/*! 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 60 The president of a university cl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 60

The president of a university claims that the mean time spent partying by all students at this university is not more than 7 hours per week. A random sample of 40 students taken from this university showed that they spent an average of \(9.50\) hours partying the previous week with a standard deviation of \(2.3\) hours. Test at the \(2.5 \%\) significance level whether the president's claim is true. Explain your conclusion in words.

Short Answer

Expert verified
Based on the statistical evidence, we reject the claim of the university president that the mean time spent studying by all students at this university is not more than 7 hours per week. There is sufficient evidence at the 2.5% significance level to suggest that it is more than 7 hours a week.

Step by step solution

01

Set up the hypotheses

The null hypothesis \(H_0\) is that the mean time spent partying is not more than 7 hours per week. So, \(H_0: \mu \leq 7\). The alternative hypothesis \(H_1\) is that the mean time spent partying is more than 7 hours per week, i.e., \(H_1: \mu > 7\).
02

Determine the test statistic

We calculate the t test statistic using the formula: \(t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}\), where \(\bar{X}\) is the sample mean (9.5), \(\mu_0\) is the hypothesized population mean (7), \(s\) is the sample standard deviation (2.3), and \(n\) is sample size (40). From calculation, the test statistic is \(t = 5.89\)
03

Determine the reference value

Using the t-distribution table, the reference value for one-tailed t-test with degree of freedom \(df = n-1 = 39\) at \(2.5\%\) significance level is roughly around 2.022.
04

Making a decision

Since the calculated test statistic (5.89) is greater than the reference value (2.022), we reject the null hypothesis. This suggests at 2.5% significance level, the data provides enough evidence to say the mean time students spent partying is significantly greater than 7 hours.

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.

Null Hypothesis
In statistics, the null hypothesis is a foundational concept used in hypothesis testing. It represents a statement that there is no effect or no difference. In most cases, the null hypothesis is what researchers aim to test against and potentially reject.
  • The null hypothesis, often symbolized as \(H_0\), serves as a starting assumption.
  • In the context of the university example, the null hypothesis was formulated as \(H_0: \mu \leq 7\), which suggests the mean time students spend on partying does not exceed 7 hours weekly.
  • Generally, rejecting the null hypothesis implies that there is sufficient statistical evidence suggesting an observed effect or difference.
Understanding the null hypothesis is crucial because it helps determine if deviations observed in the sample data are likely due to random chance or indicate a significant finding.
t-test
A t-test is a type of inferential statistic used to determine if there is a significant difference between the means of two groups, which may be related in certain features. It is a handy method in testing hypotheses about population means.
  • The test compares the sample mean to the hypothesized population mean.
  • In the given example, a t-test was applied to challenge the claim about students鈥� weekly partying hours. The test checks if the sample mean of 9.5 hours is statistically different from the 7 hours assumed under the null hypothesis.
  • The formula for the t-statistic is \(t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}}\), which compares the difference between the sample mean (\(\bar{X}\)) and null hypothesis mean (\(\mu_0\)) relative to the standard error.
This test is an essential tool in statistics, as it helps verify if a sample accurately reflects the general population by assessing mean differences.
Significance Level
The significance level, denoted by \(\alpha\), is a critical concept in hypothesis testing that measures the threshold for rejecting the null hypothesis. It defines the probability of making a Type I error, which occurs when the null hypothesis is incorrectly rejected.
  • Common significance levels include 0.05, 0.01, and in this example, 0.025 (2.5%).
  • A significance level of 2.5% indicates that there is a 2.5% risk of concluding that a difference exists when there is no actual difference.
  • The selected significance level determines the critical value or cutoff point beyond which the null hypothesis will be rejected.
In the case of the university study, setting an alpha level at 2.5% provided a stringent criterion for determining if the mean partying time significantly differs from the president's claim. The lower the alpha, the less likely random chance alone explains observed data.
Sample Mean
The sample mean is the average value from a sample set of data, representing an estimate of the population mean. It is the sum of all observations divided by the number of observations. Understanding the sample mean is vital when conducting tests involving inference about populations.
  • To calculate, add all sample values and divide by the sample size.
  • In the university example, the sample mean was 9.5 hours, indicating students spent more time on average partying than what was hypothesized under the null hypothesis.
  • The sample mean acts as a primary metric in t-tests and provides insight into trends within the sampled subset.
The sample mean is instrumental for researchers and analysts because it offers a practical approximation of the entire population's behaviors or characteristics.

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 random sample of 200 observations produced a sample proportion equal to \(.60 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.01\). a. \(H_{0}: p=.63\) versus \(H_{1}: p<.63\) b. \(H_{0}: p=.63\) versus \(H_{1}: p \neq .63\)

Consider \(H_{0}: p=.45\) versus \(H_{1}: p<.45\). a. A random sample of 400 observations produced a sample proportion equal to .42. Using \(\alpha=.025\), would you reject the null hypothesis? b. Another random sample of 400 observations taken from the same population produced a sample proportion of .39. Using \(\alpha=.025\), would you reject the null hypothesis?

Make the following hypothesis tests about \(p\). a. \(H_{0}: p=.57, \quad H_{1}: p \neq .57, \quad n=800, \quad \hat{p}=.50, \quad \alpha=.05\) b. \(H_{0}: p=.26, \quad H_{1}: p<.26, \quad n=400, \quad \hat{p}=.23, \quad \alpha=.01\) c. \(H_{0}: p=.84, \quad H_{1}: p>.84, \quad n=250, \quad \hat{p}=.85, \quad \alpha=.025\)

The Bath Heritage Days, which take place in Bath, Maine, have been popular for, among other things, an eating contest. In 2009, the contest switched from blueberry pie to a Whoopie Pie (www.timesrecord.com), which consists of two large, chocolate cake-like cookies filled with a large amount of vanilla cream. Sixty-five randomly selected adults are chosen to eat a 1 -pound Whoopie Pie, and the average time for 59 adults (out of these 65 ) is \(127.10\) seconds. Based on other Whoopie Pie-eating contests throughout the United States, suppose that the standard deviation of the times taken by all adults to consume 1-pound Whoopie pies are known to be \(23.80\) seconds. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the mean time to eat a 1 -pound Whoopie Pie is more than 2 minutes. Will you reject the null hypothesis at \(\alpha=.01\) ? Explain. What if \(\alpha=.02\) ? b. Test the hypothesis of part a using the critical-value approach. Will you reject the null hypothesis at \(\alpha=.01\) ? What if \(\alpha=.02\) ?

In a 2009 nonscientific poll on the Web site of the Daily Gazette of Schenectady, New York, readers were asked the following question: "Are you less inclined to buy a General Motors or Chrysler vehicle now that they have filed for bankruptcy?" Of the respondents, \(56.1 \%\) answered "Yes" (http://www. dailygazette.com/polls/2009/jun/Bankruptcy/). In a recent survey of 1200 adult Americans who were asked the same question, 615 answered "Yes." Can you reject the null hypothesis at the \(1 \%\) significance level in favor of the alternative that the percentage of all adult Americans who are less inclined to buy a General Motors or Chrysler vehicle since the companies filed for bankruptcy is different from \(56.1 \%\) ? Use both the \(p\) -value and the critical-value approaches.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.