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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?

Short Answer

Expert verified
Using hypothesis testing, calculate the test statistic using the given formula and compare it with the critical value obtained from the z-table. If the test statistic is greater, reject the null hypothesis; if less or equal, don't reject. The answer will depend on the calculated value of the test statistic.

Step by step solution

01

Formulate the Hypothesis

The null hypothesis (H0) is that the population proportion (p) equals 0.2 (20%). The alternative hypothesis (Ha) is that the percentage is greater than 0.2; in symbols, we write this as:\n H0: p = 0.2\n Ha: p> 0.2
02

Calculate the Test Statistic

Use the formula for the test statistic in a one-sample z test for a proportion: \n Z = (p̂ - p0) / (sqrt [(p0(1-p0))/n]) \n where p̂ is the sample proportion, p0 is the proportion in the null hypothesis and n is the sample size. In this case, p̂ = 210/1000 = 0.21, p0 = 0.2 and n = 1000. Substituting these values, the test statistic Z is computed.
03

Get the Z critical value and Draw Conclusion

Since the problem involves claiming that MORE than 20% population believes in it, it's a one-sided z-test. For 95% confidence level, the critical value for one-sided test from z-table is 1.645 (approx). If the test statistic computed is higher than the critical value, then we reject the null hypothesis, meaning there is enough evidence to say more than 20% of Americans believe the lottery is the best way to accumulate wealth. If the test statistic is less or equal than the critical value, then we can't reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is a foundational concept in hypothesis testing. It represents a statement we aim to test and typically suggests no effect or no difference. In the context of this problem, the null hypothesis claims that the proportion of adult Americans who think playing the lottery is the best way to gather \(\$200,000\) in wealth equals 20%. Formally, it is expressed as:
  • \(H_0: p = 0.2\)
Establishing a clear null hypothesis is crucial as it provides a baseline against which we compare observed data. Rejecting or failing to reject the null hypothesis is central in drawing meaningful conclusions.
This hypothesis acts as a default assumption that the sampled data must convincingly contradict for any claims suggesting an alternative be considered valid.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_a\), stands in opposition to the null hypothesis. It represents a new claim we want to test, generally suggesting a specific effect or a difference. In our survey exercise, the alternative hypothesis posits that the proportion of adult Americans believing in the lottery as a key to wealth exceeds 20%.
  • \(H_a: p > 0.2\)
This one-sided hypothesis reflects a directionally specific inquiry, looking to demonstrate an increase over the hypothesized 20%.
Choosing between one-sided and two-sided hypotheses is important; it guides the direction and nature of statistical tests we apply. In this case, the alternative hypothesis implies a belief in the prevalence of lottery-playing preference among Americans beyond what's initially assumed.
One-Sample Z Test
The one-sample z-test is a statistical test used to determine if there is a significant difference between a sample proportion and a known population proportion. It is particularly useful when the population standard deviation is known or when the sample size is reasonably large, typically over 30. In this case, we are comparing the survey proportion to the hypothesized value of 20%.
Using the formula for the z-test statistic:
  • \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)
Where:
  • \( \hat{p} = \frac{210}{1000} = 0.21 \), the sample proportion
  • \( p_0 = 0.2 \), the null hypothesis proportion
  • \( n = 1000 \), sample size
Calculate the test statistic, and compare it against a critical z-value to determine if the null hypothesis can be rejected. Critical values depend on the confidence level chosen; here, for a 95% confidence level, it is approximately 1.645 for a one-sided test. A computed z-value greater than this suggests significant evidence for the alternative hypothesis.
Survey Analysis
Survey analysis plays a pivotal role in understanding public opinion or behavior trends. In this exercise, a survey of 1000 adult Americans provides a snapshot of beliefs regarding wealth accumulation strategies. Statistical tests, like the one-sample z-test utilized here, help determine the broader implications of this sample data for the entire population.
Key considerations in survey analysis include:
  • Sample size: Larger samples typically provide more reliable estimates of population parameters.
  • Random sampling: Ensures every member of the population has an equal chance of being selected, reducing bias.
  • Data Interpretation: Statistical results must be translated into meaningful insights about population behaviors and beliefs.
By interpreting the survey data accurately and using hypothesis tests, researchers can arrive at evidence-backed conclusions. This assists in painting a clear picture of societal trends and tendencies, like beliefs about the lottery as a means to wealth.

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Most popular questions from this chapter

10.48 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.

In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA Today. March 4, 2010 ). The sample was selected in a way that was designed to produce a sample that was representative of women in the targeted age group. a. Do the sample data provide convincing evidence that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money? Test the relevant hypotheses using \(\alpha=.01\). b. Would it be reasonable to generalize the conclusion from Part (a) to all working women? Explain why or why not.

A researcher speculates that because of differences in diet, Japanese children may have a lower mean blood cholesterol level than U.S. children do. Suppose that the mean level for U.S. children is known to be 170 . Let \(\mu\) represent the mean blood cholesterol level for all Japanese children. What hypotheses should the researcher test?

The true average diameter of ball bearings of a certain type is supposed to be 0.5 inch. What conclusion is appropriate when testing \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq\) 0.5 inch each of the following situations: a. \(n=13, t=1.6, \alpha=.05\) b. \(n=13, t=-1.6, \alpha=.05\) c. \(n=25, t=-2.6, \alpha=.01\) d. \(\quad n=25, t=-3.6\)

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.

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