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Public Agenda conducted a survey of 1379 parents and 1342 students in grades \(6-12\) regarding the importance of science and mathematics in the school curriculum (Assodated Press, February 15,2006 ). It was reported that \(50 \%\) of students thought that understanding science and having strong math skills are essential for them to succeed in life after school, whereas \(62 \%\) of the parents thought it was crucial for today's students to learn science and higher-level math. The two samples -parents and students-were selected independently of one anorher. Is there sufficient evidence to conclude that the proportion of parents who regard science and mathematics as crucial is different than the corresponding proportion for students in grades \(6-12\) ? 'Test the relevant hypotheses using a significance level of .05.

Short Answer

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Step by step solution

01

Write down the given information

First note down the information given in the exercise. We are told that \(50 \%\) of students thought understanding science and math is crucial, this is \(p_s = 0.5 \) or the proportion of students. Similarly, the exercise states that \(62 \%\) of parents thought it was crucial, so the proportion of parents is \(p_p = 0.62\). The survey sizes were \(n_s = 1342\) for students and \(n_p = 1379\) for parents. Our significance level, \(\alpha\), is \(0.05\).
02

Define the Null and Alternative Hypothesis

The null hypothesis \(H_0\) is that there is no difference between the proportions of students and parents who consider math and science as necessary, thus \(H_0 : p_s = p_p \). The alternative hypothesis \(H_1\) is that there is a difference, so \(H_1 : p_s \neq p_p \).
03

Comparison of two sample proportions

Establish a pooled proportion using the formula: \(p = ( (n_s * p_s) + (n_p * p_p) ) / (n_s + n_p)\) which accounts for the total number of 'successes' over the total sample size.
04

Calculate Test Statistic

The test statistic for the difference between two proportions is given by \(z = (p_s - p_p) / \sqrt{p * (1-p) * [ (1/n_s) + (1/n_p) ]}\). This will give us the observed difference between the proportions relative to what one would expect if the null hypothesis was true.
05

Decision Rule

Using a significance level of \(\alpha = 0.05\) for a two-sided test, the critical value from the standard normal distribution is \(\pm 1.96\). If the calculated \(|z|\) is greater than 1.96, then reject the null hypothesis.
06

Conclusion

Based on the achieved z score and the decision rule, we can then make our conclusion about the null hypothesis. If we reject the null, it would suggest that there is a difference between the proportions. Otherwise, there isn't enough evidence to suggest a difference.

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Key Concepts

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

Statistical Significance
Understanding statistical significance is crucial for interpreting the results of a hypothesis test in statistics. It helps to determine whether the observed effect is due to chance or if it reflects a real difference or effect in the population being studied. In the context of our exercise, statistical significance is assessed at a certain threshold known as the significance level, denoted by \( \alpha \). The commonly used significance level is 0.05, implying a 5% risk of concluding that a difference exists when there is none—in other words, a 5% chance of making a Type I error.

When we compute the test statistic—in this case, the z-score for the comparison of two sample proportions—we compare that statistic to a critical value from the standard normal distribution. If the absolute value of our test statistic, or \( |z| \), is greater than this critical value, the result is considered statistically significant. This means we would reject the null hypothesis, as the likelihood of the observed difference occurring by chance is less than 5%. Statistical significance does not measure the size or importance of an effect, only whether the evidence suggests it is unlikely to be due to random variation alone.
Null and Alternative Hypothesis
In hypothesis testing, we begin by clearly stating two competing hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). The null hypothesis represents the default position or status quo. It posits that there is no effect or no difference—in this exercise, it's the assertion that there's no difference between the proportion of students and parents who consider math and science essential for future success, which we express as \( H_0 : p_s = p_p \).

The alternative hypothesis represents what we aim to prove or suggest. It is the hypothesis that there is an effect or a difference, \( H_1 : p_s eq p_p \) in our case. It specifies that the proportions are not equal, meaning that one group views math and science as crucial at a different rate than the other.

Our goal is to determine if the observed data provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis. A common misunderstanding involves thinking a 'non-rejection' of the null means it is true. However, it simply means there isn't sufficient evidence against it given the data and the significance level.
Comparison of Two Sample Proportions
Comparing two sample proportions is a common statistical method used to determine if there is a significant difference between the proportions of two independent groups. In our problem, we're comparing the proportion of students (\( p_s \)) who think understanding science and math is crucial to success after school to the proportion of parents (\( p_p \)) who believe the same.

To do this, we calculate a pooled proportion which takes into account combined successes and sample sizes from both groups. This pooled proportion \( p \) is used under the assumption that the null hypothesis is true, and it serves as the basis for calculating the standard error and, subsequently, the test statistic \( z \).

If the test statistic falls outside the range defined by our critical values (determined by our significance level of 0.05), we have enough evidence to suggest a significant difference between the two sample proportions. This process is not only foundational in hypothesis testing but also extremely relevant in a variety of practical scenarios, such as polling differences, medical treatment effectiveness, market research, and much more.

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Most popular questions from this chapter

Are college students who take a freshman orientation course more or less likely to stay in college than those who do not take such a course? The article "A Longitudinal Study of the Retention and Academic Performance of Participants in Freshmen Orientation Courses" (journal of College Student Development [1994]: \(444-\) 449) reported that 50 of 94 randomly selected students who did not participate in an orientation course returned for a second year. Of 94 randomly selected students who did take the orientation course, 56 returned for a second year. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\), the difference in the proportion returning for students who do not take an orientation course and those who do. Give an interpretation of this interval.

Some commercial airplanes recirculate approximately \(50 \%\) of the cabin air in order to increase fuel efficiency. The authors of the paper "Aircraft Cabin Air Redrculation and Symptoms of the Common Cold" (journal of the American Medical Association [2002]: 483-486) studied 1100 airline passengers who flew from San Francisco to Denver between January and April 1999\. Some passengers traveled on airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 517 passengers who flew on planes that did not recirculate air, 108 reported post-flight respiratory symptoms, while 111 of the 583 passengers on planes that did recirculate air reported such symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-flight respiratory symptoms differs for planes that do and do not recirculate air? Test the appropriate hypotheses using \(\alpha=.05\). You may assume that it is reasonable to regard these two samples as being independently selected and as representative of the two populations of interest.

The paper "If It's Hard to Read, It's Hard to Do" (Psychological Science [20081: 986-988) described an interesting study of how people perceive the effort required to do certain tasks. Each of 20 students was randomly assigned to one of two groups. One group was given instructions for an exercise routine that were printed in an easy-to-read font (Arial). The other group received the same set of instructions, but printed in a font that is considered difficult to read (Brush). After reading the instructions, subjects estimated the time (in minutes) they thought it would take to complete the exercise routine. Summary statistics are given below. The authors of the paper used these data to carry out a two-sample \(t\) test, and concluded that at the 10 significance level, there was convincing evidence that the mean estimated time to complete the exercise routine was less when the instructions were printed in an easy-to-read font than when printed in a difficult-to-read font. Discuss the appropriateness of using a two-sample \(t\) test in this situation.

An electronic implant that stimulates the auditory nerve has been used to restore partial hearing to a number of deaf people. In a study of implant acceptability (Los Angeles Times. January 29,1985 ), 250 adults born deaf and 250 adults who went deaf after learning to speak were followed for a period of time after receiving an implant. Of those deaf from birth, 75 had removed the implant, whereas only 25 of those who went deaf after learning to speak had done so. Does this suggest that the true proportion who remove the implants differs for those who were born deaf and those who went deaf after learning to speak? 'Test the relevant hypotheses using a .01 significance level.

The press release referenced in the previous exercise also included data from independent surveys of teenage drivers and parents of teenage drivers. In response to a question asking if they approved of laws banning the use of cell phones and texting while driving, \(74 \%\) of the teens surveyed and \(95 \%\) of the parents surveyed said they approved. The sample sizes were not given in the press release, but for purposes of this exercise, suppose that 600 teens and 400 parents of teens responded to the surveys and that it is reasonable to regard these samples as representative of the two populations. Do the data provide convincing evidence that the proportion of teens that approve of cell- phone and texting bans while driving is less than the proportion of parents of teens who approve? 'Test the relevant hypotheses using a significance level of \(.05\).

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