/*! 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 16 Based on information from The De... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 16

Based on information from The Denver Post, a random sample of \(n_{1}=12\) winter days in Denver gave a sample mean pollution index of \(\bar{x}_{1}=43\). Previous studies show that \(\sigma_{1}=21\). For Englewood (a suburb of Denver), a random sample of \(n_{2}=14\) winter days gave a sample mean pollution index of \(\bar{x}_{2}=36 .\) Previous studies show that \(\sigma_{2}=15\). Assume the pollution index is normally distributed in both Englewood and Denver. Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a \(1 \%\) level of significance.

Short Answer

Expert verified
There is not enough evidence to show a difference in pollution indices between Denver and Englewood.

Step by step solution

01

State the Hypotheses

We need to set up the null and alternative hypotheses. The null hypothesis \( H_0 \) asserts that there is no difference in the average pollution index between Denver and Englewood: \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) claims that there is a difference: \( \mu_1 eq \mu_2 \).
02

Determine the Significance Level

The significance level given in the problem is \( \alpha = 0.01 \) or \( 1\% \). This level will be used to decide whether to reject the null hypothesis.
03

Calculate the Test Statistic

The test statistic for comparing two means with known standard deviations is given by the formula:\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } } \].Substituting the given values, \( z = \frac{43 - 36}{\sqrt{\frac{21^2}{12} + \frac{15^2}{14} } } \). Calculating this, we find \( z \approx 0.9908 \).
04

Determine the Critical Value

For a two-tailed test with a significance level of \( 0.01 \), the critical \( z \)-values are \( \pm 2.576 \). These values define the rejection region for the null hypothesis.
05

Compare the Test Statistic to the Critical Value

The calculated test statistic \( z \approx 0.9908 \) is within the interval \(-2.576 < z < 2.576\). It does not exceed the critical values, so it falls within the range where we fail to reject the null hypothesis.
06

Conclusion

Since the test statistic does not lie in the rejection region, we fail to reject the null hypothesis at the \( 1\% \) significance level. Therefore, there is not enough statistical evidence to suggest a difference in the mean pollution indices of Denver and Englewood during winter.

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.

Understanding Statistical Significance
Statistical significance is a critical concept in hypothesis testing. It helps us determine if the observed result is likely due to chance. When testing a hypothesis, we choose a significance level, often represented by the Greek letter alpha (\( \alpha \)). Common choices are 0.05 (5%) or 0.01 (1%). A smaller significance level means a stricter criterion for rejecting the null hypothesis. In this exercise, a 1% significance level was used. This means we accept a 1% risk of concluding that a difference exists when there is none.
  • The lower the alpha (\( \alpha \)), the less likely we are to make a Type I error, which is rejecting a true null hypothesis.
  • The significance level determines the critical values, which are used to compare against the test statistic to decide whether to reject or fail to reject the null hypothesis.
Understanding this concept helps determine the strength and reliability of your findings.
The Basics of Normal Distribution
The normal distribution is an essential concept in statistics, often referred to as the "bell curve" due to its shape. Many natural phenomena, including test scores, heights, and pollution indices, tend to follow a normal distribution.
  • In a normal distribution, data is symmetrically distributed with most observations clustering around the central peak. This peak is the mean, median, and mode of the distribution.
  • The further you move from the mean, the fewer data points you will find. Understanding normal distribution allows for the application of various statistical tests such as the z-test. In this exercise, it is assumed that the pollution indices are normally distributed, which is crucial for applying the z-test effectively in hypothesis testing.
What is a Z-Test?
A z-test is a statistical test used to determine if there is a significant difference between the means of two groups. It is especially useful when the standard deviations of both populations are known and the sample size is reasonably large.
  • The test statistic in a z-test is calculated using a formula that involves the means and standard deviations of the samples, as well as their sizes. Specifically, \( z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} } } \), where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, and \( \sigma_1 \) and \( \sigma_2 \) are the standard deviations.
  • For this test, we compare the calculated z-value with critical values, determined by the chosen significance level, to decide on the null hypothesis.
The z-test's power lies in its ability to make inferences about population means based on sample data, and it was used effectively in this exercise to assess pollution indices.
Null and Alternative Hypothesis Explained
In hypothesis testing, formulating the null and alternative hypotheses is a starting point. The null hypothesis (\( H_0 \)) always represents a status quo or no effect situation.
  • In the provided exercise, the null hypothesis states that there is no difference in the mean pollution index between Denver and Englewood: \( \mu_1 = \mu_2 \).
  • The alternative hypothesis (\( H_a \)) is what you seek to prove. It suggests that there is indeed a difference (\( \mu_1 eq \mu_2 \)).
  • Testing these hypotheses involves determining if the data provides sufficient evidence against \( H_0 \), which can be checked using statistical tests like the z-test.
The correct setup of these hypotheses is crucial as it guides the entire analysis and decision-making process in statistical testing.

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

Consider a hypothesis test of difference of means for two independent populations \(x_{1}\) and \(x_{2} .\) What are two ways of expressing the null hypothesis?

Consider a set of data pairs. What is the first step in processing the data for a paired differences test? What is the formula for the sample test statistic \(t ?\) Describe each symbol used in the formula.

Weatherwise is a magazine published by the American Meteorological Society. One issue gives a rating system used to classify Nor'easter storms that frequently hit New England and can cause much damage near the ocean. A severe storm has an average peak wave height of \(\mu=16.4\) feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. Peak wave heights are usually measured from land (using binoculars) off fixed cement piers. Suppose that a reading of 36 waves showed an average wave height of \(\bar{x}=17.3\) feet. Previous studies of severe storms indicate that \(\sigma=3.5\) feet. Does this information suggest that the storm is (perhaps temporarily) increasing above the severe rating? Use \(\alpha=0.01\).

Based on information from Harper's Index, \(r_{1}=37\) people out of a random sample of \(n_{1}=100\) adult Americans who did not attend college believe in extraterrestrials. However, out of a random sample of \(n_{2}=100\) adult Americans who did attend college, \(r_{2}=47\) claim that they believe in extraterrestrials. Does this indicate that the proportion of people who attended college and who believe in extraterrestrials is higher than the proportion who did not attend college but believe in extraterrestrials? Use \(\alpha=0.01\).

To test \(\mu\) for an \(x\) distribution that is mound-shaped using sample size \(n \geq 30\), how do you decide whether to use the normal or the Student's \(t\) distribution?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

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.