/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Some commercial airplanes recirc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Some commercial airplanes recirculate approximately \(50 \%\) of the cabin air in order to increase fuel efficiency. The authors of the paper "Aircraft Cabin Air Recirculation 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.

Short Answer

Expert verified
The short answer would be determined based on the p-value and the level of significance. If the p-value is less than .05, there is enough evidence to reject the null hypothesis and conclude that the proportion of passengers with post-flight respiratory symptoms differs for planes that do and do not recirculate air. If the p-value is greater than .05, there is not enough evidence to reject the null hypothesis, meaning we can't conclude that there is a difference in the proportions.

Step by step solution

01

State the hypotheses

The null hypothesis (\(H_0\)): The proportion of passengers with post-flight respiratory symptoms on planes that do and don't recirculate air is the same. The alternative hypothesis (\(H_1\)): The two proportions are not equal.
02

Calculate sample proportions

Calculate the proportion of passengers with post-flight respiratory symptoms for each group. For the non-recirculating air group, the proportion (\(p_1\)) is 108/517. For the recirculating air group, the proportion (\(p_2\)) is 111/583.
03

Calculate the standard error

Calculate the standard error using the formula \(SE = \sqrt{p(1 - p)(1/n_1 + 1/n_2)}\) where \(p\) is the pooled sample proportion, calculated as \((x_1 + x_2) / (n_1 + n_2)\) (x - number of successes in a sample, n - sample size).
04

Compute the test statistic

Calculate the test statistic using the formula \(Z = (p_1 - p_2) / SE\).
05

Calculate the p-value

The p-value is the probability of observing a sample statistic as extreme as the test statistic. Since the alternative hypothesis is not directional (it doesn't state which proportion is larger), use the two-tail approach for calculating the p-value. Refer to the standard normal distribution table to find the probability associated with the calculated Z.
06

Conclude the hypothesis test

If the p-value is less than the level of significance (\(\alpha = .05\)), reject the null hypothesis. If the p-value is greater than \(\alpha\), do not reject the null hypothesis. Based on this decision, conclude if there is enough evidence to say that the proportion of passengers with post-flight respiratory symptoms differs for planes that do and do not recirculate air.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
The null hypothesis, often symbolized as \(H_0\), is a statement that suggests there is no effect or no difference in a given context. In hypothesis testing, it's the claim that there's no change or that a certain parameter equals a specific value.
In the context of airplane cabin air recirculation, and its connection with post-flight respiratory symptoms, our null hypothesis posits that the proportion of travelers experiencing symptoms is the same, whether or not the aircraft recirculates air.
This hypothesis serves as the starting point for our analysis. It's the idea we aim to test against, using statistical methods, to determine if observed data like collected samples significantly differ, warranting a rejection of this hypothesis.
Alternative Hypothesis
The alternative hypothesis, represented as \(H_1\), suggests that there is an effect or a difference. It's what researchers hope to support through their analysis.
For the airline study, the alternative hypothesis proposes that the proportion of passengers with respiratory symptoms does vary between those on aircrafts with recirculated air versus those without. By identifying this hypothesis, researchers set out to find if the proportions differ, indicating that air recirculation could influence symptom incidence.
The alternative hypothesis is pivotal, as it gives direction to the test, either affirming a change or highlighting a difference when compared to the null hypothesis.
Standard Error
The standard error (SE) measures the accuracy with which a sample represents a population. It is calculated using sample proportions and provides insight into the variability of the data.
In the context of our exercise, we compute the standard error to assess how much the sample proportions of respiratory symptoms among passengers could differ due to random sampling.
The formula for the standard error in comparing two proportions is: \[ SE = \sqrt{p(1 - p)(\frac{1}{n_1} + \frac{1}{n_2})} \] where \( p \) is the pooled sample proportion, \( n_1 \) is the sample size for non-recirculating planes, and \( n_2 \) for recirculating planes.
This computation is crucial for determining if observed differences are statistically significant.
P-Value
The p-value is a probability measure that helps determine the significance of results obtained from a statistical hypothesis test.
A low p-value indicates that the observed data is highly unlikely under the assumption that the null hypothesis is true. Conversely, a high p-value suggests that observed data is plausible when the null hypothesis holds.
In our scenario, we conduct a two-tailed test because our alternative hypothesis only states that the proportions are not equal, without specifying which is greater. We then calculate the p-value using the standard normal distribution table and our test statistic \( Z \).
This value will tell us if the deviations we observe in our test are just due to random chance or if they suggest real differences between the groups.
Z-Test
The Z-test is a statistical method used to determine if there is a significant difference between sample and population means, or among two sample means. It's appropriate when data follows a normal distribution, and we know the population variance. In our case, we compare proportions.
The Z-test formula used to find the test statistic is: \[ Z = \frac{(p_1 - p_2)}{SE} \] where \( p_1 \) and \( p_2 \) are the respective sample proportions of passengers experiencing symptoms in the respective groups, and \( SE \) is the computed standard error.
Once calculated, this Z-value indicates how far our sample's observation deviates from the null hypothesis. We use this in conjunction with the p-value to decide whether to reject the null hypothesis, providing insights into our original research question.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A researcher at the Medical College of Virginia conducted a study of 60 randomly selected male soccer players and concluded that frequently "heading" the ball in soccer lowers players' IQs (USA Today, August 14 1995). The soccer players were divided into two groups, based on whether they averaged 10 or more headers per game. Mean IQs were reported in the article, but the sample sizes and standard deviations were not given. Suppose that these values were as given in the accompanying table. $$ \begin{array}{l|ccc} & & \text { Sample } & \text { Sample } \\ & n & \text { Mean } & \text { sd } \\ \hline \text { Fewer Than 1O Headers } & 35 & 112 & 10 \\ 10 \text { or More Headers } & 25 & 103 & 8 \\ \hline \end{array} $$ Do these data support the researcher's conclusion? Test the relevant hypotheses using \(\alpha=.05 .\) Can you conclude that heading the ball causes lower \(\mathrm{IQ}\) ?

"Doctors Praise Device That Aids Ailing Hearts" (Associated Press, November 9,2004 ) is the headline of an article that describes the results of a study of the effectiveness of a fabric device that acts like a support stocking for a weak or damaged heart. In the study, 107 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs plus surgery to install the stocking. After two years, \(38 \%\) of the 57 patients receiving the stocking had improved and \(27 \%\) of the patients receiving the standard treatment had improved. Do these data provide convincing evidence that the proportion of patients who improve is higher for the experimental treatment than for the standard treatment? Test the relevant hypotheses using a significance level of \(.05 .\)

Common Sense Media surveyed 1000 teens and 1000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show, Tell Too Much Online," San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen in a way that makes it reasonable to regard them as representative of American teens and parents of American teens. a. When asked if they check their online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked his or her site more than 10 times a day, 40 said yes. Use a significance level of .01 to carry out a hypothesis test to determine if there is convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day. b. The article also reported that 390 of the teens surveyed said they had posted something on their networking site that they later regretted. Would you use the two-sample \(z\) test of this section to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted? Explain why or why not. c. Using an appropriate test procedure, carry out a test of the hypothesis given in Part (b). Use \(\alpha=.05\) for this test.

Two different underground pipe coatings for preventing corrosion are to be compared. The effect of a coating (as measured by maximum depth of corrosion penetration on a piece of pipe) may vary with depth, orientation, soil type, pipe composition, etc. Describe how an experiment that filters out the effects of these extraneous factors could be carried out.

Do certain behaviors result in a severe drain on energy resources because a great deal of energy is expended in comparison to energy intake? The article "The Energetic Cost of Courtship and Aggression in a Plethodontid Salamander" (Ecology [1983]: 979-983) reported on one of the few studies concerned with behavior and energy expenditure. The accompanying table gives oxygen consumption \((\mathrm{mL} / \mathrm{g} / \mathrm{hr})\) for male-female salamander pairs. (The determination of consumption values is rather complicated. It is partly for this reason that so few studies of this type have been carried out.) $$ \begin{array}{lccc} \text { Behavior } & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Sample } \\ \text { Mean } \end{array} & \begin{array}{c} \text { Sample } \\ \text { sd } \end{array} \\ \hline \text { Noncourting } & 11 & .072 & .0066 \\ \text { Courting } & 15 & .099 & .0071 \\ \hline \end{array} $$ a. The pooled \(t\) test is a test procedure for testing \(H_{0}: \mu_{1}-\mu_{2}=\) hypothesized value when it is reasonable to assume that the two population distributions are normal with equal standard deviations \(\left(\sigma_{1}=\right.\) \(\sigma_{2}\) ). The test statistic for the pooled \(t\) test is obtained by replacing both \(s_{1}\) and \(s_{2}\) in the two-sample \(t\) test statistic with \(s_{p}\) where \(s_{p}=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}\) When the population distributions are normal with equal standard deviations and \(H_{0}\) is true, the resulting pooled \(t\) statistic has a \(t\) distribution with \(\mathrm{df}=n_{1}+\) \(n_{2}-2 .\) For the reported data, the two sample standard deviations are similar. Use the pooled \(t\) test with \(\alpha=.05\) to determine whether the mean oxygen consumption for courting pairs is higher than the mean oxygen consumption for noncourting pairs. b. Would the conclusion in Part (a) have been different if the two-sample \(t\) test had been used rather than the pooled \(t\) test?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.