/*! 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 26 Personal Data and Privacy. There... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Personal Data and Privacy. There is widespread concern among the general public about how their personal data are used. A Pew Internet survey in 2019 asked a sample of U.S. adults whether they were concerned about how much information law enforcement might know about them. Of the 2887 White non- Hispanics in the survey, 1617 said they were concerned. Of the 445 Black non- Hispanics in the survey, 325 said they were concerned. \(\underline{23}\) Is there good evidence that the proportions of White non-Hispanics and Black non- Hispanics who were concerned about how much information law enforcement might know about them differ?

Short Answer

Expert verified
The proportions differ significantly.

Step by step solution

01

Define the Hypotheses

We need to determine if there is a significant difference between the proportions of two groups. We set up the null hypothesis as \( H_0: p_1 = p_2 \) and the alternative hypothesis as \( H_a: p_1 eq p_2 \), where \( p_1 \) is the proportion of concerned White non-Hispanics and \( p_2 \) is the proportion of concerned Black non-Hispanics.
02

Calculate Sample Proportions

Calculate the sample proportions for each group. \( \hat{p}_1 = \frac{1617}{2887} \approx 0.560 \) for White non-Hispanics, and \( \hat{p}_2 = \frac{325}{445} \approx 0.730 \) for Black non-Hispanics.
03

Compute the Pooled Proportion

Under the null hypothesis that the proportions are equal, calculate the pooled proportion using both samples: \( \hat{p} = \frac{1617+325}{2887+445} = \frac{1942}{3332} \approx 0.583 \).
04

Find the Standard Error

Calculate the standard error of the difference between the two sample proportions: \( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{0.583 \times 0.417 \times \left(\frac{1}{2887} + \frac{1}{445}\right)} \approx 0.0239 \).
05

Calculate the Test Statistic

Compute the test statistic: \( z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.560 - 0.730}{0.0239} \approx -7.13 \).
06

Determine the Critical Value

For a two-tailed test with a common significance level of 5%, the critical values for a standard normal distribution are approximately \( \pm 1.96 \).
07

Make a Decision

Since the calculated test statistic \( -7.13 \) is less than \(-1.96\), we reject the null hypothesis. This indicates that there is a significant difference between the proportions.

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
In statistical hypothesis testing, a null hypothesis is a kind of assumption. It suggests that there is no effect or no difference beyond the natural variation we may observe in the data. When comparing two groups, as in this exercise examining concerns about privacy among different racial groups, the null hypothesis (\( H_0 \)) assumes that the proportions of these groups are the same.
  • For the current scenario, the null hypothesis is expressed as \( H_0: p_1 = p_2 \), indicating no difference in concern levels.
  • The alternative hypothesis (\( H_a \)) suggests a difference exists, expressed as \( H_a: p_1 eq p_2 \).
Understanding this is critical because rejecting the null hypothesis, like in this exercise, implies there could be genuine differences in the population proportions that we have observed. The decision to reject or not reject the null hypothesis depends on the results of the statistical test performed.
Sample Proportion
The sample proportion is a statistic that estimates the proportion of individuals in a population that possess a particular characteristic. In the context of this exercise, we determine the proportion of people within each racial group who expressed concern over personal data privacy.
To find these proportions, you divide the number of concerned individuals by the total number surveyed in each group:
  • For White non-Hispanics: 1617 are concerned out of 2887 surveyed, leading to a sample proportion (\( \hat{p}_1 \)) of approximately 0.560.
  • For Black non-Hispanics: 325 are concerned out of 445 surveyed, yielding a sample proportion (\( \hat{p}_2 \)) of about 0.730.
Sample proportions are fundamental in hypothesis testing as they provide the actual data estimates we compare to the hypothesized proportions, under the assumption of the null hypothesis.
Standard Error
The standard error (SE) provides an estimate of the variability of a statistic. Specifically, it measures how much the sample proportion you calculate might vary if you took many samples.
For this exercise, we calculate the standard error of the difference between two sample proportions. This is crucial because it helps us determine how close our sample estimates might be to the true population proportions.
  • To calculate the standard error: \( SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \)
  • Here, \( \hat{p} \) is the pooled proportion, representing combined data from both samples, which is \(\approx 0.583\).
  • The calculated standard error in this context is approximately 0.0239.
The smaller the standard error, the more precise our sample estimate is, and in hypothesis testing, it allows us to compute the Z-Statistic.
Z-Statistic
The Z-Statistic is the value we calculate to test our hypothesis. It quantifies how many standard deviations our sample statistic is away from the null hypothesis statistic.
In this case, the Z-Statistic helps determine if the difference between sample proportions is significant:
  • The formula for the Z-Statistic is \( z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \)
  • With our data, \( z = \frac{0.560 - 0.730}{0.0239} \approx -7.13 \)
The Z-Statistic is compared against critical values (e.g., \( \pm 1.96 \) for a 5% significance level in a two-tailed test) to decide if the null hypothesis should be rejected.
Since our calculated \( z \) is less than \(-1.96\), it leads our decision to reject the null hypothesis, indicating a statistically significant difference between the groups' proportions.

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

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

Significant Does Not Mean Important. Never forget that even small effects can be statistically significant if the samples are large. To illustrate this fact, consider a sample of 148 small businesses. During a three-year period, 15 of the 106 businesses headed by men and seven of the 42 businesses headed by women failed. 22 a. Find the proportions of failures for businesses headed by women and businesses headed by men. These sample proportions are quite close to each other. Give the \(P\). value for the \(z\) test of the hypothesis that the same proportion of women's and men's businesses fail. (Use the two-sided alternative.) The test is very far from being significant. b. Now suppose that the same sample proportions came from a sample 30 times as large. That is, 210 out of 1260 businesses headed by women and 450 out of 3180 businesses headed by men fail. Verify that the proportions of failures are exactly the same as in part (a). Repeat the \(z\) test for the new data and show that it is now significant at the \(\alpha=0.05\) level. c. It is wise to use a confidence interval to estimate the size of an effect rather than just giving a P-value. Give the large sample \(95 \%\) confidence intervals for the difference between the proportions of women's and men's businesses that fail for the settings of both parts (a) and (b). What is the effect of larger samples on the confidence interval? Do you think the size of the difference between the proportions is an important difference? In responding to Exercises \(23.26\) through \(23.36\) follow the Plan, Solve, and Conclude steps of the four-step process.

Abecedarian Early Childhood Education Program: Adult Outcomes. The Abecedarian Project is a randomized controlled study to assess the effects of intensive early childhood education on children who were at high risk based on several sociodemographic indicators. 2 The project randomly assigned some children to a treatment group that was provided with early educational activities before kindergarten and the remainder to a control group. A recent follow-up study interviewed subjects at age 30 and evaluated educational, economic, and socioemotional outcomes to learn if the positive effects of the program continued into adulthood. The follow-up study included 52 individuals from the treatment group and 49 from the control group. Out of these, 39 from the treatment group and 26 from the control group were considered "consistently" employed (working 30 + hours per week in at least 18 of the 24 months prior to the interview). Does the study provide significant evidence that subjects who had early childhood education have a higher proportion of consistent employment than those who did not? How large is the difference between the proportions in the two populations that are consistently employed? Do inference to answer both questions. Be sure to explain exactly what inference you choose to do.

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 ).

{ Protecting Skiers and Snowboarders. Most alpine }\( skiers and snowboarders do not use helmets. Do helmets reduce the risk of head injuries? A study in Norway compared skiers and snowboarders who suffered head injuries with a control group who were not injured. Of 578 injured subjects, 96 had worn a helmet. Of the 2992 in the control group, 656 wore helmets. 9 Is helmet use less common among skiers and snowboarders who have head injuries? Follow the four-step process as illustrated in Examples 23.4 and \)23.5$ (pages 524 and 525 ). (Note that this is an observational study that compares injured and uninjured subjects. An experiment that assigned subjects to helmet and no- helmet groups would be more convincing.)

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.