/*! 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 8 At the time she was hired as a s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, "You can average more than \(\$ 80\) a day in tips." Assume the standard deviation of the population distribution is 3.24 . Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was 84.85 . At the .01 significance level, can Ms. Brigden conclude that she is earning an average of more than 80 in tips?

Short Answer

Expert verified
Yes, Ms. Brigden earns more than $80 on average in tips.

Step by step solution

01

Formulate Hypotheses

Formulate the null and alternative hypotheses. The null hypothesis \(H_0\) posits that Beth's average tips are \( \leq \\(80 \). The alternative hypothesis \(H_1\) claims that her average tips are \( > \\)80 \). Thus:\[ H_0: \mu \leq 80 \]\[ H_1: \mu > 80 \]
02

Calculate the Test Statistic

Use the formula for the \(z\)-test statistic: \[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]where \(\bar{x} = 84.85\), \(\mu = 80\), \(\sigma = 3.24\), and \(n = 35\). Substituting these values in:\[ z = \frac{84.85 - 80}{3.24 / \sqrt{35}} \]Calculate the denominator:\(3.24 / \sqrt{35} \approx 0.5476\)Thus, the \(z\)-statistic:\[ z \approx \frac{4.85}{0.5476} \approx 8.86 \]
03

Determine the Critical Value

For a significance level of 0.01 and a one-tailed test, find the critical value using the standard normal distribution table. The critical \(z\)-value at \(0.01\) significance level is approximately 2.33.
04

Make a Decision

Compare the calculated \(z\)-statistic with the critical \(z\)-value. Here, \(z = 8.86\) which is much greater than the critical \(z\)-value of 2.33. Therefore, reject the null hypothesis \(H_0\).
05

Conclusion

Since we have rejected the null hypothesis, there is sufficient evidence at the \(0.01\) significance level to conclude that Ms. Brigden earns, on average, more than \(\$80\) in tips per day.

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.

z-test statistic
The z-test statistic is a crucial part of hypothesis testing, especially when we are dealing with large sample sizes or known population variance. In the context of Beth Brigden's tips, the z-test statistic helps us decide if the average amount of tips she earns is significantly different from a specified value, in this case, $80. The formula to calculate the z-test statistic is:\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]where:
  • \( \bar{x} \) is the sample mean, which is 84.85 here.
  • \( \mu \) is the population mean according to the null hypothesis, which is 80.
  • \( \sigma \) is the population standard deviation, given as 3.24.
  • \( n \) is the sample size, which is 35 days of observation.
By substituting these values, we calculate the z-value which tells us how many standard deviations the sample mean is away from the population mean under the null hypothesis. A large z-value, as seen in this problem (8.86), suggests a significant difference.
significance level
The significance level is a threshold set by the researcher to determine when to reject the null hypothesis. It's a measure of how willing we are to make a type I error, which is the incorrect rejection of a true null hypothesis. In statistics, it's often denoted by \( \alpha \). For Beth Brigden's test, \( \alpha \) is set to 0.01.Setting a low significance level, like 0.01, means we're being very cautious. It indicates that there's only a 1% risk of claiming that Beth’s average tips are more than $80 when they're actually not. The lower the significance level, the stronger the evidence must be to reject the null hypothesis.This 0.01 level also impacts the critical value used in hypothesis testing, which is explained further in later sections.
null and alternative hypotheses
In hypothesis testing, formulating the null and alternative hypotheses is a fundamental first step. These hypotheses guide the direction of the test. In Beth Brigden's scenario:
  • The null hypothesis (\( H_0 \)) posits that the average tip amount is less than or equal to \(80 (\( \mu \leq 80 \)).
  • The alternative hypothesis (\( H_1 \)) suggests that Beth's average tips are greater than \)80 (\( \mu > 80 \)).
The null hypothesis represents the status quo or the claim we would test against, while the alternative presents the outcome we want to support with our data. By rejecting or failing to reject the null hypothesis, we establish the impact of the findings. Here, rejecting \( H_0 \) allows us to assert with confidence that Beth indeed earns more than the level initially stated.
critical value
The critical value acts as a benchmark in hypothesis testing. It helps us decide when to reject the null hypothesis. For Beth Brigden's test, a critical value corresponds to a significance level of 0.01 for a one-tailed test.To find this value, we refer to the standard normal distribution (z) table. For a significance level \( \alpha = 0.01 \) and a one-tailed test, the critical z-value is approximately 2.33.In comparing the calculated z-statistic to this critical value:
  • If the z-statistic is greater than the critical value, we reject \( H_0 \).
  • If it is less, we do not reject \( H_0 \).
In Beth's case, the calculated z-value of 8.86 exceeds the critical value, leading to the rejection of the null hypothesis and supporting the conclusion that her average tips are indeed more than $80.

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

Traditionally, two percent of the citizens of the United States live in a foreign country because they are disenchanted with U.S. politics or social attitudes. In order to test if this proportion has increased since the September 11, 2001, terror attacks, U.S. consulates contacted a random sample of 400 of these expatriates. The sample yields 12 people who report they are living overseas because of political or social attitudes. Can you conclude this data shows the proportion of politically motivated expatriates has increased? Use the 0.05 significance level.

The liquid chlorine added to swimming pools to combat algae has a relatively short shelf life before it loses its effectiveness. Records indicate that the mean shelf life of a 5 -gallon jug of chlorine is 2,160 hours (90 days). As an experiment, Holdlonger was added to the chlorine to find whether it would increase the shelf life. A sample of nine jugs of chlorine had these shelf lives (in hours): $$\begin{array}{|lllllllll|}\hline 2,159 & 2,170 & 2,180 & 2,179 & 2,160 & 2,167 & 2,171 & 2,181 & 2,185 \\\\\hline\end{array}$$ At the .025 level, has Holdlonger increased the shelf life of the chlorine? Estimate the \(p\) -value.

The National Safety Council reported that 52 percent of American turnpike drivers are men. A sample of 300 cars traveling southbound on the New Jersey Turnpike yesterday revealed that 170 were driven by men. At the .01 significance level, can we conclude that a larger proportion of men were driving on the New Jersey Turnpike than the national statistics indicate?

According to the Coffee Research Organization (http://www.coffeeresearch.org) the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday. $$\begin{array}{|llllllllllll|}\hline 3.1 & 3.3 & 3.5 & 2.6 & 2.6 & 4.3 & 4.4 & 3.8 & 3.1 & 4.1 & 3.1 & 3.2 \\\\\hline\end{array}$$ At the .05 significance level does this sample data suggest there is a difference between the national average and the sample mean from senior citizens?

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.