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Chapter 10: Problem 36

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. Four hundred of these companies indicated that they monitor employees' web site visits. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Is there sufficient evidence to conclude that more than \(75 \%\) of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\). b. Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of \(.01\).

Short Answer

Expert verified
To answer part a, if the calculated Z value for part a is greater than 2.33, there is sufficient evidence to conclude that more than 75% of U.S. businesses monitor employees' website visits at a 0.01 significance level. For part b, if the calculated Z value for part b is greater than 2.33, there is sufficient evidence to conclude that more than 50% of U.S. businesses monitor employees' website visits at a 0.01 significance level.

Step by step solution

01

Identify the Null and Alternative Hypotheses for part a

The null hypothesis \(H_0\) is that 75% (or 0.75) of U.S. businesses monitor employees' web site visits. The alternative hypothesis \(H_a\) is that more than 75% (or 0.75) of U.S. businesses monitor employees' web site visits. \nSo, \(H_0: p = 0.75\) and \(H_a: p > 0.75\)
02

Calculate the Test Statistics for part a

The test statistic is given by the formula \(Z = \frac{{\hat{p} - p}}{{\sqrt{\frac{{pq}}{{n}}}}}\), where \(\hat{p}\) is the sample proportion, \(p\) and \(q\) are the hypothesised population proportion and its complement respectively, and \(n\) is the sample size. Substituting the known values, \(Z = \frac{{0.76 - 0.75}}{{\sqrt{\frac{{0.75 \times 0.25}}{{526}}}}\)
03

Make a Decision for part a

Compare the calculated Z value with the critical Z value from the Z table for a 0.01 significance level (which is 2.33 for a one-tailed test). If the calculated Z value is greater than the critical Z value, reject the null hypothesis.
04

Identify the Null and Alternative Hypotheses for part b

The null hypothesis \(H_0\) is that 50% (or 0.5) of U.S. businesses monitor employees' web site visits. The alternative hypothesis \(H_a\) is that more than 50% (or 0.5) of U.S. businesses monitor employees' web site visits. \nSo, \(H_0: p = 0.5\) and \(H_a: p > 0.5\)
05

Calculate Test Statistics for part b

Calculate the test statistic using the same formula as in part a, but replacing p with 0.5 and q with 0.5 as per the null hypothesis for part b. The test statistic becomes \(Z = \frac{{0.76 - 0.5}}{{\sqrt{\frac{{0.5 \times 0.5}}{{526}}}}\)
06

Make a Decision for part b

Just like in part a, compare the calculated Z value with the critical Z value from the Z table for a 0.01 significance level (again, which is 2.33 for a one-tailed test). If the calculated Z value is greater than the critical Z value, reject the null hypothesis.

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Key Concepts

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

Significance Level
In hypothesis testing, the significance level is crucial because it defines the threshold for deciding whether to reject the null hypothesis. It is represented by the Greek letter alpha (\( \alpha \)).
The significance level can be thought of as the probability of making a Type I error, which happens when the null hypothesis is incorrectly rejected.For example, a common significance level is 0.05, which means there is a 5% chance of rejecting the null hypothesis if it is actually true. In the given exercise, the significance level is set to 0.01. This is more stringent, meaning there is only a 1% risk of mistakenly rejecting the null hypothesis.
Choosing a lower significance level like 0.01 is often used in situations where being wrong has serious consequences or when greater statistical certainty is required. The chosen significance level guides the interpretation of the test statistic and is essential in drawing conclusions in hypothesis testing.
Null Hypothesis
The null hypothesis, usually denoted by \( H_0 \), is a default position that there is no effect or no difference. It serves as the starting assumption for statistical testing.
In hypothesis testing, the aim is often to gather evidence to support the rejection of the null hypothesis.In the context of the exercise, the null hypothesis for part a is that exactly 75% of U.S. businesses monitor employees' web site visits, meaning \( H_0: p = 0.75 \). For part b, the null hypothesis asserts that 50% monitor such visits,\( H_0: p = 0.5 \).
The null hypothesis is typically stated in such a way that it includes the possibility of equality. It's essential because it provides a benchmark to compare observed data with. By assessing the plausibility of the null hypothesis, statisticians can make informed decisions based on sample data.
Alternative Hypothesis
The alternative hypothesis, denoted by \( H_a \), represents what we aim to support. It proposes a different scenario from the null hypothesis and suggests that there is an effect or difference.
In many cases, it states that the observed data is significantly different from what the null hypothesis proposes.For example, in the given problem, the alternative hypothesis for part a states that more than 75% of U.S. businesses monitor employees' web site visits \( H_a: p > 0.75 \). Similarly, for part b, \( H_a: p > 0.5 \) implies that more than 50% monitor these visits.
The alternative hypothesis is supported if the test statistic falls into the critical region defined by the chosen significance level. When evidence strongly indicates that the null hypothesis is unlikely, researchers may support the alternative hypothesis, leading to new insights or conclusions in the study.

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Most popular questions from this chapter

Much concern has been expressed in recent years regarding the practice of using nitrates as meat preservatives. In one study involving possible effects of these chemicals, bacteria cultures were grown in a medium containing nitrates. The rate of uptake of radio-labeled amino acid was then determined for each culture, yielding the following observations: \(\begin{array}{llllllll}7251 & 6871 & 9632 & 6866 & 9094 & 5849 & 8957 & 7978\end{array}\) \(\begin{array}{lllllll}7064 & 7494 & 7883 & 8178 & 7523 & 8724 & 7468\end{array}\) Suppose that it is known that the true average uptake for cultures without nitrates is 8000 . Do the data suggest that the addition of nitrates results in a decrease in the true average uptake? Test the appropriate hypotheses using a significance level of \(.10 .\)

The city council in a large city has become concerned about the trend toward exclusion of renters with children in apartments within the city. The housing coordinator has decided to select a random sample of 125 apartments and determine for each whether children are permitted. Let \(\pi\) be the true proportion of apartments that prohibit children. If \(\pi\) exceeds . 75 , the city council will consider appropriate legislation. a. If 102 of the 125 sampled apartments exclude renters with children, would a level \(.05\) test lead you to the conclusion that more than \(75 \%\) of all apartments exclude children? b. What is the power of the test when \(\pi=.8\) and \(\alpha=.05\) ?

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" (Journal of Nutrition Education [1992]: \(179-184\) ) reported the following summary statistics on daily caffeine consumption for a random sample of adult women: \(n=47, \bar{x}=215 \mathrm{mg}, s=\) \(235 \mathrm{mg}\), and the data values ranged from 5 to 1176 . a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose that it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given information contradict this prior belief? Test the appropriate hypotheses at significance level. \(10 .\)

Researchers have postulated that because of differences in diet, Japanese children have a lower mean blood cholesterol level than U.S. children do. Suppose that the mean level for U.S. children is known to be 170 . Let \(\mu\) represent the true mean blood cholesterol level for Japanese children. What hypotheses should the researchers test?

An automobile manufacturer is considering using robots for part of its assembly process. Converting to robots is an expensive process, so it will be undertaken only if there is strong evidence that the proportion of defective installations is lower for the robots than for human assemblers. Let \(\pi\) denote the true proportion of defective installations for the robots. It is known that human assemblers have a defect proportion of 02 . a. Which of the following pairs of hypotheses should the manufacturer test: $$ H_{0}: \pi=.02 \text { versus } H_{s}: \pi<.02 $$ or $$ H_{0}: \pi=.02 \text { versus } H_{a}: \pi>.02 $$ Explain your answer. b. In the context of this exercise, describe Type I and Type II errors. c. Would you prefer a test with \(\alpha=.01\) or \(\alpha=.1 ?\) Explain your reasoning.
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