91Ó°ÊÓ

The article "Plugged In, but Tuned Out" (USA Today, January 20,2010 ) summarizes data from two surveys of kids age 8 to 18 . One survey was conducted in 1999 and the other was conducted in \(2009 .\) Data on number of hours per day spent using electronic media that are consistent with summary quantities given in the article are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, assume that it is reasonable to regard the two samples as representative of kids age 8 to 18 in each of the 2 years that the surveys were conducted. $$ \begin{array}{l|lllllllllllllll} \hline 2009 & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 & 10 & 9 & 8 \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 & 5 & 6 & 6 \\ \hline \end{array} $$ a. Because the given sample sizes are small, in order for the two-sample \(t\) test to be appropriate, what assumption must be made about the distribution of electronic media use times? Use the given data to construct graphical displays that would be useful in determining whether this assumption is reasonable. Do you think it is reasonable to use these data to carry out a two-sample \(t\) test? b. Do the given data provide convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999 ? Test the relevant hypotheses using a significance level of .01 c. 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 .

Step by step solution

01

Distribution assumption for two-sample t test

To conduct a two-sample t test, it needs to be assumed that the distribution of the data is approximately normal. This assumption can be verified graphically via a histogram or a box plot. If the data does not seem to diverge too much from normality (i.e., is unimodal and symmetric), it can be said it is safe to carry out tho two-sample t test.

02

Hypothesis testing

Null hypothesis (H0) is that the mean number of hours per day spent using electronic media in 1999 is equal to that in 2009. Alternative hypothesis (Ha) is that the mean number of electronic media usage hours in 2009 is greater than that in 1999. Use a two-sample t test at a significance level of 0.01 to test these hypotheses. If the p-value obtained is less than 0.01, reject the null hypothesis.

03

Constructing 98% confidence interval

To build a 98% confidence interval for the difference between the means, use the following formula: \( ( \bar{x}_{2009} - \bar{x}_{1999}) + t^* \times \sqrt{\frac{s_{2009}^2}{n_{2009}} + \frac{s_{1999}^2}{n_{1999}}}, ( \bar{x}_{2009} - \bar{x}_{1999}) - t^* \times \sqrt{\frac{s_{2009}^2}{n_{2009}} + \frac{s_{1999}^2}{n_{1999}}})\), where \( \bar{x}_{2009}, \bar{x}_{1999} \) are sample means, \( s_{2009}, s_{1999} \) are sample standard deviations, \( n_{2009}, n_{1999} \) are sample sizes for 2009 and 1999 respectively. The value of \( t^* \) is the t-score for the chosen level of confidence (in this case, 98%).

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Key Concepts

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

Hypothesis Testing

In the realm of statistics, hypothesis testing serves as a fundamental method to make inferences about population parameters based on sample data. When deploying hypothesis testing, researchers formulate two opposing hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). The null hypothesis typically posits no effect or no difference, essentially serving as the skeptics' stance, while the alternative hypothesis posits the presence of an effect or a difference.

In the context of the two-sample \( t \) test, as seen in our electronic media usage investigation with kids, we're comparing two separate groups (1999 vs. 2009 data). The null hypothesis could state that there is no difference in mean hours of electronic media use between these two years. Conversely, the alternative hypothesis posits that there is indeed a difference (specifically that the 2009 mean usage is greater in this scenario).

Once these hypotheses are established, we conduct the test, which yields a p-value. This p-value assists in decision-making: if it’s below our predetermined significance level (in this case, 0.01), we reject the null hypothesis, inclined to believe that there's evidence favoring the alternative hypothesis - suggestive of a change in media use habits over a decade. If the p-value is above the significance level, we do not reject the null hypothesis, indicating that the sample data hasn't provided sufficient evidence to show a difference in average media use.

Confidence Interval Estimate

A confidence interval estimate is a range of values, derived from the sample statistics, that is likely to enclose the true population parameter with a specified level of confidence. It's more than a simple estimate of the mean; it provides a range that, ideally, captures the true mean a certain percentage of the time if the same study were to be repeated multiple times.

In the case of our study, constructing a 98% confidence interval for the difference between the means of electronic media usage in 2009 and 1999 gives us a range of values that we are 98% confident contains the true difference in mean hours. The wider the interval, the more uncertain the estimate—which is a trade-off for higher confidence. The interval is created using the mean differences between the two samples and takes into account the variability within each group, as well as the size of the groups.

Interpreting a 98% confidence interval involves acknowledging that if we were to conduct this study multiple times, and each time compute a new confidence interval estimate, 98% of these intervals would contain the true mean difference in media use across the populations from the two specified years.

Normality Assumption

The normality assumption is a key consideration when applying parametric tests such as the two-sample \( t \) test. This assumption posits that the data is distributed normally (i.e., in a bell-shaped curve) within the population. When we assume normality, we are effectively stating that the mean, median, and mode of the data are equal or very close, and that the data are symmetrically distributed about the mean.

To verify this assumption graphically, histograms or box plots can be created. These visualizations help identify whether the data is unimodal and symmetric, which would make it reasonable to proceed with the \( t \) test. In this scenario from our exercise, the graphical assessment assists us in gauging whether the electronic media usage times follow a normal distribution. Given small sample sizes, it's especially important to look at the data graphically because the central limit theorem may not provide as much 'cover' for the assumption violations as it would for larger sample sizes.

Ultimately, while real-world data frequently strays from perfect normality, slight deviations might not preclude the use of a two-sample \( t \) test, especially if the sample size is large. However, with small samples, as seen in the exercise, the normality assumption becomes more critical to the validity of the test's result.