/*! 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 20 For the United States, the mean ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 20

For the United States, the mean monthly Internet bill is \(\$ 32.79\) per household (CNBC, January 18,2006 . A sample of 50 households in a southern state showed a sample mean of \(\$ 30.63 .\) Use a population standard deviation of \(\sigma=\$ 5.60\) a. Formulate hypotheses for a test to determine whether the sample data support the conclusion that the mean monthly Internet bill in the southem state is less than the national mean of \(\$ 32.79\) b. What is the value of the test statistic? c. What is the \(p\) -value? d. \(A t a=.01,\) what is your conclusion?

Short Answer

Expert verified
The mean monthly Internet bill in the southern state is significantly less than the national mean.

Step by step solution

01

Formulate Hypotheses

The null hypothesis (H_0) is that the mean monthly Internet bill in the southern state is equal to the national mean: \( H_0: \mu = 32.79 \). The alternative hypothesis (H_a) is that the mean monthly Internet bill in the southern state is less than the national mean: \( H_a: \mu < 32.79 \). This is a one-tailed test.
02

Calculate the Test Statistic

The test statistic for a Z-test is calculated using the formula: \[ Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]where \( \bar{x} = 30.63 \) is the sample mean, \( \mu = 32.79 \) is the population mean, \( \sigma = 5.60 \) is the population standard deviation, and \( n = 50 \) is the sample size.Calculate: \[ Z = \frac{30.63 - 32.79}{5.60 / \sqrt{50}} = \frac{-2.16}{0.791} \approx -2.73 \]
03

Find the P-value

The P-value is the probability that the test statistic would take on a value as extreme as, or more extreme than, the observed value, assuming that the null hypothesis is true. Since this is a left-tailed test, we look up the value on the standard normal distribution table.For \( Z = -2.73 \), the P-value corresponds to approximately 0.0032.
04

Make a Conclusion at \( \alpha = 0.01 \)

Compare the P-value to the significance level \( \alpha = 0.01 \). Since the P-value (0.0032) is less than \( \alpha \) (0.01), we reject the null hypothesis. There is enough evidence to support the conclusion that the mean monthly Internet bill in the southern state is less than the national mean.

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 the Z-Test
A Z-test is a statistical method used to determine if there is a significant difference between sample and population means. It is especially useful when comparing a sample to a known population mean with a known standard deviation.
The Z-test formula is:
  • \( Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \)
Where:
  • \(\bar{x}\) = sample mean
  • \(\mu\) = population mean
  • \(\sigma\) = population standard deviation
  • \(n\) = sample size
Using this formula, the test statistic value helps us understand how far our sample mean is from the population mean in terms of standard deviations. A larger absolute Z-value indicates that a test statistic is farther from the null hypothesis, suggesting the sample may not be part of the population at a certain level of confidence.
Decoding the P-Value
The P-value is a crucial output of statistical tests. It indicates the probability of observing test results at least as extreme as those observed if the null hypothesis is true. In simpler terms, it helps us gauge the strength of the evidence against the null hypothesis.
  • A small P-value (typically 鈮� 0.05) suggests strong evidence against the null hypothesis, leading to its rejection.
  • A large P-value (> 0.05) suggests weak evidence against the null hypothesis, leading to its retention.
For example, in a one-tail Z-test, such as our exercise, finding a P-value of 0.0032 means there's only a 0.32% chance of observing a test statistic as extreme as the sample's, assuming the null hypothesis is true. This low P-value is compelling evidence to reject the null hypothesis, suggesting that the sample mean is indeed different from the population mean.
Exploring Statistical Significance
Statistical significance is a concept that tells us whether the result of a study or test is likely to have occurred under random chance or not. It is determined by comparing the P-value against a preset significance level, usually denoted as \(\alpha\).
  • Common significance levels are 0.05, 0.01, and 0.001.
  • If the P-value is less than or equal to \(\alpha\), the results are considered statistically significant.
  • If the P-value is greater than \(\alpha\), the results are not statistically significant.
In the context of our exercise, with an \(\alpha\) of 0.01 and a P-value of 0.0032, the results are statistically significant. This means there is enough statistical evidence to conclude that the mean monthly Internet bill for households in the southern state is less than the national mean, beyond a reasonable doubt of random chance.

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

AOL Time Warner Inc.'s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10,2003 ). Assume that for a sample of 40 days during the first half of \(2003,\) the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers. a. What are the hypotheses if CNN management would like information on any change in the CNN viewing audience? b. What is the \(p\) -value? c. Select your own level of significance. What is your conclusion? d. What recommendation would you make to CNN management in this application?

The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager wants to conduct a research study to see whether the new bonus plan increases sales volume. To collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus plan for a one-month period. a. Develop the null and alternative hypotheses most appropriate for this situation. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

The label on a 3 -quart container of orange juice states that the orange juice contains an average of 1 gram of fat or less. Answer the following questions for a hypothesis test that could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

In \(2008,46 \%\) of business owners gave a holiday gift to their employees. A 2009 survey of business owners indicated that \(35 \%\) planned to provide a holiday gift to their employees (Radio WEZV, Myrtle Beach, SC, November 11,2009 ). Suppose the survey results are based on a sample of 60 business owners. a. How many business owners in the survey planned to provide a holiday gift to their employees in \(2009 ?\) b. Suppose the business owners in the sample did as they planned. Compute the \(p\) -value for a hypothesis test that can be used to determine if the proportion of business owners providing holiday gifts had decreased from the 2008 level. c. Using a .05 level of significance, would you conclude that the proportion of business owners providing gifts decreased? What is the smallest level of significance for which you could draw such a conclusion?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\quad \bar{x}=20\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

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.