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91Ó°ÊÓ

Let \(p\) denote the proportion of students living on campus at a large university who plan to move off campus in the next academic year. For a large sample \(z\) test of \(H_{0}: p=0.70\) versus \(H_{\mathrm{a}}: p>0.70,\) find the \(P\) -value associated with each of the following values of the \(z\) test statistic. a. 1.40 b. 0.92 c. 1.85 d. 2.18 e. -1.40

Short Answer

Expert verified
The P-values for the given z-test statistics can be found with the use of a standard normal distribution table or statistical software. If the calculated P-value is less than the significance level (0.05 is commonly used), it provides strong evidence against the null hypothesis that the proportion of students who plan to move off campus the next year is 0.70.

Step by step solution

01

Understand the problem and identify P-value

The problem is asking to find the P-value for each given z test statistic. P-values are obtained by integrating the standard normal curve from the value of the z score to infinity because the test is right-tailed (i.e., \(H_{a}: p>0.70\)). The aim is to find the area under the curve for the given values of z.
02

Calculation of P-value for each z-value

The P-value is calculated by looking up the z-scores in a table of the standard normal distribution, or using a function such as norm.sf(z-score) in a statistical software or programming language that incorporates statistical functions, and then subtracting the result from 1 if necessary to get a right-tail probability. Repeat this process for each of the given z-values: 1.40, 0.92, 1.85, 2.18, and -1.40.
03

Interpret the results

The calculated P-values represent the probability of obtaining a z-score as extreme as the one observed given that the null hypothesis is true. The smallest the P-value, the strongest is the evidence against the null hypothesis in favor of the alternative. Compare the calculated P-values with the significance level (usually 0.05). If the P-value is smaller than the significance level, you would reject the null hypothesis.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to determine whether there is enough evidence in a sample of data to infer that a certain condition holds true for the entire population. The process begins by proposing two opposing hypotheses: the null hypothesis (\(H_{0}\)) and the alternative hypothesis (\(H_{a}\)). The null hypothesis typically represents a baseline or status quo condition, while the alternative hypothesis represents what the researcher is trying to provide evidence for. Hypothesis testing then uses sample data to decide whether to reject the null hypothesis in favor of the alternative, based on a calculated probability value, or P-value.

By setting a significance level, typically 0.05, researchers have a cutoff for deciding whether to reject the null hypothesis. If the P-value is lower than the significance level, it suggests that the observed data is highly unlikely under the assumption that the null hypothesis is true, supporting the alternative hypothesis.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution that has a mean of zero and a standard deviation of one. This distribution is symmetrical around the mean, and it has a bell-shaped curve, reflecting how random variables are expected to be distributed in many natural and social phenomena. When dealing with hypothesis testing and P-values, the z test statistic, derived from our sample data, is often compared against this standard normal distribution.

Since it's standardized, it allows us to transform scores from a normal distribution into a comparable form called z-scores. Z-scores indicate how many standard deviations an element is from the mean and can be used to calculate probabilities and P-values for hypothesis tests.
Statistical Significance
In hypothesis testing, statistical significance is determined by the P-value, which measures the strength of the evidence against the null hypothesis. If the P-value is low, the test results are statistically significant, meaning the difference found in the data is unlikely to be due to chance alone. A common threshold for statistical significance is a P-value of less than 0.05, which means there is less than a 5% probability that the observed results are a consequence of random variation.
Null Hypothesis
The null hypothesis, denoted as \(H_{0}\), constitutes the default statement that there is no effect or no difference, and any observed effect is due to sampling or experimental error. For example, in the provided exercise, the null hypothesis claims that the proportion of students living on campus who plan to move off campus is 70%, represented as \(H_{0}: p=0.70\). When conducting a z test, the null hypothesis becomes a benchmark to determine if the data provides enough evidence to support the alternative hypothesis.
Alternative Hypothesis
Opposing the null hypothesis is the alternative hypothesis, represented as \(H_{a}\) or \(H_{1}\), which asserts that there is a statistically significant effect or difference that the research aims to detect. In the context of our exercise, the alternative hypothesis suggests that more than 70% of students plan to leave campus housing, written as \(H_{a}: p>0.70\). If the P-value obtained from the z test is low enough, it supports the alternative hypothesis, suggesting a movement of the student proportion beyond the 70% indicated by the null hypothesis.
Normal Distribution Tables
To find the P-value from a z test statistic, one can use normal distribution tables, also known as z-tables. These tables provide the cumulative probability associated with a given z-score, which typically represents the area under the curve to the left of that z-score on the standard normal distribution. For the right-tailed test in our exercise (where \(H_{a}: p>0.70\)), the P-value can be found by subtracting the table value from 1. Since z-tables are based on a standardized distribution, they provide a convenient way to translate z-scores into probabilities without doing complex calculations.

It's important to note that the P-value is not the absolute probability of the null hypothesis being true or false but rather measures the probability of obtaining a result at least as extreme as the one observed, given that the null hypothesis is true.

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Most popular questions from this chapter

The authors of the article "Perceived Risks of Heart Disease and Cancer Among Cigarette Smokers" (Journal of the American Medical Association [1999]: \(1019-1021\) ) expressed the concern that a majority of smokers do not view themselves as being at increased risk of heart disease or cancer. A study of 737 current smokers found that only 295 believe they have a higher than average risk of cancer. Do these data suggest that \(p,\) the proportion of all smokers who view themselves as being at increased risk of cancer, is less than \(0.5,\) as claimed by the authors of the paper? For purposes of this exercise, assume that this sample is representative of the population of smokers. Test the relevant hypotheses using \(\alpha=0.05\)

A television manufacturer states that at least \(90 \%\) of its TV sets will not need service during the first 3 years of operation. A consumer group wants to investigate this statement. A random sample of \(n=100\) purchasers is selected and each person is asked if the set purchased needed repair during the first 3 years. Let \(p\) denote the proportion of all sets made by this manufacturer that will not need service in the first 3 years. The consumer group does not want to claim false advertising unless there is strong evidence that \(p<0.9\). The appropriate hypotheses are then \(H_{0}: p=0.9\) versus \(H_{a}: p<0.9\). a. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=\) 0.01 or one that uses \(\alpha=0.10 ?\) Explain.

Consider the following quote from the article "Review Finds No Link Between Vaccine and Autism" (San Luis Obispo Tribune, October 19,2005 ): "We found no evidence that giving MMR causes Crohn's disease and/or autism in the children that get the MMR,' said Tom Jefferson, one of the authors of The Cochrane Review. 'That does not mean it doesn't cause it. It means we could find no evidence of it." (MMR is a measles-mumps-rubella vaccine.) In the context of a hypothesis test with the null hypothesis being that MMR does not cause autism, explain why the author could not conclude that the MMR vaccine does not cause autism.

In a survey of 1,005 adult Americans, \(46 \%\) indicated that they were somewhat interested or very interested in having Web access in their cars (USA Today, May 1,2009 ). Suppose that the marketing manager of a car manufacturer claims that the \(46 \%\) is based only on a sample and that \(46 \%\) is close to half, so there is no reason to believe that the proportion of all adult Americans who want car Web access is less than \(0.50 .\) Is the marketing manager correct in his claim? Provide statistical evidence to support your answer. For purposes of this exercise, assume that the sample can be considered representative ofp adult Americans.

Step 2 of the five-step process for hypothesis testing is selecting an appropriate method. What is involved in completing this step?

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