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Chapter 13: Problem 46

The article "Plugged In, but Tuned Out" (USA Today, January 20,2010 ) summarizes data from two surveys of kids ages 8 to 18 . One survey was conducted in 1999 and the other was conducted in \(2009 .\) Data on the number of hours per day spent using electronic media, consistent with summary quantities given in the article, are given in the following table (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years the surveys were conducted. Construct and interpret a \(98 \%\) confidence interval estimate of the difference between the mean number of hours per day spent using electronic media in 2009 and \(1999 .\) $$ \begin{array}{llllllllllllllll} 2009 & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 & 10 & 9 & 8 \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 & 5 & 6 & 6 \end{array} $$

Short Answer

Expert verified
The 98% confidence interval estimate of the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 can be obtained as \(\bar{x1} - \bar{x2} \pm Z*\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}\) after calculating the necessary statistics.

Step by step solution

01

Calculate Means

First, find the means of the samples: For 2009, the sample mean (\(\bar{x1}\)) can be calculated as \(\frac{5+9+5+8+7+6+7+9+7+9+6+9+10+9+8}{15} = 7.60.\)For 1999, the sample mean (\(\bar{x2}\)) can be calculated as \(\frac{4+5+7+7+5+7+5+6+5+6+7+8+5+6+6}{15} = 6.07.\)
02

Calculate Standard Deviations

Next, calculate the sample standard deviations:For 2009, calculate the standard deviation \(s1\) by finding the square root of the average of the squared deviations from the mean.For 1999, calculate the standard deviation \(s2\) similarly.
03

Find the Z-Value

The Z-value corresponds to the desired level of confidence. A 98% confidence level corresponds to a Z-value of 2.33 (found in Z-tables).
04

Calculate the Confidence Interval

Finally, calculate the confidence interval using the formula:\(\bar{x1} - \bar{x2} \pm Z*\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}\)With \(n1\) and \(n2\) being the sample sizes (15 for each year). Calculate the standard errors \(\frac{s1^2}{n1}\) and \(\frac{s2^2}{n2}\) and their sum, get the square root of this sum and multiply it with the Z-value. Then add and subtract this term from the difference of the means (\(\bar{x1} - \bar{x2}\)). This will give the lower and upper bounds of the 98% confidence interval.

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Key Concepts

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

Electronic Media Usage
Electronic media usage refers to the amount of time individuals spend interacting with various electronic devices. This includes activities such as watching TV, browsing the internet, playing video games, and using other digital media. In this context, a survey of kids aged 8 to 18 was conducted to analyze how their electronic media usage changed from 1999 to 2009. This data helps in understanding the habits and impacts of digital consumption on younger generations. Changes in such patterns can have significant implications for education, health, and social behaviors.
Sample Mean
The sample mean is a statistical measure that provides an average of a particular dataset. It is calculated by adding up all the individual data points and then dividing by the total number of points. In our exercise, the sample mean for electronic media usage in 2009 was computed by dividing the total hours by the number of days observed, resulting in 7.60 hours. Similarly, for 1999, it was calculated to be 6.07 hours. This computation gives a clear sense of the average media consumption for each period, serving as a key component for further analysis.
Standard Deviation
Standard deviation is a crucial concept in statistics that indicates the amount of variation or dispersion in a set of values. It tells us how much the values differ from the mean. To find the standard deviation, we first calculate the mean, and then determine the average of the squared differences from this mean. Taking the square root of this average gives us the standard deviation. In the given survey, both 2009 and 1999 datasets are analyzed to find their respective standard deviations. This helps to understand how consistent or varied the media usage was in each year.
Z-Value
A Z-value, often found in Z-tables, represents the number of standard deviations a data point is from the mean. It is used especially in the context of confidence intervals and hypothesis testing. For a 98% confidence interval, the corresponding Z-value is 2.33. This means that we are looking at a range where there is only a 2% chance that the actual mean would fall outside this interval. In the exercise, multiplying the Z-value by the standard error allows us to determine how far the confidence interval stretches from the calculated mean difference. This reveals the precision of our estimate of the change in electronic media usage.

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Most popular questions from this chapter

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