha = 0.01 \), the test is more stringent.
  • For \( \alpha = 0.05 \), it's moderately stringent.
  • For \( \alpha = 0.10 \), the test is less stringent.
These levels reflect how confident we need to be about our evidence against the null hypothesis before we reject it.
Sample Size
Sample size, denoted as \( n \), is the number of observations in a dataset. It plays a significant role in statistical tests, including the chi-square tests. In hypothesis testing for variance, sample size affects the degrees of freedom and thus impacts the outcome of the test.

In the original exercise, the sample sizes given are 20, 12, and 15, and they are directly used to determine the degrees of freedom for our chi-square distribution. The formula is \( u = n - 1 \), which means higher sample size increases degrees of freedom, leading to more accurate approximations of the chi-square distribution.

Consider the sample size:\
  • For \( n = 20 \), we have more data points leading to \( u = 19 \).
  • For \( n = 12 \), less data with \( u = 11 \).
  • For \( n = 15 \), moderate data with \( u = 14 \).
Each size influences the critical value used in testing.
Degrees of Freedom
Degrees of freedom, abbreviated as \( u \), are an integral aspect of any statistical test. They can be thought of as the number of values in the final calculation of a statistic that are free to vary. In the context of the chi-square test for variance in the original exercise, they depend solely on the sample size.

The formula used is \( u = n - 1 \), reflecting that with each additional observation, you gain one degree of freedom. For example, with a sample of 20, you have 19 degrees of freedom. This is because once you know 19 values, the 20th value is predetermined if you're maintaining a fixed sum.

Higher degrees of freedom generally imply a more reliable approximation to the chi-square distribution. Each calculation in the exercise utilizes this concept to find the correct critical value from a chi-square distribution table, affecting the confidence in the test's results.
Critical Value
The critical value in a chi-square test acts as a threshold, which the test statistic must exceed to reject the null hypothesis. It is determined using the significance level and degrees of freedom, serving as a cut-off point in the chi-square distribution.

When conducting a test, the critical value is what you'll compare to the calculated chi-square statistic. If the statistic is below the critical value in a lower tail test, like in the original problem, you reject the null hypothesis \( H_0 \).

The specific values in the exercise show how for different \( \alpha \) levels and sample sizes (and hence degrees of freedom), the critical value changes:
  • For \( \alpha = 0.01 \) and \( u = 19 \), the critical value is 7.63.
  • For \( \alpha = 0.05 \) and \( u = 11 \), it is 4.57.
  • For \( \alpha = 0.10 \) and \( u = 14 \), it is 6.57.
Understanding how to find and use the critical value ensures a correct decision is made regarding the null hypothesis in statistical tests.

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 textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis \(H_{0}: \mu=12\) against \(H_{1}: \mu<12,\) using a random sample of four specimens. (a) What is the type I error probability if the critical region is defined as \(\bar{x}<11.5\) kilograms? (b) Find \(\beta\) for the case in which the true mean elongation is 11.25 kilograms. (c) Find \(\beta\) for the case in which the true mean is 11.5 kilograms.

Suppose that eight sets of hypotheses about a population proportion of the form $$H_{0}: p=0.3 \quad H_{1}: p>0.3$$ have been tested and that the \(P\) -values for these tests are \(0.15,0.83,\) \(0.103,0.024,0.03,0.07,0.09,\) and 0.13. Use Fisher's procedure to combine all of these \(P\) -values. Is there sufficient evidence to con- clude that the population proportion exceeds \(0.30 ?\)

A quality-control inspector is testing a batch of printed circuit boards to see whether they are capable of performing in a high temperature environment. He knows that the boards that will survive will pass all five of the tests with probability \(98 \% .\) They will pass at least four tests with probability \(99 \%,\) and they always pass at least three. On the other hand, the boards that will not survive sometimes pass the tests as well. In fact, \(3 \%\) pass all five tests, and another \(20 \%\) pass exactly four. The rest pass at most three tests. The inspector decides that if a board passes all five tests, he will classify it as "good." Otherwise, he'll classify it as "bad." (a) What does a type I error mean in this context? (b) What is the probability of a type I error? (c) What does a type II error mean here? (d) What is the probability of a type II error?

The mean weight of a package of frozen fish must equal 22 oz. Five independent samples were selected, and the statistical hypotheses about the mean weight were tested. The \(P\) -values that resulted from these tests were \(0.065,0.0924,0.073,0.025,\) and \(0.021 .\) Is there sufficient evidence to conclude that the mean package weight is not equal to 22 oz?

The proportion of adults living in Tempe, Arizona, who are college graduates is estimated to be \(p=0.4 .\) To test this hypothesis, a random sample of 15 Tempe adults is selected. If the number of college graduates is between 4 and 8 , the hypothesis will be accepted; otherwise, you will conclude that \(p \neq 0.4\). (a) Find the type I error probability for this procedure, assuming that \(p=0.4\) (b) Find the probability of committing a type II error if the true proportion is really \(p=0.2\).

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.