/*! 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 27 Personal Data and Privacy. The s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Personal Data and Privacy. The survey in Exercise \(23.26\) also looked at possible differences in the proportions of White non- Hispanics and Black non-Hispanics who were concerned about how much information their friends and family might know about them. Of the 2887 White nonHispanics in the survey, 1010 said they were concerned. Of the 445 Black non-Hispanics in the survey, 271 said they were concerned. Is there evidence of a difference between the proportions of White non-Hispanics and Black non- Hispanics who were concerned about how much information their friends and family might know about them?

Short Answer

Expert verified
Yes, there is evidence of a difference if the p-value < 0.05.

Step by step solution

01

Define Hypotheses

Let's define the null and alternative hypotheses. The null hypothesis \( H_0 \) assumes that there is no difference in the proportions. Mathematically, \( p_1 = p_2 \), where \( p_1 \) is the proportion of concerned White non-Hispanics and \( p_2 \) is the proportion of concerned Black non-Hispanics. The alternative hypothesis \( H_a \) assumes that there is a difference: \( p_1 eq p_2 \).
02

Calculate Sample Proportions

Calculate the sample proportions. For White non-Hispanics, the proportion \( \hat{p}_1 = \frac{1010}{2887} \). For Black non-Hispanics, the proportion \( \hat{p}_2 = \frac{271}{445} \).
03

Calculate Pooled Proportion

Calculate the pooled proportion \( \hat{p} \) which combines both groups. It is calculated as \( \hat{p} = \frac{1010 + 271}{2887 + 445} \).
04

Compute Standard Error

Using the pooled proportion, calculate the standard error (SE) of the difference in proportions \( SE = \sqrt{ \hat{p} (1 - \hat{p}) \left( \frac{1}{2887} + \frac{1}{445} \right) } \).
05

Calculate Z-Score

The Z-score is calculated to find the difference between the two proportions in terms of standard errors. Compute \( Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \).
06

Find P-value

Using the Z-score from Step 5, find the p-value to determine the statistical significance of the result. A Z-test table or statistical software can be used for this.
07

Conclusion from P-value

Compare the p-value to the significance level, typically \( \alpha = 0.05 \). If the p-value is less than \( \alpha \), we reject the null hypothesis, meaning there is evidence of a difference. If not, we fail to 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 91Ó°ÊÓ!

Key Concepts

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

Proportions
In statistics, the concept of proportions is crucial when comparing different groups. A proportion represents a part of a whole and is often expressed as a fraction or percentage. In the context of hypothesis testing, we use proportions to understand the characteristics of particular groups within a set population.
In this exercise, we are dealing with two specific proportions:
  • For White non-Hispanics, the proportion of those concerned about privacy is calculated as \( \hat{p}_1 = \frac{1010}{2887} \).
  • For Black non-Hispanics, the proportion is \( \hat{p}_2 = \frac{271}{445} \).
These proportions help identify if there is a meaningful difference in concern between the two ethnic groups regarding their privacy.
Null Hypothesis
The null hypothesis, symbolized as \( H_0 \), is a statement that assumes there is no effect or no difference in the context of the problem. In hypothesis testing involving proportions, this hypothesis is designed to be tested against the data.
In our exercise:
  • The null hypothesis states that the proportions of White non-Hispanics and Black non-Hispanics who are concerned about privacy are equal.
  • This is formulated mathematically as \( p_1 = p_2 \).
The null hypothesis serves as a starting point to test whether there is indeed a statistically significant difference between the two groups we are comparing.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \), is the opposite of the null hypothesis. It suggests that there is an effect or a difference that contrasts with the null hypothesis. This hypothesis is what researchers typically aim to support.
In this specific problem:
  • The alternative hypothesis proposes that the proportions of the two groups are not equal, indicating a difference in concerns over privacy.
  • This can be expressed as \( p_1 eq p_2 \).
The alternative hypothesis is what would lead to rejecting the null hypothesis, possibly leading us to a meaningful conclusion about changes between two proportions.
Z-Test
A Z-test is a statistical test that determines whether there is a significant difference between the means or proportions of two groups. It's a useful tool when dealing with large sample sizes or known standard deviations.
In this exercise, a Z-test helps determine if there's a real difference between the proportions of concerned individuals in both ethnic groups.
  • We compute the Z-score, which tells us the number of standard errors the observed difference is away from the hypothesized difference (if the null hypothesis were true).
  • The formula for Z-score is \( Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \), where SE is the standard error computed from our data.
The Z-test culminates in a Z-score, which is then used to find the p-value, indicating how likely our data would appear under the null hypothesis.
Statistical Significance
Statistical significance is a key concept in hypothesis testing. It helps us determine if the results observed in our study are due to chance or if they reflect a true difference in the population.
This is achieved by comparing the p-value to a predefined significance level, often \( \alpha = 0.05 \).
  • The p-value is derived from the Z-score and represents the probability of observing the obtained result, or one more extreme, assuming the null hypothesis is true.
  • If the p-value is less than \( \alpha \), we reject the null hypothesis, which means the results are potentially statistically significant.
  • If the p-value is greater or equal to \( \alpha \), we fail to reject the null hypothesis.
This step is crucial in making informed decisions about whether the differences between groups are meaningful.

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

I'll Stop Smoking-Soon! Chantix is different from most other quit-smoking products in that it targets nicotine receptors in the brain, attaches to them, and blocks nicotine from reaching them. A randomized, double-blind, placebocontrolled clinical trial on Chantix was conducted with a 24week treatment period. Participants in the study were cigarette smokers who were either unwilling or unable to quit smoking in the next month but were willing to reduce their smoking and make an attempt to quit within the next three months. Subjects received either Chantix or a placebo for 24 weeks, with a target of reducing the number of cigarettes smoked by \(50 \%\) or more by week \(4,75 \%\) or more by week 8 , and a quit attempt by 12 weeks. The primary outcome measured was continuous abstinence from smoking during weeks 15 through 24. Of the 760 subjects taking Chantix, 244 abstained from smoking during weeks 15 through 24 , whereas 52 of the 750 subjects taking the placebo abstained during this same time period. 24 a. Give a \(99 \%\) confidence interval for the difference (treatment minus placebo) in the proportions of smokers who would abstain from smoking during weeks 15 through 24 . b. Pfizer, the company that manufactures Chantix, claims that Chatix is proven to help smokers quit. Does your confidence interval support this claim? Discuss.

Avobenzone is one of the active ingredients in several commercially available sunscreens. It can be absorbed into the bloodstream when sunscreen is applied to the skin. The Food and Drug Administration has expressed concern about the safety of absorbing too much avobenzone. Amounts less than or equal to \(0.5 \mathrm{ng} / \mathrm{mL}\) (nanograms absorbed per milliliter applied) are considered acceptable. Researchers recruited 12 healthy volunteers to investigate avobenzone absorption for two different commercially available sunscreens: a spray and a cream. Subjects were randomly assigned to one of the two sunscreens, with 6 subjects for each. Subjects had 2 milligrams of sunscreen per \(1 \mathrm{~cm}^{2}\) applied to \(75 \%\) of their body surface area (area outside of normal swimwear). The amount of avobenzone absorbed into the bloodstream after 6 hours was then measured for each subject. Four of the six subjects receiving the spray had avobenzone levels exceeding \(0.5 \mathrm{ng} / \mathrm{mL}\), and one of the six subjects receiving the cream had levels exceeding \(0.5 \mathrm{ng} / \mathrm{mL} .16\) The \(z\) test for "no difference" in the two proportions exceeding \(0.5 \mathrm{ng} / \mathrm{mL}\) against "the two proportions differ" has a. \(z=1.76, P<0.05\). b. \(z=1.84, P<0.055\). c. \(z=1.76,0.05

Take \(p_{9}\) and \(p_{12}\) to be the proportions of all ninth- and 12 thgraders who ate breakfast daily. The hypotheses to be tested are a. \(H_{0}: p_{9}=p_{12}\) versus \(H_{a}: p_{9} \neq p_{12}\). b. \(H_{0}: p_{9}=p_{12}\) versus \(H_{a}: p_{9}>p_{12}\). c. \(H_{0}: p_{9}=p_{12}\) versus \(H_{a}: p_{9}

Smoking among Seniors. Since the 1990 s, daily cigarette smoking among high school seniors has dropped from around \(12 \%\) to slightly over \(3 \%\). In 2017, random samples of 1725 female and 1564 male high school seniors found that \(3.7 \%\) of the females and \(3.1 \%\) of the males had smoked cigarettes daily in the 30 days before the survey. \(\underline{2}\) Give a \(95 \%\) confidence interval for the difference between the proportions of the populations of male and female high school seniors who had smoked cigarettes daily in the 30 days before the survey. Follow the four-step process as illustrated in Examples \(23.2\) and \(23.3\) (pages 518 and 520 ).

Aerosolized Vaccine for Measles. An aerosolized vaccine for measles was developed in Mexico and has been used on more than 4 million children since 1980 . Aerosolized vaccines have the advantages of being able to be administered by people without clinical training and do not cause injectionassociated infections. Despite these advantages, data about efficacy of the aerosolized vaccines against measles compared to subcutaneous injection of the vaccine have been inconsistent. Because of this, a large randomized controlled study was conducted using children in India. The primary outcome was an immune response to measles measured 91 days after the treatments. Among the 785 children receiving the subcutaneous injection, 743 developed an immune response, while among the 775 children receiving the aerosolized vaccine, 662 developed an immune response. 3 a. Compute the proportion of subjects experiencing the primary outcome for both the aerosol and injection groups. b. Can we safely use the large-sample confidence interval for comparing the proportion of children who developed an immune response to measles in the aerosol and injection groups? Explain. c. Give a \(95 \%\) confidence interval for the difference between the proportion of children in the aerosol and injection groups who experienced the primary outcome. d. The study described is an example of a noninferiority clinical trial intended to show that the effect of a new treatment, the aerosolized vaccine, is not worse than that of the standard treatment by more than a specified margin. 4 Specifically, is the percentage of children who developed an immune response for the aerosol treatment more than \(5 \%\) below the percentage for the subcutaneous injected vaccine? The 5percentage-point difference was based on previous studies and the fact that with a bigger difference, the aerosolized vaccine would not provide the levels of protection necessary to achieve herd immunity. Using your answer in part (c), do you feel the investigators demonstrated the noninferiority of the aerosolized vaccine? Explain.

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.