/*! 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 25 According to a recent survey, Am... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to a recent survey, Americans get a mean of 7 hours of sleep per night. A random sample of 50 students at West Virginia University revealed the mean length of time slept last night was 6 hours and 48 minutes \((6.8\) hours \()\). The standard deviation of the sample was 0.9 hour. At the \(5 \%\) level of significance, is it reasonable to conclude that students at West Virginia sleep less than the typical American? Compute the \(p\) -value.

Short Answer

Expert verified
No, the data does not show enough evidence to conclude that WVU students sleep less than typical Americans.

Step by step solution

01

Hypotheses Formation

First, formulate the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the mean sleep duration for the students is equal to 7 hours, i.e., \( \mu = 7 \). The alternative hypothesis \( H_a \) claims that the mean sleep duration for the students is less than 7 hours, i.e., \( \mu < 7 \).
02

Identify Sample Statistics

Extract the relevant statistics from the problem. The sample mean is \( \bar{x} = 6.8 \) hours, the sample standard deviation is \( s = 0.9 \) hours, and the sample size is \( n = 50 \).
03

Calculate the Test Statistic

Use the formula for the test statistic for a single sample mean: \[ z = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] Substitute \( \bar{x} = 6.8 \), \( \mu = 7 \), \( s = 0.9 \), and \( n = 50 \) into the formula to calculate \( z \). This gives us \[ z = \frac{6.8 - 7}{0.9 / \sqrt{50}} = \frac{-0.2}{0.1273} \approx -1.57 \].
04

Determine the p-value

Find the p-value associated with the calculated z-value \(-1.57\) using the standard normal distribution table. The p-value for \( z = -1.57 \) corresponds to the area to the left of \( z \), which is approximately 0.0582.
05

Make a Decision

Compare the p-value to the level of significance \( \alpha = 0.05 \). Since the p-value \( 0.0582 \) is greater than \( \alpha \), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that students at West Virginia University sleep less than the typical American.

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.

Null Hypothesis
In hypothesis testing, the null hypothesis is a starting point. It's a statement that there is no effect or difference, and it is typically denoted as \(H_0\). In this exercise, the null hypothesis claims that the average sleep duration of students at West Virginia University is equal to 7 hours. Basically, it's saying that there's no difference between the students' sleeping habits and the national average.

For statistical tests, the null hypothesis sets a baseline that the data will either support or contradict. It's crucial in determining whether a perceived effect is statistically significant or just due to random variation. Remember, failing to reject the null hypothesis doesn't prove it true; it simply suggests insufficient evidence against it.

  • It serves as the default or initial assumption.
  • Testing aims to either reject or fail to reject it, not to accept it outright.
Alternative Hypothesis
The alternative hypothesis is the counterpart to the null hypothesis, denoted as \(H_a\). It represents what the researcher actually wants to prove. In this context, the alternative hypothesis asserts that the students sleep less than the average American, that is, their mean sleep duration is less than 7 hours.

This hypothesis is crucial for scientific inquiry since it forms the basis for research questions and can lead to new discoveries.

When conducting a hypothesis test, we look for statistical evidence strong enough to reject the null hypothesis in favor of the alternative. Keep in mind that the alternative hypothesis is what we hope to find if the data shows a significant deviation from the null.

  • It represents a new effect or finding.
  • Typically assumed true only if the data strongly suggests it.
p-value
The p-value is a critical concept in hypothesis testing that helps determine the strength of the evidence against the null hypothesis. It represents the probability of observing test results at least as extreme as those obtained, given that the null hypothesis is true.

In our exercise, the p-value is calculated based on a test statistic derived from the observed data, and it was found to be approximately 0.0582. This value tells us how inconsistent the observed data is with the null hypothesis. A smaller p-value indicates stronger evidence against \(H_0\).

Comparing the p-value with a predefined significance level \(\alpha\), typically 0.05, helps in decision making:

  • If p-value \(< \alpha\), we reject \(H_0\).
  • If p-value \(\geq \alpha\), we fail to reject \(H_0\).
Test Statistic
The test statistic is a standardized value used in hypothesis testing to decide whether to support or reject the null hypothesis. This statistic measures the extent of deviation of the sample data from what is expected under the null hypothesis.

In the given problem, the test statistic is a \(z\)-value calculated using the sample mean, population mean, sample standard deviation, and sample size. The calculation formula is:

\[ z = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

For this example, substituting the known values yields \(z \approx -1.57\). The negative sign of the \(z\)-value indicates that the sample mean is less than the hypothesized population mean.

The \(z\)-value helps us find the p-value, which in turn guides the decision-making process. It's a vital component in summarizing how far the data is from \(H_0\).

  • A large absolute \(z\)-value indicates the data is far from what is expected.
  • It's typically compared with critical values to make a statistical decision.

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

According to the Census Bureau, 3.13 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.86 residents. The standard deviation of this sample was 1.20 residents. At the .05 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.13 persons?

A new weight-watching company, Weight Reducers International, advertises that those who join will lose an average of 10 pounds after the first two weeks. The standard deviation is 2.8 pounds. A random sample of 50 people who joined the weight reduction program revealed a mean loss of 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers will lose less than 10 pounds? Determine the \(p\) -value.

Rutter Nursery Company packages its pine bark mulch in 50 -pound bags. From a long history, the production department reports that the distribution of the bag weights follows the normal distribution, and the standard deviation of the packaging process is 3 pounds per bag. At the end of each day, Jeff Rutter, the production manager, weighs 10 bags and computes the mean weight of the sample. Below are the weights of 10 bags from today's production. \(\begin{array}{lllllllll}45.6 & 47.7 & 47.6 & 46.3 & 46.2 & 47.4 & 49.2 & 55.8 & 47.5 & 48.5\end{array}\) a. Can Mr. Rutter conclude that the mean weight of the bags is less than 50 pounds? Use the .01 significance level. b. In a brief report, tell why Mr. Rutter can use the \(z\) distribution as the test statistic. c. Compute the \(p\) -value.

For Exercises 5-8: (a) State the null hypothesis and the alternate hypothesis. (b) State the decision rule. (c) Compute the value of the test statistic. (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The manufacturer of the \(X-15\) steel-belted radial truck tire claims that the mean mileage the tire can be driven before the tread wears out is 60,000 miles. Assume the mileage wear follows the normal distribution and the standard deviation of the distribution is 5,000 miles. Crosset Truck Company bought 48 tires and found that the mean mileage for its trucks is 59,500 miles. Is Crosset's experience different from that claimed by the manufacturer at the .05 significance level?

A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Its records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, the agency believes that the mean selling time is now greater than 90 days. A statewide survey of 100 recently sold farms revealed a mean selling time of 94 days, with a standard deviation of 22 days. At the . 10 significance level, has there been an increase in selling time?

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.