/*! 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 118 A 2008 study performed by career... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 118

A 2008 study performed by careerbuilder.com entitled \(\mathrm{No}\), Really, Your Excuse is Totally Believable! notes that \(11 \%\) of workers who call in sick do so to catch up on housework. Suppose that in a survey of 675 male workers who have called in sick, 61 did so to have time to catch up on housework. At the \(2 \%\) significance level, can you conclude that the proportion of all male workers who call in sick do so to catch up on housework is different from \(11 \%\) ?

Short Answer

Expert verified
Based on the statistical analysis, there is insufficient evidence to conclude that the proportion of all male workers who call in sick to catch up on housework is different from 11%.

Step by step solution

01

Set up the hypotheses

The null hypothesis is that the proportion of all male workers who call in sick do so to catch up on housework is equal to 11% (0.11), denoted as \(H_0: p = 0.11\). The alternative hypothesis is that the proportion is different from 11%, denoted as \(H_a: p \neq 0.11\).
02

Calculate the sample proportion

The sample proportion (\(\hat{p}\)) is calculated as the number of favorable outcomes divided by the total number of outcomes. In this case, \(\hat{p}\) = 61/675 = 0.0904.
03

Calculate the test statistic

The test statistic (z) in a proportion hypothesis test is calculated using the formula: \(z = (\hat{p} - p_0) / \sqrt{p_0(1 - p_0)/n}\), where \(p_0\) is the proportion under the null hypothesis and \(n\) is the sample size. So \(z = (0.0904 - 0.11) / \sqrt{(0.11)(0.89)/675} = -1.96.
04

Determine the critical value

Since the hypothesis test is two-tailed (i.e., \(H_a: p \neq 0.11\)), the critical value corresponds to the 2% significance level divided by 2. From the standard normal distribution table, the critical value for 1% in each tail is 卤2.58.
05

Make a decision and draw a conclusion

Compare the test statistic with the critical value. Since -1.96 does not fall in the critical region of 卤2.58, we fail to reject the null hypothesis. Hence, there is insufficient evidence to conclude that the proportion of all male workers who call in sick to catch up on housework is different from 11%.

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
The null hypothesis is the starting point of hypothesis testing. It represents the assumption that there is no difference or effect. In this case, the null hypothesis asserts that the proportion of male workers who call in sick to catch up on housework is exactly 11% (0.11). We denote it as \(H_0: p = 0.11\).

It acts as a baseline for comparison, allowing us to measure whether observed data deviate significantly from this assumed value. If evidence from the data strongly contradicts it, we may reject the null hypothesis. Otherwise, we accept that our data are consistent with it.
Alternative Hypothesis
Contrary to the null, the alternative hypothesis suggests there is a difference or effect. It claims that the proportion of male workers engaging in housework while calling in sick is different from 11%, denoted as \(H_a: p eq 0.11\).

The alternative hypothesis opens the possibility for variation or change from the status quo. It is a two-tailed hypothesis in this context, meaning we are interested in deviations in both directions鈥攈igher or lower than 11%.
Significance Level
The significance level, often denoted by \(\alpha\), is the threshold at which we decide whether to reject the null hypothesis. In this exercise, the significance level is set at 2% or 0.02.

This level represents the probability of rejecting the null hypothesis when it is actually true (Type I error). A lower significance level means we require strong evidence to reject the null, making our test more stringent. By splitting this into two tails for a two-tailed test, each tail has a 1% significance level.
Test Statistic
The test statistic provides a standardized value that tells us how far our sample statistic is from the hypothesized population parameter under the null. In this problem, we calculate the z-score using the formula:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]
where \(\hat{p}\) is the sample proportion (0.0904), \(p_0\) is the null hypothesis proportion (0.11), and \(n\) is the sample size (675). For this exercise, the z-score calculated is -1.96.

This score helps us determine where our observed data falls within the standard normal distribution.
Critical Value
The critical value is the cutoff point that defines the boundaries of the rejection region based on the significance level. For a 2% significance level, we split it equally into two tails, using 1% per tail.

From the standard normal distribution table, the critical values for this level are 卤2.58.

Comparing the test statistic (-1.96) with these critical values helps us decide: if the test statistic falls beyond 卤2.58, we reject the null hypothesis. In this case, since -1.96 falls within the range, we do not have enough evidence to reject the null.

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

Consider the null hypothesis \(H_{0}: \mu=100\). Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of \(.01\), show the rejection and nonrejection regions and find the critical value(s) of \(t\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 100\) b. \(H_{1}: \mu>100\) c. \(H_{1}: \mu<100\)

Before a championship football game, the referee is given a special commemorative coin to toss to decide which team will kick the ball first. Two minutes before game time, he receives an anonymous tip that the captain of one of the teams may have substituted a biased coin that has a \(70 \%\) chance of showing heads each time it is tossed. The referee has time to toss the coin 10 times to test it. He decides that if it shows 8 or more heads in 10 tosses, he will reject this coin and replace it with another coin. Let \(p\) be the probability that this coin shows heads when it is tossed once. a. Formulate the relevant null and alternative hypotheses (in terms of \(p\) ) for the referee's test. b. Using the referee's decision rule, find \(\alpha\) for this test.

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.

Consider \(H_{0}: \mu=80\) versus \(H_{1}: \mu \neq 80\) for a population that is normally distributed. a. A random sample of 25 observations taken from this population produced a sample mean of 77 and a standard deviation of 8 . Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 25 observations taken from the same population produced a sample mean of 86 and a standard deviation of \(6 .\) Using \(\alpha=.01\), would you reject the null hypothesis?

For each of the following examples of tests of hypothesis about the population proportion, show the rejection and nonrejection regions on the graph of the sampling distribution of the sample proportion. a. A two-tailed test with \(\alpha=.10\) b. A left-tailed test with \(\alpha=.01\) c. A right-tailed test with \(\alpha=.05\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Pure 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.