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In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the sample observations was \(12.7\) hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of \(.05\) to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a signifi-

Short Answer

Expert verified
a. With a standard deviation of 5 hours, there is not enough evidence to conclude that the mean time Canadians spend on the internet is more than 12.5 hours. b. With a standard deviation of 2 hours, there is enough evidence to support the alternative hypothesis.

Step by step solution

01

Formulate the Hypotheses

First, state the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_A\)). \(H_0: \mu = 12.5\) and \(H_A: \mu > 12.5\).
02

Calculate the Test Statistic with Standard Deviation as 5

The test statistic is calculated using the formula \(t = \frac{(\overline{x} - \mu)}{(\sigma/\sqrt{n})}\) where \(\overline{x}\) is the sample mean, \(\mu\) is the population mean (from the null hypothesis), \(\sigma\) is the standard deviation, and \(n\) is the number of observations in the sample. Here, \(\overline{x} = 12.7\), \(\mu = 12.5\), \(\sigma = 5\), and \(n = 1000\). Substituting the values gives \(t = 0.8\).
03

Find the Critical Value and Decide

Using a t-table, find the critical value for a two-tailed test with significance level \(\alpha = 0.05\) and \(df = n - 1 = 999\). If the calculated t-score is greater than the critical value, reject the null hypothesis. Here, the critical value is approximately \(1.96\), so we fail to reject the null hypothesis, since \(0.8 < 1.96\). There is not enough evidence to conclude that Canadians spend more than 12.5 hours on the internet.
04

Calculate the Test Statistic with Standard Deviation as 2

Repeat the second step, this time with \(\sigma = 2\). This gives a new t-score of \(2.83\).
05

Find the Critical Value and Decide Again

This time, we have \(2.83 > 1.96\). Therefore, we reject the null hypothesis and conclude that there is enough evidence that Canadians spend more than 12.5 hours on the internet on average when the standard deviation is 2 hours.

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Key Concepts

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

Null and Alternative Hypothesis
Understanding the null and alternative hypothesis is crucial when embarking on hypothesis testing. The null hypothesis (\(H_0\)) is a statement of no effect or no difference and acts as a default assumption for the test. For instance, if we are testing whether Canadians spend more than 12.5 hours on the Internet, the null hypothesis is that they spend exactly 12.5 hours, notated as \(H_0: \(\mu = 12.5\)\).

Conversely, the alternative hypothesis (\(H_A\) or \(H_1\)) is what you aim to support, suggesting that there is an effect or a difference. In our example, the alternative hypothesis posits that Canadians spend more than 12.5 hours on the Internet, expressed as \(H_A: \(\mu > 12.5\)\). The outcome of the hypothesis test will lead to either the rejection of the null hypothesis in favor of the alternative or a failure to reject the null, thus maintaining our initial assumption.
Test Statistic Calculation
The test statistic is a standardized value used to determine the probability of the sample data if the null hypothesis were true. Calculating a test statistic involves several parameters, including the sample mean (\(\overline{x}\)), hypothesized population mean (\(\mu\)), sample standard deviation (\(\sigma\)), and the sample size (\(n\)). In our case, the formula \(t = \frac{(\overline{x} - \mu)}{(\sigma/\sqrt{n})}\) is used. With a sample mean of 12.7 hours and assuming a standard deviation of 5 hours for 1000 Internet users, the test statistic is calculated as 0.8.

It's the comparison of the test statistic to a critical value from the t-distribution that provides the basis for a decision regarding the null hypothesis. If the test statistic falls into the region beyond the critical value, the null hypothesis is rejected.
Significance Level
The significance level, typically denoted by \(\alpha\), is a threshold for how much evidence you require to claim that an observed effect in your sample is present in the population. It's the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. A common choice for \(\alpha\) is 0.05, meaning there is a 5% risk of concluding that a difference exists when there is none.

A crucial step in hypothesis testing is to use this significance level to determine the critical value from a statistical distribution, which defines the boundary for rejecting the null hypothesis. If our test statistic exceeds the critical value corresponding to \(\alpha = 0.05\), we reject the null hypothesis on the grounds that our result would be very unlikely if \(H_0\) were true.
Sample Standard Deviation
The sample standard deviation (\(\sigma\)) is a measure of the amount of variation or dispersion in a set of sample values. A low standard deviation indicates that the values tend to be close to the mean of the sample, while a high standard deviation indicates that the values are spread out over a wider range. The sample standard deviation affects the calculation of the test statistic, influencing the result of the hypothesis test. For example, with a standard deviation of 5 hours, we concluded there wasn't enough evidence to reject the null hypothesis, but reducing the standard deviation to 2 hours resulted in a test statistic large enough to reject the null hypothesis - suggesting that the mean time Canadians spend on the Internet is indeed greater than 12.5 hours. This illustrates how the precision of data (as captured by \(\sigma\)) changes the outcome of hypothesis tests.

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Most popular questions from this chapter

Pairs of \(P\) -values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\) -value leads to rejection of \(H_{0}\) at the given significance level. a. \(\quad P\) -value \(=.084, \alpha=.05\) b. \(\quad P\) -value \(=.003, \alpha=.001\) c. \(\quad P\) -value \(=.498, \alpha=.05\) d. \(\quad P\) -value \(=.084, \alpha=.10\) e. \(P\) -value \(=.039, \alpha=.01\) f. \(\quad P\) -value \(=.218, \alpha=.10\)

An automobile manufacturer who wishes to advertise that one of its models achieves \(30 \mathrm{mpg}\) (miles per gallon) decides to carry out a fuel efficiency test. Six nonprofessional drivers are selected, and each one drives a car from Phoenix to Los Angeles. The resulting fuel efficiencies (in miles per gallon) are: \(\begin{array}{llllll}27.2 & 29.3 & 31.2 & 28.4 & 30.3 & 29.6\end{array}\) Assuming that fuel efficiency is normally distributed under these circumstances, do the data contradict the claim that true average fuel efficiency is (at least) \(30 \mathrm{mpg}\) ?

In a national survey of 2013 adults, 1590 responded that lack of respect and courtesy in American society is a serious problem, and 1283 indicated that they believe that rudeness is a more serious problem than in past years (Associated Press, April 3, 2002 ). Is there convincing evidence that less than three-quarters of U.S. adults believe that rudeness is a worsening problem? Test the relevant hypotheses using a significance level of \(.05\).

Use the definition of the \(P\) -value to explain the following: a. Why \(H_{0}\) would be rejected if \(P\) -value \(=.0003\) b. Why \(H_{0}\) would not be rejected if \(P\) -value \(=.350\)

The international polling organization Ipsos reported data from a survey of 2000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24, 2006). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. Suppose that the sample mean and standard deviation were \(\$ 9.15\) and \(\$ 7.60\), respectively. (These values are consistent with a histogram of the sample data that appears in the report.) Do these data provide convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be appropriate is less than \(\$ 10\) ? Carry out a hypothesis test with a significance level of \(.01\).

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