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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 number of hours per day spent using electronic media, consistent with summary quantities in the article, are given below (the actual sample sizes for the two surveys were much larger). For purposes of this exercise, you can assume that the two samples are representative of kids ages 8 to 18 in each of the 2 years when the surveys were conducted. $$ \begin{array}{lllllllllllll} \mathbf{2 0 0 9} & 5 & 9 & 5 & 8 & 7 & 6 & 7 & 9 & 7 & 9 & 6 & 9 \\ & 10 & 9 & 8 & & & & & & & & & \\ 1999 & 4 & 5 & 7 & 7 & 5 & 7 & 5 & 6 & 5 & 6 & 7 & 8 \\ & 5 & 6 & 6 & & & & & & & & & \\ & & & & & & & & & & & \end{array} $$ a. Because the given sample sizes are small, what assumption must be made about the distributions of electronic media use times for the two-sample \(t\) test to be appropriate? 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 0.01 .

Short Answer

Expert verified
To answer the exercise fully, a two-sample t-test should be carried out using the data for 1999 and 2009. Graphs should be created to validate the normal distribution assumption. The hypothesis test will determine if there's a significant increase in media usage in 2009 compared to 1999 at a 0.01 significance level.

Step by step solution

01

Assumptions for the two-sample t-test

For the t-test to be appropriate, we need to assume that the time spent on electronic media for the two years follows a normal distribution. This assumption is necessary because of the small sample size. To test this assumption, we can use graphical methods, such as histograms and normal probability plots. If the data doesn't greatly deviate from the shape of a typical normal distribution, we can believe the assumption to be reasonable.
02

Hypothesis formation for the t-test

The next step is to form the null and alternate hypotheses for the test. The null hypothesis (H0) is: There is no difference in the mean number of hours kids spent using electronic media in 1999 and 2009. Alternatively, the alternate hypothesis (Ha) is: The mean number of hours kids spent using electronic media in 2009 is greater than in 1999.
03

Conducting the t-test

Utilize the given data to run the t-test. First, calculate the sample means and standard deviations for both years. Thereafter, calculate the test statistic using the formula for the two-sample t-test: \( t = \frac{ \bar{x}_{2009} - \bar{x}_{1999} } { \sqrt{ \frac{s_{2009}^2}{n_{2009}} + \frac{s_{1999}^2}{n_{1999}} } } \) where \(\bar{x}\) is the sample mean, \(s^2\) is the sample variance, and \(n\) is the sample size.
04

Test decision

Determine the critical value for the t-distribution with \(n_{2009} + n_{1999} - 2\) degrees of freedom at a significance level of 0.01. If the computed t-value is greater than the critical value, reject the null hypothesis and conclude that the mean number of hours kids spent using electronic media in 2009 was greater than in 1999. If not, fail to reject the null hypothesis and conclude that there is not enough evidence to show a significant increase in media use.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a core concept in statistics used to determine if there is enough evidence to support a specific hypothesis about a population parameter, such as the population mean. The process begins by proposing two contrasting statements: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). The null hypothesis typically represents a statement of 'no effect' or 'no difference.' In the context of our exercise, \( H_0 \) suggests that the mean number of hours children spent using electronic media in 1999 and 2009 is the same. Conversely, the alternative hypothesis \( H_a \) posits that these means are different, specifically that media usage has increased by 2009.

To test these hypotheses, we use a test statistic that, under the null hypothesis, follows a known distribution. For small samples, like in our exercise, we often use the t-distribution. If the test statistic falls into a critical region, we reject the null hypothesis in favor of the alternative. This decision is made with an associated significance level, which defines the probability of committing a Type I error—rejecting the null hypothesis when it is, in fact, true.
Sample Mean and Variance
The sample mean and variance are key measures that describe the center and spread of data. The sample mean \( \bar{x} \) is computed by summing all observations and dividing by the number of observations. For example, to calculate the mean number of hours spent using electronic media for 2009, we add the daily hours and divide by the number of children surveyed.

The sample variance \( s^2 \) measures the variability within the data by averaging the squared differences between each data point and the sample mean. It provides insight into how dispersed the measurements are in relation to the mean. Both the mean and variance are integral to the computation of the t-statistic for the two-sample t-test and ultimately help us determine if the observed difference between two sample means is statistically significant.
Normal Distribution Assumption
A crucial assumption in many statistical tests, including the two-sample t-test, is that the data follow a normal distribution, especially with small sample sizes. If the sample is large, the Central Limit Theorem helps mitigate the need for this assumption, as the distribution of sample means will be approximately normal regardless of the underlying distribution of the data.

In our case, the sample sizes are small, so the normal distribution assumption is essential. We can assess this assumption visually using histograms or Q-Q plots. For each sample (1999 and 2009), these graphical displays should show a bell-shaped curve similar to that of a normal distribution. If the assumption is violated, the results of the two-sample t-test may not be reliable. Therefore, verifying normality is a critical step before proceeding with the hypothesis test.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold set by the researcher which determines how much evidence is required to reject the null hypothesis. Common significance levels include 0.05, 0.01, and 0.10. In the exercise, a significance level of 0.01 is used, indicating we require very strong evidence before we conclude that the average media usage has changed.

The choice of significance level influences the chance of a Type I error. At a 0.01 significance level, there is just a 1% chance of incorrectly rejecting the null hypothesis when it is true. As such, the selection of \( \alpha \) affects the rigor of the study—lower \( \alpha \) values make the test more conservative. Ultimately, if our computed t-statistic exceeds the critical value associated with our chosen \( \alpha \) level, we have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis, consistent with the procedures of hypothesis testing.

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Most popular questions from this chapter

Each person in a random sample of 228 male teenagers and a random sample of 306 female teenagers was asked how many hours he or she spent online in a typical week (Ipsos, January 25,2006 ). The sample mean and standard deviation were 15.1 hours and 11.4 hours for the males and 14.1 hours and 11.8 hours for the females. a. The standard deviation for each of the samples is large, indicating a lot of variability in the responses to the question. Explain why it is not reasonable to think that the distribution of responses would be approximately normal for either the population of male teenagers or the population of female teenagers. b. Given your response to Part (a), would it be appropriate to use the two- sample \(t\) test to test the null hypothesis that there is no difference in the mean number of hours spent online in a typical week for male teenagers and female teenagers? Explain why or why not. c. If appropriate, carry out a test to determine if there is convincing evidence that the mean number of hours spent online in a typical week is greater for male teenagers than for female teenagers. Use \(\alpha=0.05\).

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Head movement evaluations are important because disabled individuals may be able to operate communications aids using head motion. The paper "Constancy of Head Turning Recorded in Healthy Young Humans" (Journal of Biomedical Engineering [2008]\(: 428-436)\) reported the accompanying data on neck rotation (in degrees) both in the clockwise direction (CL) and in the counterclockwise direction (CO) for 14 subjects. For purposes of this exercise, you may assume that the 14 subjects are representative of the population of adult Americans. Based on these data, is it reasonable to conclude that mean neck rotation is greater in the clockwise direction than in the counterclockwise direction? Carry out a hypothesis test using a significance level of 0.01 . $$ \begin{array}{lccccccc} \text { Subject: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { CL: } & 57.9 & 35.7 & 54.5 & 56.8 & 51.1 & 70.8 & 77.3 \\ \text { CO: } & 44.2 & 52.1 & 60.2 & 52.7 & 47.2 & 65.6 & 71.4 \\ \text { Subject: } & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \text { CL: } & 51.6 & 54.7 & 63.6 & 59.2 & 59.2 & 55.8 & 38.5 \\ \text { CO: } & 48.8 & 53.1 & 66.3 & 59.8 & 47.5 & 64.5 & 34.5 \end{array} $$

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