/*! 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 18 Pathological Gamblers Pathologic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Pathological Gamblers Pathological gambling is an impulse-control disorder. The American Psychiatric Association lists 10 characteristics that indicate the disorder in its DSM-IV manual. The National Gambling Impact Study Commission randomly selected 2417 adults and found that 35 were pathological gamblers. Is there evidence to conclude that more than \(1 \%\) of the adult population are pathological gamblers at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
There is not enough evidence to conclude that more than \(1\%\) of the adult population are pathological gamblers at the \(0.05\) significance level.

Step by step solution

01

Define the Hypotheses

State the null and alternative hypotheses. Let \(p\) represent the population proportion of pathological gamblers. Null hypothesis: \(H_0: p = 0.01\) Alternative hypothesis: \(H_1: p > 0.01\)
02

Identify the Significance Level

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

Calculate the Test Statistic

Use the sample proportion \( \hat{p} = \frac{x}{n} \) where \(x\) is the number of pathological gamblers and \(n\) is the total number of individuals sampled. \( \hat{p} = \frac{35}{2417} \approx 0.0145 \) Then apply the formula for the test statistic for a single proportion: \( z = \frac{\hat{p} - p_0}{\sqrt{ \frac{p_0 (1 - p_0)}{n} }} \) where \( p_0 = 0.01 \). Thus, \( z = \frac{0.0145 - 0.01}{\sqrt{ \frac{0.01 (1 - 0.01)}{2417} }} \approx 1.52 \)
04

Determine the Critical Value and Compare

For a right-tailed test at \(\alpha = 0.05\), the critical value is \(1.645\) from the standard normal distribution. Compare the calculated test statistic to the critical value: If \(z > 1.645\), reject the null hypothesis; otherwise, do not reject it. Since \(1.52 < 1.645\), we fail to reject the null hypothesis.
05

Conclusion

Based on the statistical analysis, there is not enough evidence at the \(0.05\) level of significance to conclude that more than \(1\%\) of the adult population are pathological gamblers.

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
The null hypothesis, denoted as \( H_0 \), is a statement that suggests there is no effect or no difference in a given situation. It is the default or original assumption that any observed effect or difference is due to random variation. In the context of our exercise, the null hypothesis is \( H_0: p = 0.01 \), which means we assume initially that the proportion of pathological gamblers in the adult population is exactly 1%. By setting up a null hypothesis, we create a baseline that allows us to determine if there's enough evidence to suggest an alternative scenario. Always remember, the null hypothesis serves as a straw man for statistical testing—it is what we try to disprove.
Alternative Hypothesis
On the other hand, the alternative hypothesis, denoted as \( H_1 \), is a statement that contradicts the null hypothesis. It represents what we seek to provide evidence for in our testing. For the pathological gamblers problem, the alternative hypothesis is \( H_1: p > 0.01 \). This suggests that the proportion of pathological gamblers is greater than 1%. The alternative hypothesis helps guide the direction of our test and specifies what we are testing against the status quo.
Significance Level
The significance level, represented by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true—also known as the Type I error rate. It sets the threshold for determining whether an observed effect is statistically significant. In our exercise, the significance level is set to 0.05. This means there is a 5% risk that we will reject the null hypothesis incorrectly. Choosing a significance level is essential because it balances the sensitivity and specificity of our test. For most research, a standard level of 0.05 is commonly used.
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis. In this problem, we use the z-statistic for a population proportion, which is calculated as follows:
\[ z = \frac{\hat{p} - p_0}{\sqrt{ \frac{p_0 (1 - p_0)}{n} }} \]
here:
  • \( \hat{p} \) is the sample proportion
  • \( p_0 \) is the null hypothesis proportion
  • \( n \) is the sample size
For our sample, \( \hat{p} \) was determined to be approximately 0.0145. Plugging in the values, we get a z-statistic of approximately 1.52. By comparing this to the critical value from the standard normal distribution, we decide whether to reject the null hypothesis. Remember, the test statistic is a crucial part of making objective, data-driven decisions in hypothesis testing.

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

Test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{aligned}&H_{0}: p=0.9 \text { versus } H_{1}: p \neq 0.9\\\&n=500 ; x=440 ; \alpha=0.05\end{aligned}$$

Conduct the appropriate test. A simple random sample of size \(n=19\) is drawn from a population that is normally distributed. The sample mean is found to be \(0.8,\) and the sample standard deviation is found to be 0.4. Test whether the population mean is less than 1.0 at the \(\alpha=0.01\) level of significance.

To test \(H_{0}: \mu=100\) versus \(H_{1}: \mu \neq 100,\) a simple random sample of size \(n=23\) is obtained from a population that is known to be normally distributed. (a) If \(\bar{x}=104.8\) and \(s=9.2,\) compute the test statistic. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.01\) level of significance, determine the critical values. (c) Diaw a \(t\) -distribution that depicts the critical region. (d) Will the researcher reject the null hypothesis? Why?

In \(2001,\) the mean household expenditure for energy was \(\$ 1493,\) according to data obtained from the U.S. Energy Information Administration. An economist wanted to know whether this amount has changed significantly from its 2001 level. In a random sample of 35 households, he found the mean expenditure (in 2001 dollars) for energy during the most recent year to be \(\$ 1618\), with standard deviation \(\$ 321 .\) (a) Do you believe that the mean expenditure has changed significantly from the 2001 level at the \(\alpha=0.05\) level of significance? (b) Construct a \(95 \%\) confidence interval about the mean energy expenditure. What does the interval imply?

A poll conducted by CNN, USA Today, and Gallup reported the following results: "According to the most recent CNN/USA Today/Gallup poll, conducted June 28-July 1, a majority of Americans (52\%) approve of the job Bush is doing as president." The poll results were obtained by conducting simple random sample of 1014 adults aged 18 years old or older, with a margin of error of ±3 percentage points. State what is wrong with the conclusions presented by the pollsters.

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.