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"Most Like it Hot" is the title of a press release issued by the Pew Research Center (March 18, 2009 , www.pewsocialtrends.org). The press release states that "by an overwhelming margin, Americans want to live in a sunny place." This statement is based on data from a nationally representative sample of 2260 adult Americans. Of those surveyed, 1288 indicated that they would prefer to live in a hot climate rather than a cold climate. Do the sample data provide convincing evidence that a majority of all adult Americans prefer a hot climate over a cold climate? Use the nine-step hypothesis testing process with \(\alpha=.01\) to answer this question.

Short Answer

Expert verified
The sample data provides convincing evidence at the 0.01 significance level that a majority of all adult Americans prefer a hot climate over a cold climate. Further, we are 99% confident that the true population proportion preferring a hot climate is between 55% and 59%.

Step by step solution

01

State the Hypotheses

The null hypothesis (H0) asserts that the majority of adult Americans do not prefer hot weather, i.e., the proportion is less than or equal to 0.5. The alternative hypothesis (H1) proposes that the majority of adult Americans prefer hot weather, i.e., the proportion is more than 0.5.So, \(H_0: p \leq 0.5\)And \(H_1: p > 0.5\)
02

Formulate an Analysis Plan

For this analysis, the test statistic is a z-score (z). The significance level is 0.01.
03

Analyze Sample Data

Using sample data, we calculate the pooled sample proportion (\(p\)) and standard error (\(SE\)). Using those measures, we compute the value of the test statistic (z). The number of successes is 1288, the number of observations is 2260. \(p = 1288/2260 = 0.57\). And, \(SE = sqrt{ [ p * ( 1 - p ) / n ] } = sqrt{ [0.57 * ( 1 - 0.57 ) / 2260 ] }=0.01.\)
04

Compute the Test Statistic

Next, we find the z-score by subtracting the null hypothesized value from the observed sample proportion and then dividing by the standard error: \(z = (p - 0.5) / SE = (0.57 - 0.5) / 0.01 = 7.\)
05

Find the P-Value

The P-value associated with a 7.0 z-score for a one-sided test is 0, because a 7.0 score is so high that the area in the z-distribution above 7.0 is virtually 0.
06

Interpret the Results

Since the P-value (0) is less than the significance level (0.01), we reject the null hypothesis.
07

State the Conclusion

We reject the null hypothesis and conclude that the data provides convincing evidence that a majority of all adult Americans prefer a hot climate over a cold climate, at a 1% significance level.
08

Confidence Interval

Even though this step isn't required by the problem, it is a good practice to report a confidence interval for the unknown parameter. A confidence interval for a single population proportion is given by \(p ± z * SE\), where z is the z*-value for the chosen level of confidence. For the 99% level of confidence, the z*-value is 2.58. Thus, the confidence interval is \(0.57 ± 2.58 * 0.01 = (0.55, 0.59)\). This is the range in which we expect the true population proportion of adults preferring hot climate lies, with 99% confidence.
09

Validation

The sample size is large enough (n > 30), so the Central Limit Theorem ensures the sampling distribution of the sample proportion is approximately normally distributed. This validates our z-test.

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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 the starting point. It is the default assumption that there is no effect or no difference. In our specific case, the null hypothesis asserts that 50% or fewer of adult Americans prefer a hot climate over a cold one. This is formulated as \(H_0: p \leq 0.5\), where \(p\) represents the true proportion of adults preferring a hot climate. The null hypothesis is essentially saying that any observed preference for hot weather in the sampled data is due to random chance rather than being a true reflection of the population's preference.
Before analyzing the data, researchers assume the null hypothesis is true. It serves as the hypothesis to be tested and potentially refuted through statistical analysis. Only when there is significant evidence against it will it be rejected in favor of the alternative hypothesis. This foundation allows researchers to assess if the data provides enough evidence to make claims beyond what randomness would suggest.
Alternative Hypothesis
The alternative hypothesis (denoted as \(H_1\)) directly contrasts the null hypothesis. It reflects the outcome the researcher expects or aims to prove. In our example, the alternative hypothesis is that more than 50% of adult Americans prefer a hot climate. It is expressed as \(H_1: p > 0.5\). This statement assumes that the majority do have a preference for warm climates, suggesting a significant shift from the null hypothesis.
For hypothesis testing, the alternative hypothesis is "accepted" if the data provides strong enough evidence to reject the null hypothesis. It represents the possibility that the effect or difference being investigated really does exist in the population, as opposed to being a random occurrence in the sample data.
This hypothesis is crucial for decision-making because it helps guide the statistical procedures used and defines the nature of the claim being tested.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold used to determine if the observed data is statistically significant. Often set before the analysis, it is the probability of rejecting the null hypothesis when it is actually true. In our example, the significance level is set at 0.01, or 1%. This means that there is a 1% chance of concluding that more than half of adult Americans prefer a hot climate, even if this is not the case.

The choice of the significance level is critical. A lower \(\alpha\) level indicates a more stringent test, reducing the likelihood of a Type I error (false positive). The significance level depends on the context of the study and how serious the consequences of a false positive might be. With a significance level of 0.01, the probability of mistakenly rejecting the null hypothesis is kept very low.
P-Value
The P-value helps in making conclusions from hypothesis test results. It quantifies how likely it is to observe the data, or something more extreme, assuming the null hypothesis is true. In the example, the calculated P-value is practically 0 due to the extremely high z-score of 7.0. This low P-value indicates that observing such a strong preference for hot weather among the sample is nearly impossible if the majority of Americans do not actually prefer a hot climate.

A P-value is compared to the significance level, \(\alpha\). If the P-value is less than \(\alpha\), the null hypothesis is rejected. Here, because the P-value (0) is less than the significance level of 0.01, we can safely reject the null hypothesis and conclude that there is strong evidence that a majority of adult Americans prefer a hot climate over a cold one.

The P-value provides a way to measure the strength of evidence against the null hypothesis, helping to validate results.
Confidence Interval
A confidence interval offers a range of values that likely include the true population parameter. It is calculated from the sample data and provides a degree of certainty about where the population parameter lies. In this example, the confidence interval is built around the sample proportion of adults preferring hot weather, which is 0.57. For a 99% confidence level, we calculate the interval using the z*-value of 2.58, providing the interval (0.55, 0.59).

This implies that there is a 99% probability that the true proportion of Americans who prefer hot weather is between 55% and 59%.
  • Confidence intervals are useful because they give more than a simple, yes-or-no hypothesis test.
  • They allow us to estimate the parameter with some acknowledged level of uncertainty.
Incorporating confidence intervals in hypothesis testing offers a clearer picture of what the data suggests about the population.

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Most popular questions from this chapter

Assuming a random sample from a large population, for which of the following null hypotheses and sample sizes \(n\) is the large-sample \(z\) test appropriate: a. \(H_{0}: p=.2, n=25\) b. \(H_{0}: p=.6, n=210\) c. \(H_{0}: p=.9, n=100\) d. \(H_{0}: p=.05, n=75\)

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?

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. The relevant hypotheses are \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10 .\) a. If \(t=-2.3\) and \(\alpha=.05\) is selected, what conclusion is appropriate? b. If \(t=-1.83\) and \(\alpha=.01\) is selected, what conclusion is appropriate? c. If \(t=0.47,\) what conclusion is appropriate?

A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was 4.09 ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005\()\). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards that the students said they carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.2 . Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09 ?\)

The paper "Debt Literacy, Financial Experiences and Over-Indebtedness" (Social Science Research Network, Working paper W14808, 2008 ) included analysis of data from a national sample of 1000 Americans. One question on the survey was: "You owe \(\$ 3000\) on your credit card. You pay a minimum payment of \(\$ 30\) each month. At an Annual Percentage Rate of \(12 \%\) (or \(1 \%\) per month), how many years would it take to eliminate your credit card debt if you made no additional charges?" Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never-you will continue to be in debt; (e) don't know; and (f) prefer not to answer. a. Only 354 of the 1000 respondents chose the correct answer of never. For purposes of this exercise, you can assume that the sample is representative of adult Americans. Is there convincing evidence that the proportion of adult Americans who can answer this question correctly is less than \(.40(40 \%) ?\) Use \(\alpha=.05\) to test the appropriate hypotheses. b. The paper also reported that \(37.8 \%\) of those in the sample chose one of the wrong answers \((a, b,\) and \(c)\) as their response to this question. Is it reasonable to conclude that more than one-third of adult Americans would select a wrong answer to this question? Use \(\alpha=.05\).

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