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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

Expert verified
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

01

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.
02

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.
03

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 \( T = \frac{14.6 - 10}{11.6/ \sqrt{534}}\).
04

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

The article "Fewer Parolees Land Back Behind Bars" (Associated Press, April 11,2006 ) includes the following statement: "Just over 38 percent of all felons who were released from prison in 2003 landed back behind bars by the end of the following year, the lowest rate since 1979." Explain why it would not be necessary to carry out a hypothesis test to determine if the proportion of felons released in 2003 was less than \(.40\).

When a published article reports the results of many hypothesis tests, the \(P\) -values are not usually given. Instead, the following type of coding scheme is frequently used: \({ }^{*} p=.05,{ }^{* *} p=.01,{ }^{* * *} p=.001,{ }^{* * * *} p=.0001\). Which of the symbols would be used to code for each of the following \(P\) -values? a. \(.037\) c. 072 b. \(.0026\) d. \(.0003\)

A manufacturer of hand-held calculators receives large shipments of printed circuits from a supplier. It is too costly and time-consuming to inspect all incoming circuits, so when each shipment arrives, a sample is selected for inspection. Information from the sample is then used to test \(H_{0}=\pi=.05\) versus \(H_{a}: \pi>.05\), where \(\pi\) is the true proportion of defective circuits in the shipment. If the null hypothesis is not rejected, the shipment is accepted, and the circuits are used in the production of calculators. If the null hypothesis is rejected, the entire shipment is returned to the supplier because of inferior quality. (A shipment is defined to be of inferior quality if it contains more than \(5 \%\) defective circuits.) a. In this context, define Type I and Type II errors. b. From the calculator manufacturer's point of view, which type of error is considered more serious? c. From the printed circuit supplier's point of view, which type of error is considered more serious?

Let \(\mu\) denote the true average lifetime for a certain type of pen under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\) will be based on a sample of size 36. Suppose that \(\sigma\) is known to be \(0.6\), from which \(\sigma_{x}=0.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq\) \(-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8 ?\) when \(\mu=9.5 ?\)

The power of a test is influenced by the sample size and the choice of significance level. a. Explain how increasing the sample size affects the power (when significance level is held fixed). b. Explain how increasing the significance level affects the power (when sample size is held fixed).

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