Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Public Agenda conducted a survey of 1,379 parents and 1,342 students in grades \(6-12\) regarding the importance of science and mathematics in the school curriculum (Associated 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 these were essential. The two samples - parents and students- were independently selected random samples. Using a confidence level of \(95 \%\) estimate the difference between the proportion of students who think that understanding science and having math skills are essential and this proportion for parents.

Short Answer

Expert verified
The 95% confidence interval for the difference of proportions between students and parents who believe that understanding science and having strong math skills is essential.

Step by step solution

01

Find the Proportions

First of all, compute the proportions for both students and parents. Given, the survey was conducted among 1,379 parents and 1,342 students. It is said that \(50 \% \) of students and \( 62 \% \) of parents believe that understanding science and having math skills are essential. Then, Proportion of students \(p_1 = 0.50 \) and proportion of parents \(p_2 = 0.62 \). Also, the number of students \(n_1 = 1342 \) and the number of parents \(n_2 = 1379 \).
02

Compute Difference of Proportions

Subtract \(p_1\) from \(p_2\) to find the difference between the two proportions. \(p = p_2 - p_1 = 0.62-0.50 = 0.12\).
03

Find the Standard Error

Compute the standard error (SE) which is the square root of \( p_1(1-p_1)/n_1 + p_2(1-p_2) /n_2 \), giving \( SE = \sqrt{0.5*(1-0.5)/1342 + 0.62*(1-0.62)/1379} \).
04

Calculate the Confidence Interval

Given a 95% confidence level, the Z value is 1.96 (from the standard normal distribution table). The 95% Confidence Interval for the difference of proportions is \( p \pm Z * SE \), where Z is the Z-value. Calculating this gives the Confidence Interval.

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 Vaia!

Key Concepts

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

Confidence Interval
Understanding a confidence interval is critical when analyzing the results of a survey like the one conducted on students and parents regarding the importance of science and math skills. A confidence interval offers a range of possible values for the parameter of interest — in this case, the difference in the proportions of students and parents who believe these skills are essential. In the provided exercise, a 95% confidence interval was used. This means we can be 95% confident that the true difference in proportions lies within this calculated interval. To find this interval, we take the difference between the two proportions and adjust it by the amount of uncertainty, which is determined by the standard error and a z-value reflecting our confidence level.
Standard Error
The concept of standard error (SE) is essential in survey data analysis as it helps measure the variability or precision of the sample statistic. It's why the term 'error' should not be misunderstood as a mistake but rather as an indicator of uncertainty or variation. In the context of our exercise, the standard error quantifies the variability in the estimated difference between the two proportions being compared. Importantly, a lower standard error suggests that our estimate of the difference is more precise.

The SE is computed using the formula:
\[ SE = \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}} \]
where \(p_1\) and \(p_2\) are the sample proportions, and \(n_1\) and \(n_2\) are the sample sizes. Notably, the larger the sample size, the smaller the standard error, indicating a more reliable estimate.
Survey Data Analysis
Analyzing survey data often requires different statistical tools to interpret the collected data effectively. In the exercise, the analysis involves comparing two proportions from independent samples — students and parents. When processing survey data, we must ensure the samples are representative and that random sampling is applied, as this affects the validity of our conclusions. After validating our data collection, we use statistical methods, such as the calculation of proportions, standard errors, and confidence intervals, to infer about the entire population from our sample.
Proportion Calculation
The proportion calculation is quite foundational in statistics, as seen in our exercise. It refers to the ratio of the number of times an outcome occurs to the total number of trials or sample size. In the context of the survey, the proportion helps us determine what fraction of the respondents — students and parents — views science and math skills as essential. To calculate each proportion, divide the number of individuals with the concerned opinion by the total number surveyed, giving us \(p_1\) and \(p_2\). These proportions serve as estimates of the population parameters and form the basis for further statistical analysis, like evaluating the difference between the two groups.
Z-Value
The z-value plays a pivotal role when constructing confidence intervals within survey data analysis. It is a statistic that represents how many standard deviations an element is from the mean. Specifically, for a given confidence level—95% in the exercise—the z-value maps to that probability on the standard normal distribution. Often known as a 'z-score', this number is used to multiply the standard error in the confidence interval formula, signifying the number of standard errors away from the observed difference you must go to achieve the desired confidence in your interval estimate. For a 95% confidence interval, the z-value is commonly 1.96.
Statistics Education
Imparting the fundamentals of statistics education is invaluable, especially when tackling real-world problems such as analyzing survey data. Creating an educational approach that demystifies complex concepts and formulas into straightforward, digestible pieces helps students understand and apply statistical methods with confidence. In the statistics education process, exercises like the one given provide a practical application, reinforcing concepts such as proportion calculations, standard error, and the interpretation of confidence intervals, all of which are everyday tools in a statistician's arsenal.

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 report referenced in the previous exercise also stated that the proportion who thought their parents would help with buying a house or renting an apartment for the sample of young adults was 0.37 . For the sample of parents, the proportion who said they would help with buying a house or renting an apartment was 0.27 . Based on these data, can you conclude that the proportion of parents who say they would help with buying a house or renting an apartment is significantly less than the proportion of young adults who think that their parents would help?

"Mountain Biking May Reduce Fertility in Men, Study Says" was the headline of an article appearing in the San Luis Obispo Tribune (December 3,2002 ). This conclusion was based on an Austrian study that compared sperm counts of avid mountain bikers (those who ride at least 12 hours per week) and nonbikers. Ninety percent of the avid mountain bikers studied had low sperm count, compared to \(26 \%\) of the nonbikers. Suppose that these percentages were based on independent samples of 100 avid mountain bikers and 100 nonbikers and that these samples are representative of avid mountain bikers and nonbikers. a. Using a confidence level of \(95 \%,\) estimate the difference between the proportion of avid mountain bikers with low sperm count and the proportion for nonbikers. b. Is it reasonable to conclude that mountain biking 12 hours per week or more causes low sperm count? Explain.

"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 than those without a high school degree-43\% versus \(25 \% . "\) Suppose that these percentages were based on independent random samples of size 200 from each of the two groups of interest (college graduates and those without a high school degree). a. Are the sample sizes large enough to use the largesample confidence interval for a difference in population proportions? b. Estimate the difference in the proportion of people with a college degree who reported sunburn and the corresponding proportion for those without a high school degree using a \(90 \%\) confidence interval. c. Is zero included in the confidence interval? What does this suggest about the difference in the two population proportions? d. Interpret the confidence interval in the context of this problem.

The news release referenced in the previous exercise also included data from independent samples 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 news release, but suppose that 600 teens and 400 parents of teens were surveyed and that these samples are representative of the two populations. Do the data provide convincing evidence that the proportion of teens who approve of banning cell phone and texting while driving is less than the proportion of parents of teens who approve? Test the relevant hypotheses using a significance level of \(0.05 .\)

The authors of the paper "Adolescents and MP3 Players: Too Many Risks, Too Few Precautions" (Pediatrics [2009]:e953-e958) concluded that more boys than girls listen to music at high volumes. This conclusion was based on data from independent random samples of 764 Dutch boys and 748 Dutch girls ages 12 to \(19 .\) Of the boys, 397 reported that they almost always listen to music at a high volume setting. Of the girls, 331 reported listening to music at a high volume setting. Do the sample data support the authors' conclusion that the proportion of Dutch boys who listen to music at high volume is greater than this proportion for Dutch girls? Test the relevant hypotheses using a 0.01 significance level.

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.

Sign-up for free