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91Ó°ÊÓ

In a representative sample of 2,013 American adults, 1,590 indicated that lack of respect and courtesy in American society is a serious problem (Associated Press, April 3,2002 ). Is there convincing evidence that more than three- quarters of American adults believe that lack of respect and courtesy is a serious problem? Test the relevant hypotheses using a significance level of 0.05 .

Short Answer

Expert verified
The answer will depend on the calculated p-value compared with the significance level 0.05. If the P-value is less than 0.05, one can conclude there is convincing evidence that more than 75% of American adults believe lack of respect and courtesy is a serious problem.

Step by step solution

01

Formulating the Hypotheses

The null hypothesis \(H_0\) would assume that the proportion p (representing the American adults who believe lack of respect and courtesy is a problem) is equal to 0.75 (75%). The alternative hypothesis \(H_1\) would propose that the proportion is greater than 0.75. Formally, this can be represented as: \(H_0:p=0.75\) and \(H_1:p>0.75\).
02

Performing the Z-test

We need to calculate the test statistic which is a Z-score in this case. The formula is \(Z = \frac {(\hat{p} - p_{0})}{\sqrt{\frac{p_{0}*(1-p_{0})}{n}}}\). Here, \( \hat{p}\) is the sample proportion, \(p_0\) is the proportion in the null hypothesis, and n is the sample size. The sample proportion \(\hat{p}\) = 1590/2013 ≈ 0.79. Substituting these values into the formula, we can calculate the Z-score.
03

Calculating the P-value

The P-value represents the probability of observing a sample statistic as extreme as the test statistic. Since the claim is that more than three-quarters of the population have noticed the problem, this is a right-tailed test. To determine the P-value, we need to find the probability that a Z-score is greater than the calculated test statistic under the standard normal distribution.
04

Comparing the P-value with the Significance Level

The null hypothesis is rejected if the P-value is less than the significance level. The significance level is given as 0.05. If the calculated P-value is less than the significance level, we can conclude there is enough evidence to reject the null hypothesis and support that more than three-quarters of American adults believe that lack of respect and courtesy is a serious problem.
05

Conclusion

Based on the P-value and the significance level, we conclude whether or not there is convincing evidence that more than 75% of American adults believe lack of respect and courtesy is a serious problem.

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Key Concepts

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

Null and Alternative Hypotheses in Hypothesis Testing
Understanding the null and alternative hypotheses is the foundation of hypothesis testing in statistics. The null hypothesis (\( H_0 \) is a statement of no effect or no difference and serves as a default or starting assumption. In context of the provided example, the null hypothesis posits that the proportion of American adults who believe lack of respect and courtesy is a serious problem is equal to 75% (\( p = 0.75 \)).

The alternative hypothesis (\( H_1 \) contradicts the null hypothesis. It is what the researcher aims to prove, and in our example, it posits that the proportion is greater than 75% (\( p > 0.75 \)). This formulation is crucial as it defines the direction of the testing and what evidence is required to reject the null hypothesis in favor of the alternative hypothesis.

A well-formulated null and alternative hypothesis sets the stage for meaningful analysis. Without a clearly defined null and alternative hypothesis, the rest of the hypothesis testing process does not hold much significance.
Z-test Statistic Calculation
Calculating the Z-test statistic is an integral step in determining if the sample data provides sufficient evidence to reject the null hypothesis. The Z-test is specifically used when comparing a sample proportion to a known population proportion, assuming the sample size is large enough for the central limit theorem to apply.

The formula for the Z-test statistic is given by: \( Z = \frac {(\hat{p} - p_{0})}{{\sqrt{\frac{p_{0}*(1-p_{0})}{n}}}} \). Here, \( \hat{p} \) is the sample proportion, \( p_0 \) is the hypothesized population proportion, and \( n \) is the sample size. The Z-score measures how many standard deviations the sample proportion is away from the hypothesized population proportion.

In our exercise, by inputting the values into this formula, we obtain a Z-score that quantifies the discrepancy between the observed data and the null hypothesis. A higher Z-score indicates that the observed sample proportion is much higher than the expected under the null hypothesis, suggesting stronger evidence against it.
P-value Significance
The P-value is a vital concept in hypothesis testing and measures the strength of the evidence against the null hypothesis. It quantifies the probability of obtaining a sample statistic as extreme as the one observed (or more extreme), assuming that the null hypothesis is true.

For a given significance level (\( \alpha \) the P-value allows us to make a decision regarding the null hypothesis. If the P-value is less than \( \alpha \) we reject the null hypothesis, concluding that our sample provides sufficient evidence to support the alternative hypothesis. In our example, the significance level is 0.05. If the computed P-value based on the Z-score is below this threshold, it implicates that there's statistically significant evidence to suggest that more than three-quarters of American adults view the lack of respect and courtesy as a serious problem.

The interpretation of the P-value is pivotal in drawing conclusions: a small P-value indicates that our observed result would be very unlikely under the assumption that the null hypothesis is true, paving the way to support the alternative hypothesis with statistical evidence.

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Most popular questions from this chapter

Suppose that for a particular hypothesis test, the consequences of a Type I error are not very serious, but there are serious consequences associated with making a Type II error. Would you want to carry out the test using a small significance level \(\alpha\) (such as 0.01 ) or a larger significance level (such as 0.10 )? Explain the reason for your choice.

Every year on Groundhog Day (February 2), the famous groundhog Punxsutawney Phil tries to predict whether there will be 6 more weeks of winter. The article "Groundhog Has Been Off Target" (USA Today, Feb. 1,2011 ) states that "based on weather data, there is no predictive skill for the groundhog." Suppose that you plan to take a random sample of 20 years and use weather data to determine the proportion of these years the groundhog's prediction was correct. a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for samples of size 20 if the groundhog has only a \(50-50\) chance of making a correct prediction. b. Based on your answer to Part (a), what sample proportion values would convince you that the groundhog's predictions have a better than \(50-50\) chance of being correct?

In a representative sample of 1,000 adult Americans, only 430 could name at least one justice who was currently serving on the U.S. Supreme Court (Ipsos, January 10,2006 ). Using a significance level of \(0.01,\) determine if there is convincing evidence in support of the claim that less than half of adult Americans can name at least one justice currently serving on the Supreme Court.

A survey of 1,000 adult Americans ("Military Draft Study," AP-Ipsos, June 2005 ) included the following question: "If the military draft were reinstated, would you favor or oppose drafting women as well as men?" Forty-three percent responded that they would favor drafting women if the draft were reinstated. Using the five-step process for hypothesis testing \(\left(\mathrm{HMC}^{3}\right)\) and a 0.05 significance level, determine if there is convincing evidence that less than half of adult Americansp favor drafting women.

The article "The Benefits of Facebook Friends: Social Capital and College Students' Use of Online Social Network Sites" (Journal of Computer-Mediated Communication [2007]: \(1143-1168\) ) describes a study of \(n=286\) undergraduate students at Michigan State University. Suppose that it is reasonable to regard this sample as a random sample of undergraduates at Michigan State. You want to use the survey data to decide if there is evidence that more than \(75 \%\) of the students at this university have a Facebook page that includes a photo of themselves. Let \(p\) denote the proportion of all Michigan State undergraduates who have such a page. (Hint: See Example 10.10\()\) a. Describe the shape, center, and spread of the sampling distribution of \(\hat{p}\) for random samples of size 286 if the null hypothesis \(H_{0}: p=0.75\) is true. b. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.83\) for a sample of size 286 if the null hypothesis \(H_{0}: p=0.75\) were true? Explain why or why not. c. Would you be surprised to observe a sample proportion as large as \(\hat{p}=0.79\) for a sample of size 286 if the null hypothesis \(H_{0}: p=0.75\) were true? Explain why or why not. d. The actual sample proportion observed in the study was \(\hat{p}=0.80 .\) Based on this sample proportion, is there convincing evidence that the null hypothesis \(H_{0}: p=\) 0.75 is not true, or is \(\hat{p}\) consistent with what you would expect to see when the null hypothesis is true? Support your answer with a probability calculation.

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