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According to the Magazine Publishers of America (www.magazine.org), the average visit at the magazines' Web sites during the fourth quarter of 2007 lasted \(4.145\) minutes. Forty-six randomly selected visits to magazine's Web sites during the fourth quarter of 2009 produced a sample mean visit of \(4.458\) minutes, with a standard deviation of \(1.14\) minutes. Using the \(10 \%\) significance level and the critical value approach, can you conclude that the length of an average visit to these Web sites during the fourth quarter of 2009 was longer than \(4.145\) minutes? Find the range for the \(p\) -value for this test. What will your conclusion be using this \(p\) -value range and \(\alpha=.10\) ?

Short Answer

Expert verified
After computing the Z-score in step 2, comparing this Z value with the critical value in step 3 will help us decide whether to reject or fail to reject the null hypothesis. The exact p-value depends on the computed Z score. The decision of whether to reject or fail to reject the null hypothesis in step 5 is based on a comparison of this p-value with \(\alpha = .10\).

Step by step solution

01

Set up the null and alternative hypotheses

The null hypothesis \((H_0)\) is that the mean visit time in 2009 is equal to that of 2007. In mathematical terms, \(H_0: \mu = 4.145\) minutes. The alternative hypothesis \((H_1)\) is that the mean visit time in 2009 is longer than that of 2007, that is \(H_1: \mu > 4.145\) minutes.
02

Calculate the test statistic

The test statistic in a hypothesis test based on a normal distribution is a Z-score. This can be calculated as \(Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}\) where \(\bar{X}\) is the sample mean (4.458 minutes), \(\mu_0\) is the value under the null hypothesis (4.145 minutes), \(\sigma\) is the sample standard deviation (1.14 minutes), and \(n\) is the sample size (46). Calculate the Z value.
03

Determine the critical value and make decision

The critical value for a 10% significance level in a one-tailed test (because the alternative hypothesis involves a 'greater than') from standard normal distribution table is 1.28. If the absolute value of the Z-score from Step 2 is greater than 1.28, we reject the null hypothesis. Otherwise, we don't reject the null hypothesis.
04

Find the range for the p-value

The p-value is the smallest level of significance at which we can still reject the null hypothesis, given the observed sample data. Use the standard normal distribution table to correlate with the calculated Z score from step 2. Because this is a one-tailed test, the p-value is the area under the standard normal curve to the right of the computed Z.
05

Conclusion using p-value and alpha

If our calculated p-value is less than or equal to the significance level \(\alpha = .10\), then we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is a starting assumption for statistical tests. When we conduct a hypothesis test, we start with the null hypothesis. It usually proposes no effect or no difference in the context of the experiment.

In our exercise, the null hypothesis suggests that the mean time people spent on magazine websites in 2009 is the same as in 2007, specifically \(\mu = 4.145\) minutes. Establishing this hypothesis helps us set a baseline, allowing any statistical analysis to be clearly compared against this statement.

Hypothesis testing effectively addresses whether our sample data sufficiently shows that the actual preference or mean has changed from the null hypothesis. This is crucial for determining the validity of our experimental hypothesis.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\), states what we are seeking evidence for in our hypothesis test. It reflects the point where there's a perceived difference or effect compared to the null hypothesis.

In the exercise, our alternative hypothesis is that visitors spend more time on magazine websites in 2009 than they did in 2007. We express this as \(H_1: \mu > 4.145\) minutes.

Here, the alternative hypothesis leads us away from the assumption of no change, particularly focusing on increasing visit duration. This hypothesis is key to validating research questions that hypothesize a change or improvement from a known standard or measurement.
Significance Level
The significance level, often illustrated by \(\alpha\), represents the threshold at which we decide whether a statistical finding has enough evidence to reject the null hypothesis. It is set by the researcher before analyzing the data.

In this study, the significance level is 10%, or 0.10. This means we're willing to risk a 10% chance of incorrectly rejecting the null hypothesis when it's true, referred to as a Type I error.

This decision threshold is critical since it affects the rigor and conclusions drawn from statistical testing. A lower significance level signifies more stringent criteria for claiming a result is significant.
Z-Score
The Z-Score is a statistic that tells us how many standard deviations a data point is from the mean. In hypothesis testing, the Z-score helps us compare our sample mean to the null hypothesis mean, considering the sample size and variability.

For this exercise, the Z-score is calculated using the formula \(Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}\), where:
  • \(\bar{X}\) is the sample mean of 4.458 minutes,
  • \(\mu_0\) is 4.145 minutes,
  • \(\sigma\) is the standard deviation of 1.14 minutes,
  • \(n\) is the sample size of 46.


The Z-score quantifies the difference between our sample mean and the null hypothesis mean, adjusted for standard deviations, providing a standardized approach to testing hypotheses.
P-Value
The p-value is a critical component of hypothesis testing, showing the probability of observing test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is true.

In our case, the p-value helps us determine if our sample mean of 4.458 minutes significantly deviates from the assumed population mean of 4.145 minutes under the null hypothesis.

To find the p-value, we consider the area under the normal distribution curve to the right of our computed Z-score. A smaller p-value indicates that the observed data would be unusual if the null hypothesis were true, leading us to possibly reject \(H_0\) if the p-value is less than or equal to our significance level of 0.10. This process helps make statistically sound conclusions from empirical data.

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Most popular questions from this chapter

A computer company that recently introduced a new software product claims that the mean time it takes to learn how to use this software is not more than 2 hours for people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Test at the \(1 \%\) significance level whether the company's claim is true. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

In an observational study at Turner Field in Atlanta, Georgia, \(43 \%\) of the men were observed not washing their hands after going to the bathroom (see Exercise \(7.80\) ). Suppose that in a random sample of 95 men who used the bathroom at Camden Yards in Baltimore, Maryland, 26 did not wash their hands. a. Using the critical-value approach and \(\alpha=.10\), test whether the percentage of all men at Camden Yards who use the bathroom and do not wash their hands is less than \(43 \%\). b. How do you explain the Type I error in part a? What is the probability of making this error in part a? c. Calculate the \(p\) -value for the test of part a. What is your conclusion if \(\alpha=.10 ?\)

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A business school claims that students who complete a 3 -month typing course can type, on average, at least 1200 words an hour. A random sample of 25 students who completed this course typed, on average, 1125 words an hour with a standard deviation of 85 words. Assume that the typing speeds for all students who complete this course have an approximately normal distribution. a. Suppose the probability of making a Type I error is selected to be zero. Can you conclude that the claim of the business school is true? Answer without performing the five steps of a test of hypothesis. b. Using the \(5 \%\) significance level, can you conclude that the claim of the business school is true? Use both approaches.

Consider \(H_{0}: \mu=40\) versus \(H_{1}: \mu>40\) a. A random sample of 64 observations taken from this population produced a sample mean of 43 and a standard deviation of \(5 .\) Using \(\alpha=.025\), would you reject the null hypothesis? b, Another random sample of 64 observations taken from the same population produced a sample mean of 41 and a standard deviation of 7 . Using \(\alpha=.025\), would you reject the null hypothesis?

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