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Chapter 10: Problem 68

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 \(\alpha=.05\) ?

Short Answer

Expert verified
a. Yes, a level 0.05 test would lead to the conclusion that more than 75% of all apartments exclude children. b. The power of the test when \(\pi=.8\) and \(\alpha=.05\) is approximately 0.9865.

Step by step solution

01

Perform Hypothesis Testing

Firstly, it is important to set up the null and alternative hypothesis before performing the hypothesis testing. Null hypothesis, \(\(H_0: \pi <= 0.75\) and alternative hypothesis, \(\(H_a: \pi > 0.75\). The sample proportion \(\(p\)\) is 102/125 = 0.816. We can calculate the standard deviation of the estimate as \(\( \sqrt{ (p*(1-p)/n) }\) where n is the sample size. Therefore, \(\(SE = \sqrt{ (0.816*(1-0.816)/125)}\), approximately equals 0.0343.
02

Calculate Test Statistic

The test statistic z can be calculated as \(( \(\p\) - \(\pi\) )/SE.\), giving us \( (0.816-0.75)/0.0343 \) which is approximately 1.924.
03

Determine p-value and Conclusion

The p-value corresponding to z=1.924 with a right-tailed test is approximately 0.0273. Since this p-value is less than the significance level of 0.05, we would reject the null hypothesis and conclude that more than 75% of all apartments exclude children.
04

Calculate Power of Test

The power of the test is the probability of correctly rejecting the null hypothesis. We first find the z value for \(\alpha = 0.05\) which is 1.645 (for a one-tailed test). The z value indicating the boundary for rejecting \(\H_0\) when \(\ \pi = 0.8\) is (0.8-0.75)/\(\sqrt{ (0.75*(1-0.75)/125)} \), approximately equals 2.216. The power of the test is the area to the right of this z value, which is approximately 0.9865 using standard normal tables.

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Key Concepts

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

Proportion Test
When conducting a proportion test, we're checking if a specific proportion exceeds or falls below a certain threshold. In our example scenario, we're determining whether the proportion of apartments that exclude children (\( \pi \) exceeds 0.75 or not. To do this, we often set up two hypotheses:
  • Null Hypothesis (\( H_0 \) ): \( \pi \leq 0.75 \)
  • Alternative Hypothesis (\( H_a \) ): \( \pi > 0.75 \)
The sample proportion (\( p \)) is calculated, which in our scenario is the number of apartments excluding children (102) divided by the total sampled apartments (125), resulting in 0.816. This sample proportion is compared to the threshold in our hypothesis through statistical calculations using the standard error.
Significance Level
The significance level, often denoted by \( \alpha \), plays a crucial role in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. Commonly set at 0.05, this means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
In our example, if our calculated p-value is less than \( \alpha = 0.05 \), we reject the null hypothesis. This decision indicates that the probability of observing our sample proportion, assuming the null hypothesis is true, is low enough to reject it. Hence, in the apartment problem, a p-value of 0.0273 confirms that more than 75% of apartments are likely excluding children.
Power of a Test
The power of a test gauges its ability to correctly reject a false null hypothesis. This is crucial in determining how effective our test is. A higher power indicates a greater likelihood of detecting an effect if there is one.
To calculate power, we determine the probability of rejecting the null hypothesis when an alternative population parameter is true. For the apartment problem, if \( \pi = 0.8 \) and \( \alpha = 0.05 \), we use statistical tools like standard normal tables to find the z-value for \( \alpha \), approximately 1.645.
Then, with \( \pi \) as 0.8, recalculating gives a z-value around 2.216, resulting in a test power of approximately 0.9865. Thus, there is a high probability that the test will correctly identify when \( \pi \) has exceeded 0.75.
Standard Deviation
Standard deviation is key in evaluating how much variability exists around a sample statistic. In the context of proportion tests, it helps calculate the standard error, which measures the expected variation in the sample proportion.
For our sample of apartments, the standard deviation for sample proportion (\( SE \)) is determined using the formula:\[SE = \sqrt{ \frac{ p(1-p) }{ n }}\]where \( p \) is the sample proportion (0.816) and \( n \) is the sample size (125). This results in an \( SE \) of approximately 0.0343, providing insight into the precision of our sample proportion and informing the subsequent test statistic calculation. Standard deviation thus aids in understanding and interpreting the test's results.

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Most popular questions from this chapter

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: $$ H_{0}: \pi=.02 \text { versus } H_{a}: \pi<.02 $$ 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.

Are young women delaying marriage and marrying at a later age? This question was addressed in a report issued by the Census Bureau (Associated Press, June 8 , 1991). The report stated that in 1970 (based on census results) the mean age of brides marrying for the first time was \(20.8\) years. In 1990 (based on a sample, because census results were not yet available), the mean was \(23.9\). Suppose that the 1990 sample mean had been based on a random sample of size 100 and that the sample standard deviation was \(6.4\). Is there sufficient evidence to support the claim that in 1990 women were marrying later in life than in 1970 ? Test the relevant hypotheses using \(\alpha=.01\). (Note: It is probably not reasonable to think that the distribution of age at first marriage is normal in shape.)

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 \(.01\).

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 \(.01\).

Medical personnel are required to report suspected cases of child abuse. Because some diseases have symptoms that mimic those of child abuse, doctors who see a child with these symptoms must decide between two competing hypotheses: \(H_{0}\) : symptoms are due to child abuse \(H_{a^{*}}\) symptoms are due to disease (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) The article "Blurred Line Between Illness, Abuse Creates Problem for Authorities" (Macon Telegraph, February 28,2000 ) included the following quote from a doctor in Atlanta regarding the consequences of making an incorrect decision: "If it's disease, the worst you have is an angry family. If it is abuse, the other kids (in the family) are in deadly danger." a. For the given hypotheses, describe Type I and Type II errors. b. Based on the quote regarding consequences of the two kinds of error, which type of error does the doctor quoted consider more serious? Explain.
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