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

The article "Portable MP3 Player Ownership Reaches New High" (Ipsos Insight, June 29,2006 ) reported that in \(2006,20 \%\) of those in a random sample of 1112 Americans age 12 and older indicated that they owned an MP3 player. In a similar survey conducted in 2005 , only \(15 \%\) reported owning an MP3 player. Suppose that the 2005 figure was also based on a random sample of size 1112 . Estimate the difference in the proportion of Americans age 12 and older who owned an MP3 player in 2006 and the corresponding proportion for 2005 using a 95\% confidence interval. Is zero included in the interval? What does this tell you about the change in this proportion from 2005 to \(2006 ?\)

Short Answer

Expert verified
To provide a precise short answer, the standard error - SE needs to be calculated first. After that one could state the confidence interval and whether zero was included in it or not, and explain what the implications of that are.

Step by step solution

01

Calculation of Sample Proportions

The number of participants who reported owning an MP3 player in each year can be calculated by multiplying the total sample size (1112) by the respective proportions. For 2005, the number is \(1112 \times 0.15 = 167\) and for 2006, it is \(1112 \times 0.20 = 222\).
02

Estimation of the Difference in Proportions

The difference in the reported proportions between 2005 and 2006 can be estimated as \( p_{2006} - p_{2005} = 0.20 - 0.15 = 0.05\).
03

Calculation of Standard Error

The standard deviation of the difference (standard error) can be estimated using the formula \( SE = \sqrt{ \frac{p_{2005} \times (1 - p_{2005})}{n_{2005}} + \frac{p_{2006} \times (1 - p_{2006})}{n_{2006}} } \). Substituting with the given values, we get \( SE = \sqrt{ \frac{0.15 \times 0.85}{1112} + \frac{0.20 \times 0.80}{1112} } \).
04

Calculation of the 95% Confidence Interval

A 95% confidence interval for the difference in proportions can be calculated as \( (p_{2006} - p_{2005}) \pm Z \times SE \), where Z = 1.96, which is the Z score for a 95% confidence interval. Substituting with the estimated values from the earlier steps, we get \( CI = 0.05 \pm 1.96 \times SE \).
05

Interpretation of the Confidence Interval

If zero is not included in the confidence interval, it means that there was a significant change in the proportion of Americans owning MP3 players from 2005 to 2006. If zero is included in the confidence interval, it means that the change was not significant.

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.

Statistics
In the context of the given exercise, statistics provide us with methods and tools to analyze and interpret data. The process begins with data collection, which in this case is the survey of American MP3 player ownership over two different years.

Statistical analysis then enables us to draw conclusions from this data, such as estimating proportions and comparing them across groups or over time. By using statistical techniques, we can infer population trends or preferences based on sample data, such as determining if the popularity of MP3 players has significantly increased from one year to the next. A 95% confidence interval, as used in this exercise, is a key statistical concept which gives us a range in which we can say, with a certain level of confidence, that the true difference in population proportions lies.

If this interval does not include the value zero, we infer that there is a statistically significant difference, indicating a real change in MP3 player ownership between the years studied. Understanding this concept underscores the importance of statistical knowledge in interpreting survey results and making informed decisions or predictions.
Proportion Estimation
Proportion estimation is an essential area of statistics when we deal with categorical data. It involves estimating the proportion of a population that has a specific characteristic, based on sample data. In our exercise example, we estimate the proportion of Americans age 12 and older who owned an MP3 player based on a sample.

For accurate proportion estimation, the sample must be representative of the population, which is why a random sample is critical. The steps provided in the solution detail calculating the sample proportions and using these to estimate differences. By calculating a confidence interval around this estimated difference, we can measure the reliability of the estimation.

The narrower the confidence interval, the more precise our estimation is considered to be. This precision is influenced by the sample size and the variability within the data. Hence, understanding the principles behind proportion estimation is vital for interpreting and validating the results of data analysis, especially in surveys and studies that use sampling methods.
Data Analysis
Data analysis is the process of systematically applying statistical or logical techniques to describe and illustrate, condense and recap, and evaluate data. In the setting of our problem, data analysis involves estimating population parameters such as the ownership of MP3 players, comparing those estimates across different time points, and determining the significance of the observed changes.

The step-by-step solution we have guides us through calculating the estimated proportion of owners in each year, evaluating the standard error of the difference in proportions (which is a measure of uncertainty in our estimation), and constructing a confidence interval. Analyzing this interval helps us determine whether the difference observed in the sample is likely reflective of a real difference in the wider population.

Data analysis is integral in various fields, not just in statistics but also in business, science, and more. Mastery of data analysis techniques leads to better decision-making based on empirical evidence. In educational contexts, enhancing students’ understanding of data analysis equips them with the ability to critically evaluate and derive meaning from data which is ubiquitous in today’s information-driven world.

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

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 teachers find their work rewarding and satisfying? The article "Work- Related Attitudes" (Psychological Reports \([1991]: 443-450)\) reported the results of a survey of random samples of 395 elementary school teachers and 266 high school teachers. Of the elementary school teachers, 224 said they were very satisfied with their jobs, whereas 126 of the high school teachers were very satisfied with their work. Based on these data, is it reasonable to conclude that the proportion very satisfied is different for elementary school teachers than it is for high school teachers? Test the appropriate hypotheses using a \(.05\) significance level.

"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 counts, as 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 it is reasonable to view these samples as representative of Austrian avid mountain bikers and nonbikers. a. Do these data provide convincing evidence that the proportion of Austrian avid mountain bikers with low sperm count is higher than the proportion of Austrian nonbikers? b. Based on the outcome of the test in Part (a), is it reasonable to conclude that mountain biking 12 hours per week or more causes low sperm count? Explain.

An individual can take either a scenic route to work or a nonscenic route. She decides that use of the nonscenic route can be justified only if it reduces true average travel time by more than \(10 \mathrm{~min}\). a. If \(\mu_{1}\) refers to the scenic route and \(\mu_{2}\) to the nonscenic route, what hypotheses should be tested? b. If \(\mu_{1}\) refers to the nonscenic route and \(\mu_{2}\) to the scenic route, what hypotheses should be tested?

In a study of memory recall, eight students from a large psychology class were selected at random and given 10 min to memorize a list of 20 nonsense words. Each was asked to list as many of the words as he or she could remember both \(1 \mathrm{hr}\) and \(24 \mathrm{hr}\) later, as shown in the accompanying table. Is there evidence to suggest that the mean number of words recalled after \(1 \mathrm{hr}\) exceeds the mean recall after \(24 \mathrm{hr}\) by more than 3 ? Use a level \(.01\) test. \(\begin{array}{lrrrrrrrr}\text { Subject } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\ 1 \text { hr later } & 14 & 12 & 18 & 7 & 11 & 9 & 16 & 15 \\ 24 \text { hr later } & 10 & 4 & 14 & 6 & 9 & 6 & 12 & 12\end{array}\)

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