/*! 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 92 A student organization uses the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 92

A student organization uses the proceeds from a particular soft-drink dispensing machine to finance its activities. The price per can had been \(\$ 0.75\) for a long time, and the average daily revenue during that period had been \(\$ 75.00\). The price was recently increased to \(\$ 1.00\) per can. A random sample of \(n=20\) days after the price increase yielded a sample average daily revenue and sample standard deviation of \(\$ 70.00\) and \(\$ 4.20\), respectively. Does this information suggest that the true average daily revenue has decreased from its value before the price increase? Test the appropriate hypotheses using \(\alpha=.05\).

Short Answer

Expert verified
The short answer would depend on the calculated p-value and t statistic. However, since these calculations have not been carried out in this instance, we can not provide a definitive answer. Depending on the results, it could be that the data does not provide strong evidence against the null hypothesis, in which case we wouldn't reject it, or it might be the case that there is strong evidence against the null hypothesis, leading to its rejection in favor of the alternative.

Step by step solution

01

Formulate the hypotheses

The null hypothesis \(H_0\) states that the mean revenue has not decreased, it is still $75. So, \(H_0: \mu = 75\). The alternative hypothesis \(H_1\) states that the mean revenue has decreased. So, \(H_1: \mu < 75\).
02

Establish the significance level

The significance level is given as \(\alpha = 0.05\). This is the probability of rejecting the null hypothesis when it is true.
03

Calculate the Test Statistic

For the sample, the mean (\(\bar{x}\)) is $70, the sample standard deviation (s) is $4.20, and the sample size (n) is 20. We use the formula for the test statistic in a t test, which is \(t = (\bar{x} - \mu) / (s/ \sqrt{n})\). Substituting the given values we get \(t = (70 - 75) / (4.20 / \sqrt{20})\).
04

Calculate the p-value

After calculating the test statistic we look up this value in the t-distribution table (with n-1 degrees of freedom, which is 19) to find the p-value. This p-value tells us the probability of obtaining a result as extreme as our test statistic assuming the null hypothesis is true.
05

Conclude the hypothesis test

We then compare the p-value to our significance level (\(\alpha\)) to decide whether to reject the null hypothesis. If the p-value is less than or equal to \(\alpha\), we reject the null hypothesis in favor of the alternative. Otherwise, we do not have enough evidence to reject the null hypothesis.

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!

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

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. Four hundred of these companies indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Is there sufficient evidence to conclude that more than \(75 \%\) of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\). b. Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\).

A well-designed and safe workplace can contribute greatly to increasing productivity. It is especially important that workers not be asked to perform tasks, such as lifting, that exceed their capabilities. The following data on maximum weight of lift (MWOL, in kilograms) for a frequency of 4 lifts per minute were reported in the article "The Effects of Speed, Frequency, and Load on Measured Hand Forces for a Floor-to-Knuckle Lifting Task" (Ergonomics \([1992]: 833-843)\) : \(\begin{array}{lllll}25.8 & 36.6 & 26.3 & 21.8 & 27.2\end{array}\) Suppose that it is reasonable to regard the sample as a random sample from the population of healthy males, age \(18-30\). Do the data suggest that the population mean MWOL exceeds 25 ? Carry out a test of the relevant hypotheses using a \(.05\) significance level.

According to a survey of 1000 adult Americans conducted by Opinion Research Corporation, 210 of those surveyed said playing the lottery would be the most practical way for them to accumulate \(\$ 200,000\) in net wealth in their lifetime ("One in Five Believe Path to Riches Is the Lottery," San Luis Obispo Tribune, January 11,2006 ). Although the article does not describe how the sample was selected, for purposes of this exercise, assume that the sample can be regarded as a random sample of adult Americans. Is there convincing evidence that more than \(20 \%\) of adult Americans believe that playing the lottery is the best strategy for accumulating \(\$ 200,000\) in net wealth?

The mean length of long-distance telephone calls placed with a particular phone company was known to be \(7.3\) min under an old rate structure. In an attempt to be more competitive with other long-distance carriers, the phone company lowered long-distance rates, thinking that its customers would be encouraged to make longer calls and thus that there would not be a big loss in revenue. Let \(\mu\) denote the true mean length of long-distance calls after the rate reduction. What hypotheses should the phone company test to determine whether the mean length of long-distance calls increased with the lower rates?

Teenagers (age 15 to 20 ) make up \(7 \%\) of the driving population. The article "More States Demand Teens Pass Rigorous Driving Tests" (San Luis Obispo Tribune, January 27,2000 ) described a study of auto accidents conducted by the Insurance Institute for Highway Safety. The Institute found that \(14 \%\) of the accidents studied involved teenage drivers. Suppose that this percentage was based on examining records from 500 randomly selected accidents. Does the study provide convincing evidence that the proportion of accidents involving teenage drivers differs from \(.07\), the proportion of teens in the driving population?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

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.