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The article “Kids Digital Day: Almost 8 Hours" (USA Today, January 20,2010 ) summarized results from a national survey of 2002 Americans age 8 to 18 . The sample was selected in a way that was expected to result in a sample representative of Americans in this age group. a. Of those surveyed, 1321 reported owning a cell phone. Use this information to construct and interpret a \(90 \%\) confidence interval estimate of the proportion of all Americans age 8 to 18 who own a cell phone. b. Of those surveyed, 1522 reported owning an MP3 music player. Use this information to construct and interpret a \(90 \%\) confidence interval estimate of the proportion of all Americans age 8 to 18 who own an MP3 music player. c. Explain why the confidence interval from Part (b) is narrower than the confidence interval from Part (a) even though the confidence level and the sample size used to compute the two intervals was the same.

Short Answer

Expert verified
a. The 90% confidence interval for the proportion of all Americans age 8 to 18 who own a cell phone is calculated by the formula and falls within a certain range. b. Similarly, the 90% confidence interval for the proportion of all Americans age 8 to 18 who own an MP3 player is computed, falling within a narrower range than the previous. c. The confidence interval for MP3 player ownership is narrower than for cell phone ownership because the sample proportion for MP3 players is larger than that for cell phones. As the proportion becomes more extreme (closer to 0 or 1), the confidence interval width decreases, assuming the same confidence level and sample size.

Step by step solution

01

Compute Confidence Interval for Cell Phone Ownership

First, calculate the sample proportion (\(p\)) for cell phone ownership by dividing the number who own cell phones (1321) by the total surveyed (2002). Then, we calculate the confidence interval. In other words, \(p = \frac{1321}{2002}\). The formula to calculate a confidence interval for a population proportion is \(p \pm z \sqrt{\frac{p(1 - p)}{n}}\), where \(z\) is the z-value from the standard normal distribution for the desired confidence level. For a 90% confidence level, the z-value is 1.645. So, the confidence interval would be \(p \pm 1.645 \times \sqrt{\frac{p(1-p)}{2002}}\).
02

Compute Confidence Interval for MP3 Player Ownership

Next, calculate the sample proportion (\(p\)) for MP3 player ownership by dividing the number who own MP3 players (1522) by the total surveyed (2002). So \(p = \frac{1522}{2002}\). Use the same formula as in Step 1 to calculate this confidence interval, with the only difference being in the values of \(p\). So, the confidence interval would be \(p \pm 1.645 \times \sqrt{\frac{p(1-p)}{2002}}\).
03

Compare Confidence Intervals

Observe that the confidence interval for MP3 player ownership is narrower than that for cell phone ownership, despite the same sample size and confidence level. This is because the sample proportion of MP3 player owners is larger than that of cell phone owners. The width of a confidence interval for a proportion depends on the sample proportion (\(p\), or \(1-p\), whichever is smaller), the confidence level (through the z-value), and the sample size. As the sample proportion becomes more extreme (i.e. closer to 0 or 1), the confidence interval width decreases, given the same confidence level and sample size.
04

Interpretation

Both the confidence intervals provide an estimate of the proportion of all Americans age 8 to 18 who own a cell phone or a MP3 player respectively. They estimate that the true population proportion is within the calculated range about 90% of the time. But, the MP3 player ownership confidence interval is narrower, indicating a more precise estimate, because the sample proportion is closer to one compared to the cell phone ownership.

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Key Concepts

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

Population Proportion
When exploring statistics, particularly in survey analysis, the term 'population proportion' refers to the fraction of the population that has a certain characteristic. For instance, in the USA Today article about youngsters' ownership of technology, the population proportion would represent the percentage of all Americans aged 8 to 18 who own a particular device, like a cell phone or MP3 player.

To estimate this proportion, we use data from a sample, which is a smaller group selected from the population. This sample acts as a snapshot, providing insight into the population as a whole. The reliability of the population proportion estimate depends greatly on how well the sample represents the population. If the characteristics of the sample closely match those of the population, our estimate will be more accurate.
Sample Size
The 'sample size' is the total number of individuals or observations included in the sample. Key in the reliability of a statistical estimate, the sample size impacts the precision of our confidence interval. Generally speaking, a larger sample size provides a narrower confidence interval, which translates to greater precision in our estimation. This is because, as we collect more data, the effects of random variation diminish.

In the survey from the USA Today article, the sample size is 2002 Americans aged 8 to 18. The precision of our confidence interval estimates for both cell phone and MP3 player ownership relies on this number. If the sample size were smaller, even with the same population proportions, our confidence intervals would be wider, indicating less precision and more uncertainty in our estimates.
Z-Value
A 'z-value', also known as a z-score, is a critical component when constructing confidence intervals. It represents the number of standard deviations a data point is from the mean in a standard normal distribution. The z-value corresponds to the desired confidence level; a higher confidence level will require a larger z-value.

For instance, with a 90% confidence level, the z-value is 1.645. This number is used to calculate the margin of error in a confidence interval estimate. It’s determined by looking at the standard normal distribution—a bell curve representing the expected variation in the data. The z-value thus helps to define how confident we can be that the true population proportion falls within our calculated interval.
Standard Normal Distribution
The 'standard normal distribution' is a crucial concept in statistics, often depicted as a symmetrical bell-shaped curve. It's a special case of the normal distribution that has a mean of zero and a standard deviation of one. Lots of natural phenomena and measurement errors conform to this distribution, making it a foundational tool for statistical inference.

For our purposes in creating confidence intervals, the standard normal distribution allows us to understand how the sample proportion is likely to vary from the actual population proportion. By using the z-value associated with our desired confidence level, we can determine the range in which we expect the true population proportion to fall, within certain limits of probability. This is how we end up with the margins of error used in the confidence intervals for the ownership of cell phones and MP3 players among American youth.

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Most popular questions from this chapter

"Tongue Piercing May Speed Tooth Loss. Researchers Say" is the headline of an article that appeared in the San Luis Obispo Tribune (June 5,2002 ). The article describes a study of 52 young adults with pierced tongues. The researchers found receding gums, which can lead to tooth loss, in 18 of the participants. Construct a \(95 \%\) confidence interval for the proportion of young adults with pierced tongues who have receding gums. What assumptions must be made for use of the \(z\) confidence interval to be appropriate?

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