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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

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the sample observations was 12.7 hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of .05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.5 hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of .05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than 12.5 hours. c. Explain why the null hypothesis was rejected in the test of Part (b) but not in the test of Part (a).

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed \(100 \mathrm{lb} / \mathrm{in}^{2}\). The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

The National Cancer Institute conducted a 2 -year study to determine whether cancer death rates for areas near nuclear power plants are higher than for areas without nuclear facilities (San Luis Obispo Telegram-Tribune, September 17,1990 ). A spokesperson for the Cancer Institute said, "From the data at hand, there was no convincing evidence of any increased risk of death from any of the cancers surveyed due to living near nuclear facilities. However, no study can prove the absence of an effect." a. Let \(p\) denote the proportion of the population in areas near nuclear power plants who die of cancer during a given year. The researchers at the Cancer Institute might have considered the two rival hypotheses of the form \(H_{0}: p=\) value for areas without nuclear facilities \(H_{a}: p>\) value for areas without nuclear facilities Did the researchers reject \(H_{0}\) or fail to reject \(H_{0} ?\) b. If the Cancer Institute researchers were incorrect in their conclusion that there is no increased cancer risk associated with living near a nuclear power plant, are they making a Type I or a Type II error? Explain. c. Comment on the spokesperson's last statement that no study can prove the absence of an effect. Do you agree with this statement?

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