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Chapter 10: Problem 73

A number of initiatives on the topic of legalized gambling have appeared on state ballots. Suppose that a political candidate has decided to support legalization of casino gambling if he is convinced that more than twothirds of U.S. adults approve of casino gambling. Suppose that 1523 adults (selected at random from households with telephones) were asked whether they approved of casino gambling. The number in the sample who approved was \(1035 .\) Does the sample provide convincing evidence that more than two-thirds approve?

Short Answer

Expert verified
No, the sample does not provide convincing evidence that more than two-thirds of U.S adults approve of casino gambling.

Step by step solution

01

Define the Hypotheses

The null hypothesis \( H_0 \) is that the proportion of U.S adults who approve of casino gambling, denoted \( p \), is equal to two-thirds, i.e., \( p \leq 2/3 \). The alternative hypothesis \( H_1 \) is that more than two-thirds of U.S adults approve of casino gambling, i.e., \( p > 2/3 \).
02

Calculate the Sample Proportion

The sample proportion \(\hat{p}\) is the number of adults who approve of casino gambling divided by the total number of adults in the sample. So, \(\hat{p} = 1035 / 1523 = 0.6797 \) or \(67.97\% \).
03

Conduct the Hypothesis Test

Now, we need to see if the sample data provide enough evidence to reject the null hypothesis. We calculate the test statistic which is \(Z = (\hat{p} - p_0) / \sqrt{(p_0*(1-p_0)/n)} \), where \( p_0 \) is the proportion under the null hypothesis, \(\hat{p}\) is the sample proportion, and \( n \) is the sample size. Here, \( p_0 = 2/3 = 0.67 \), \( \hat{p} = 67.97\% = 0.6797 \), and \( n = 1523 \). So, \(Z = (0.6797 - 0.67) / \sqrt{(0.67*(1-0.67)/1523)} = 1.16 \)
04

Determine the P-value and Make the Decision

The P-value is the probability of observing a sample proportion as extreme as the one measured in the sample or more extreme, given that the null hypothesis is true. Since this is a one-tailed test to the right, the P-value is the tail-area to the right of \(Z = 1.16\) in a standard Normal distribution, which is \(P(Z > 1.16) = 0.1230\). The P-value is larger than the significance level of 0.05, so we fail to reject the null hypothesis. Thus, there is insufficient evidence to conclude that the proportion of U.S adults who approve casino gambling is more than two-thirds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \( H_0 \), is a statement that assumes no effect or no difference. It's a starting point stating that nothing unusual will happen and is often contrary to our belief or research prediction.

In this case, the null hypothesis is that the proportion of U.S adults who approve of casino gambling is equal to two-thirds (\( p \leq 2/3 \)). The reason it's set up this way is that null hypotheses generally include an "equals" part to provide a baseline against which the real-world sample data are tested. So, our \( H_0 \) is that the true approval rate is not more than the baseline two-thirds proportion.

Testing a null hypothesis starts by assuming it is true. After performing calculations and statistical tests, we aim to see if there is enough evidence in the sample to reject this assumption in favor of the alternative hypothesis (\( H_1 \)), which claims more than two-thirds approve.
Sample Proportion
The sample proportion, denoted as \( \hat{p} \), is a statistic that estimates the proportion of a specified characteristic within a sample. It's like a mini-preview of the whole population's opinion.

In the exercise, the sample proportion represents the fraction of adults in the chosen sample who approve of casino gambling. Here, we take the number of approvals (\(1035\)) and divide it by the total sample size (\(1523\)), giving us a sample proportion \(\hat{p} = 1035 / 1523 = 0.6797\) or \(67.97\%\).

Calculating the sample proportion helps us understand what the collected sample tells us about the larger population. When well-chosen and large enough, a sample can accurately reflect the population鈥檚 sentiment.
P-value
The P-value in hypothesis testing quantifies the probability of obtaining a sample result at least as extreme as the one observed, assuming the null hypothesis is true.

In simpler terms, it's a measure that helps us understand how surprising or "unlikely" our sample data is under the null hypothesis. If it is very low, it suggests that what we've observed is quite unusual under the null assumption, nudging us towards questioning the null hypothesis validity.

For the example at hand, with our test statistic \(Z = 1.16\), we find a corresponding P-value of \(P(Z > 1.16) = 0.1230\). Since this P-value is greater than the typical significance level of \(0.05\), we don't have sufficient evidence to reject the null hypothesis. In essence, the observed sample isn't convincingly distinct from what the null hypothesis would predict.

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Most popular questions from this chapter

Medical personnel are required to report suspected cases of child abuse. Because some diseases have symptoms that mimic those of child abuse, doctors who see a child with these symptoms must decide between two competing hypotheses: \(H_{0}:\) symptoms are due to child abuse \(H_{a}:\) symptoms are due to disease (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) The article "Blurred Line Between Illness, Abuse Creates Problem for Authorities" (Macon Telegraph, February 28, 2000) included the following quote from a doctor in Atlanta regarding the consequences of making an incorrect decision: "If it's disease, the worst you have is an angry family. If it is abuse, the other kids (in the family) are in deadly danger." a. For the given hypotheses, describe Type I and Type II errors. b. Based on the quote regarding consequences of the two kinds of error, which type of error does the doctor quoted consider more serious? Explain.

Ann Landers, in her advice column of October 24,1994 (San Luis Obispo Telegram-Tribune), described the reliability of DNA paternity testing as follows: "To get a completely accurate result, you would have to be tested, and so would (the man) and your mother. The test is \(100 \%\) accurate if the man is not the father and \(99.9 \%\) accurate if he is." a. Consider using the results of DNA paternity testing to decide between the following two hypotheses: \(H_{0}:\) a particular man is the father \(H_{a}\) : a particular man is not the father In the context of this problem, describe Type I and Type II errors. (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) b. Based on the information given, what are the values of \(\alpha\), the probability of a Type I error, and \(\beta\), the probability of a Type II error? c. Ann Landers also stated, "If the mother is not tested, there is a \(0.8 \%\) chance of a false positive." For the hypotheses given in Part (a), what is the value of \(\beta\) if the decision is based on DNA testing in which the mother is not tested?

The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let \(p\) be the proportion of all apartments that prohibit children. If the city council is convinced that \(p\) is greater than \(0.75\), it will consider appropriate legislation. a. If 102 of the 125 sampled apartments exclude renters with children, would a level \(.05\) test lead you to the conclusion that more than \(75 \%\) of all apartments exclude children? b. What is the power of the test when \(p=.8\) and \(\alpha=.05 ?\)

An automobile manufacturer is considering using robots for part of its assembly process. Converting to robots is an expensive process, so it will be undertaken only if there is strong evidence that the proportion of defective installations is lower for the robots than for human assemblers. Let \(p\) denote the proportion of defective installations for the robots. It is known that human assemblers have a defect proportion of \(.02\). a. Which of the following pairs of hypotheses should the manufacturer test: \(H_{0}: p=.02\) versus \(H_{a}: p<.02\) or \(H_{0}: p=.02\) versus \(H_{a}: p>.02\) Explain your answer. b. In the context of this exercise, describe Type I and Type II errors. c. Would you prefer a test with \(\alpha=.01\) or \(\alpha=.1\) ? Explain your reasoning.

A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]: \(1369-1374\) ). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) at lunchtime in New York City were approached as they entered the restaurant and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburger-chain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2000 calories, the online Healthy Dining Finder (www .healthydiningfinder.com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=.01\). d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not.e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased. f. Do you think that asking a customer to provide his or her receipt before they ordered could have introduced a potential bias? Explain.
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