Question

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

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

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 \(\frac{4+5+7+7+5+7+5+6+5+6+7+8+5+6+6}{15} = 6.07.\)

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.

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).

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.