/*! 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 48 The article referenced in the pr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 48

The article referenced in the previous exercise also reported that \(24 \%\) of the males and \(16 \%\) of the females in the 2006 sample reported owning an MP3 player. Suppose that there were the same number of males and females in the sample of 1112 . Do these data provide convincing evidence that the proportion of females that owned an MP3 player in 2006 is smaller than the corresponding proportion of males? Carry out a test using a significance level of \(.01\).

Short Answer

Expert verified
In order to determine if there's a significant difference between the proportion of females and males owning MP3 players, a hypothesis test for proportions has to be performed. After calculating the pooled proportion, standard error and z-score, a P-value should be determined according to the observed z-score. Depending on the P-value, a conclusion can be drawn if it's less than the chosen significance level, 0.01, or not. The detailed solution of the z-score and the P-value are left out as they require further calculations.

Step by step solution

01

Set Up The Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of females that owned an MP3 player in 2006 is not smaller than the corresponding proportion of males, \(p_f \geq p_m\). The alternative hypothesis (\(H_1\)) is that the proportion of females that owned an MP3 player in 2006 is smaller than the corresponding proportion of males, \(p_f < p_m\).
02

Calculate Pooled Proportion

First, calculate the pooled proportion (\(p\)) which is the total number of successes (those who own MP3 players) divided by the total number of trials (total number of people). Assuming there were equal number of males and females, the pooled proportion \(p = \frac{0.24n + 0.16n}{2n} = 0.20\), where \(n\) is the number of males or females in the sample.
03

Calculate Standard Error

Next, calculate the standard error (\(SE\)), \(SE = \sqrt{p (1 - p) (\frac{1}{n_m} + \frac{1}{n_f})} = \sqrt{0.20 * 0.80 * (\frac{2}{n})}\) where \(n_m\) and \(n_f\) are the numbers of males and females respectively.
04

Calculate z-score

Compute the z-score which is the difference between the sample proportions minus the hypothesized difference (\(p_m - p_f\)) divided by the standard error. \(z = \frac{(p_m - p_f)-(p_{0m} - p_{0f})}{SE} = \frac{(0.24 - 0.16)-(0 - 0)}{SE}\).
05

Determine the P-Value

Determine the two-side P-value associated with the observed z-score. As the test is for the lower tail, we will seek the P-value for \(P(Z < z)\) if \(Z ~ N(0,1)\). If P-value is less than the chosen significance level of 0.01, we reject the null hypothesis.

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 91影视!

Key Concepts

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

Null Hypothesis
In any statistical analysis, hypothesizing is a critical first step. The null hypothesis, represented by \(H_0\), is a statement of no effect or no difference. It serves as the default assumption that we seek evidence against through our test. For instance, when examining if the proportion of female MP3 player owners is the same or greater compared to male owners in 2006, the null hypothesis is stated as \(p_f \geq p_m\). This hypothesis is what we are questioning with our statistical test. It is only rejected if the evidence strongly suggests that it's false, making it essential to set up this hypothesis carefully and clearly.
Alternative Hypothesis
Complementing the null hypothesis is the alternative hypothesis, denoted as \(H_1\) or \(H_a\). It represents the outcome that the researchers suspect or aim to support; in essence, it is the counter proposition to the null hypothesis. Reflecting our example, the alternative hypothesis posits that the proportion of females who owned an MP3 player in 2006, \(p_f\), is actually less than that of males, \(p_m\), expressed as \(p_f < p_m\). Should the test data provide sufficient evidence, this hypothesis would be favored over the null.
Pooled Proportion
When comparing two proportions, a pooled proportion is often used to increase the precision of the estimates in a hypothesis test. It combines the successes of both groups to assess one overall proportion. This procedure assumes under the null hypothesis that both groups have the same true proportion. In the exercise, the pooled proportion is calculated by taking the total individuals (having an MP3 player from both males and females) divided by the total number of participants. Here, it鈥檚 expressed as \(p = 0.20\). This value is utilized in subsequent steps to assess the underlying hypothesis.
Standard Error
The precision and variability of sampling distributions are gauged by the standard error (SE). It tells us how much the sample proportions are expected to differ by chance alone. In the context of comparing two proportions, the standard error helps to measure the sampling variability of the difference between the sample proportions, assuming that the null hypothesis is true. Using the pooled proportion and sample sizes, the standard error is calculated; accuracy in this calculation is crucial to the integrity of the hypothesis test results.
Z-Score
The z-score is a statistic that measures the number of standard errors a data point (in this case, the difference in proportions) is from the mean. In hypothesis testing, it鈥檚 a pivotal point鈥攊t quantifies the observed difference in terms of standard error and is used to determine whether this observed difference is unusual under the null hypothesis. A large absolute value of a z-score indicates that the observed data is much different from what the null hypothesis would predict, raising doubts about the validity of the null hypothesis.
Significance Level
The significance level, often denoted by \(\alpha\), is the threshold at which you decide whether to reject the null hypothesis. It's the probability of mistakenly rejecting a true null hypothesis鈥攁 type I error. The choice of significance level (e.g., 0.05, 0.01) affects the sensitivity of the hypothesis test; a smaller value of \(\alpha\) means that you require stronger evidence to reject \(H_0\) because you are allowing a smaller chance of making a type I error. In the case presented, a very stringent \(\alpha = 0.01\) is used, implying that the evidence must be quite strong to infer that the proportion of female MP3 player ownership is indeed less than that of males.
P-Value
At the heart of deciding whether or not the data contradicts the null hypothesis lies the p-value. This is the probability of observing test results at least as extreme as the results actually observed, assuming that the null hypothesis is true. Put simply, a p-value tells us how unusual our data is. A low p-value (lower than the chosen significance level \(\alpha\)) indicates that the observed data is unlikely under the null hypothesis and leads to its rejection in favor of the alternative hypothesis. Importantly, it quantifies the strength of the evidence against the null hypothesis provided by the data.

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

The article "Movement and Habitat Use by Lake Whitefish During Spawning in a Boreal Lake: Integrating Acoustic Telemetry and Geographic Information Systems" (Transactions of the American Fisheries Society, [1999]: \(939-952\) ) included the accompanying data on weights of 10 fish caught in 1995 and 10 caught in 1996 . \(\begin{array}{lllllllllll}1995 & 776 & 580 & 539 & 648 & 538 & 891 & 673 & 783 & 571 & 627\end{array}\) \(\begin{array}{llllllllllll}1996 & 571 & 627 & 727 & 727 & 867 & 1042 & 804 & 832 & 764 & 727\end{array}\) Is it appropriate to use the independent samples \(t\) test to compare the mean weight of fish for the 2 years? Explain why or why not.

In December 2001, the Department of Veterans Affairs announced that it would begin paying benefits to soldiers suffering from Lou Gehrig's disease who had served in the Gulf War (The New York Times, December 11,2001 ). This decision was based on an analysis in which the Lou Gehrig's disease incidence rate (the proportion developing the disease) for the approximately 700,000 soldiers sent to the Gulf between August 1990 and July 1991 was compared to the incidence rate for the approximately \(1.8\) million other soldiers who were not in the Gulf during this time period. Based on these data, explain why it is not appropriate to perform a formal inference procedure (such as the two-sample \(z\) test) and yet it is still reasonable to conclude that the incidence rate is higher for Gulf War veterans than for those who did not serve in the Gulf War.

The coloration of male guppies may affect the mating preference of the female guppy. To test this hypothesis, scientists first identified two types of guppies, Yarra and Paria, that display different colorations ("Evolutionary Mismatch of Mating Preferences and Male Colour Patterns in Guppies," Animal Behaviour \([1997]: 343-51\) ). The relative area of orange was calculated for fish of each type. A random sample of 30 Yarra guppies resulted in a mean relative area of \(.106\) and a standard deviation of .055. A random sample of 30 Paria guppies resulted in a mean relative area of \(.178\) and a standard deviation \(.058\). Is there evidence of a difference in coloration? Test the relevant hypotheses to determine whether the mean area of orange is different for the two types of guppies.

Gender differences in student needs and fears were examined in the article "A Survey of Counseling Needs of Male and Female College Students" (Journal of College Student Development \([1998]: 205-208) .\) Random samples of male and female students were selected from those attending a particular university. Of 234 males surveyed, \(27.5 \%\) said that they were concerned about the possibility of getting AIDS. Of 568 female students surveyed, \(42.7 \%\) reported being concerned about the possibility of getting AIDS. Is there sufficient evidence to conclude that the proportion of female students concerned about the possibility of getting AIDS is greater than the corresponding proportion for males?

"Smartest People Often Dumbest About Sunburns" is the headline of an article that appeared in the San Luis Obispo Tribune (July 19,2006 ). The article states that "those with a college degree reported a higher incidence of sunburn that those without a high school degree43 percent versus 25 percent." For purposes of this exercise, suppose that these percentages were based on random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). Is there convincing evidence that the proportion experiencing a sunburn is higher for college graduates than it is for those without a high school degree? Answer based on a test with a \(.05\) significance level.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation
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.