Question

A survey of teenagers and parents in Canada conducted by the polling organization Ipsos ("Untangling the Web: The Facts About Kids and the Internet," January \(25 .\) 2006) included questions about Internet use. It was reported that for a sample of 534 randomly selected teens, the mean number of hours per week spent online was \(14.6\) and the standard deviation was \(11.6\). a. What does the large standard deviation, \(11.6\) hours, tell you about the distribution of online times for this sample of teens? b. Do the sample data provide convincing evidence that the mean number of hours that teens spend online is greater than 10 hours per week?

Short answer

Part a. A large standard deviation of 11.6 hours indicates a high variability in the online times among the sample of teenagers, meaning their time spent online greatly differs from the average. Part b. The p-value linked to the calculated test statistic will provide the convincing evidence. If it is less than 0.05 (common threshold), we can conclude that the mean number of hours teens spend online is greater than 10 hours per week.

Step-by-step solution

Analysis of Standard Deviation

Standard deviation measures the dispersion of a dataset from its mean. A large standard deviation of \(11.6\) hours indicates that the time spent online by teenagers in the sample varies greatly from the mean value of \(14.6\) hours. This suggests a diverse range of internet usage habits among the teens.

Formulate Null and Alternative Hypotheses

Null hypothesis (H0): \(\mu = 10\) hours per week, meaning the population mean is equal to 10 hours; Alternative hypothesis (HA): \(\mu > 10\) hours per week, meaning the population mean is greater than 10 hours.

Calculate the Test Statistic

The test statistic for a one-sample t-test is given by \(T = \frac{\overline{x} - \mu}{s/ \sqrt{n}}\), where \(\overline{x}\) is the sample mean, \( \mu \) is the population mean under null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size. Substituting the given values, we get \( T = \frac{14.6 - 10}{11.6/ \sqrt{534}}\).

Find the p-value

Once the test statistic is calculated, look up the corresponding p-value. If it is less than the chosen significance level (\(\alpha\), usually 0.05), then the null hypothesis is rejected, providing evidence that teens spend on average more than 10 hours online per week.

Key concepts

Understanding Standard Deviation

Standard deviation is a statistical measure that quantifies the degree of variation or dispersion of a set of values. A low standard deviation means that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

The large standard deviation of 11.6 hours in the context of the teenagers’ online time suggests there is considerable inconsistency in internet use among them. Some teens may use the internet sparingly, perhaps just a few hours a week, while others may use it extensively, exceeding the average of 14.6 hours considerably. This variability can be due to many factors such as different interests, levels of access to technology, or varying schedules.

An important part of understanding standard deviation is to consider the context of the data. In this case, considering that the activity is 'time spent online,' a broad range of hours could be perfectly normal; recreational internet use isn't as standardized as other behaviors might be.

The Role of the Null Hypothesis in Testing

The null hypothesis is a foundational concept in hypothesis testing. It represents a statement of no effect or no difference and serves as a starting point for statistical tests. When we perform a hypothesis test, we assume the null hypothesis is true unless there is sufficient evidence from our sample data to suggest otherwise.

In the example of the survey on teenagers' online habits, the null hypothesis would be formally stated as 'Teenagers in Canada spend, on average, 10 hours per week online.' We denote this as \(H_0: \mu = 10 \). The alternative hypothesis is the opposite claim we're trying to provide evidence for, which, in this case, is 'Teenagers in Canada spend more than 10 hours per week online.' We denote this as \(H_A: \mu > 10\).

The null hypothesis is crucial because it creates a benchmark against which we measure the observed effect of our study. If we find sufficient evidence to reject the null hypothesis through various statistical tests, we can then support the alternative hypothesis with more confidence.

Analyzing Results with a One-Sample t-Test

A one-sample t-test is a statistical procedure used to determine whether the mean of a single sample is significantly different from a known or hypothesized population mean. It's especially useful when the population standard deviation is unknown and the sample size is relatively small (usually less than 30, although it can be used with larger samples as well).

The formula for the test statistic in a one-sample t-test is given by \(T = \frac{\overline{x} - \mu}{s / \sqrt{n}}\), where \(\overline{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size. It yields a value that, combined with degrees of freedom (n-1), can be used to determine the p-value.

The p-value represents the probability of observing the given result, or one more extreme, if the null hypothesis is true. A small p-value (\< 0.05) suggests that the observed data is unlikely under the null hypothesis, leading to its rejection in favor of the alternative hypothesis. Thus, in the case of the survey, if the calculated p-value is small, we can conclude with evidence that the average time spent online by teens is indeed greater than 10 hours per week.