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91Ó°ÊÓ

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 \(p\) be the proportion of all apartments that prohibit children. If the city council is convinced that \(p\) is greater than 0.75 , it 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 \(p=.8\) and \(\alpha=.05 ?\)

Short Answer

Expert verified
For part a, if the calculated test statistic is greater than the critical value or the p-value is less than 0.05, there is sufficient evidence to suggest that more than 75% of apartments exclude children. For part b, the power of the test would have to be calculated using the obtained test statistic under the assumption that \(p = 0.8\).

Step by step solution

01

Set Up Hypotheses

The null hypothesis is \(H_0: p = 0.75\), namely, 75% of apartments exclude children. The alternative hypothesis is \(H_1: p > 0.75\), implying that more than 75% of apartments exclude children.
02

Calculate Test Statistic for Part A

The test statistic is calculated by the formula \(Z = ( \hat{p} - p_0 ) / \sqrt{ p_0 * (1 - p_0) / n} \), where \(\hat{p}\) is the sample proportion, \(p_0\) is the null hypothesis proportion, and \(n\) is the sample size. Here, \(\hat{p} = 102/125 = 0.816\), \(p_0 = 0.75\), and \(n = 125\). Plug these values into the formula to get the test statistic.
03

Find Critical Value or P-Value for Part A

As a one-tailed test at the 0.05 level, the critical value can be found from a standard normal distribution table. Or, alternatively, calculate the p-value.
04

Make Decision for Part A

If the test statistic is greater than the critical value or the p-value is less than 0.05, reject the null hypothesis and conclude that there is sufficient evidence to suggest that more than 75% of apartments exclude children.
05

Calculate Test Statistic for Part B under Assumption \(p = 0.8\)

Calculate the test statistic under the assumption \(p = 0.8\) as \(Z = (p - p_0 ) / \sqrt{ p_0 * (1 - p_0) / n}\). Here,\(p = 0.8\), \(p_0 = 0.75\), and \(n = 125\). Plug these values into the formula to get the test statistic.
06

Find Power of the Test for Part B

Calculate the power of the test as the probability that you will reject the null hypothesis when \(p = 0.8\), i.e., \(1 - \beta\), where \(\beta\) is the probability of a Type II error.

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Key Concepts

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

Null and Alternative Hypotheses
In the realm of hypothesis testing, the null and alternative hypotheses are the initial key elements to any analysis. The null hypothesis, denoted as H0, is a statement of no effect or no difference and serves as the assumption that we initially accept to be true until evidence suggests otherwise.

Conversely, the alternative hypothesis, symbolized as H1 or Ha, represents what the researcher is seeking to prove. It is a statement indicating the presence of an effect or a difference. The alternative hypothesis is accepted when there is sufficient statistical evidence to reject the null hypothesis.

Using the exercise as an example, the null hypothesis (H0) posits that the proportion of all apartments that prohibit children is 75%, symbolically expressed as H0: p = 0.75. On the flip side, the alternative hypothesis (H1) suggests that this proportion is greater than 75%, written as H1: p > 0.75. This sets the stage for testing whether the evidence supports legislative consideration due to a higher exclusion rate.
Test Statistic Calculation
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's essentially a z-score: the number of standard deviations a data-point is from the mean of the sampling distribution of the statistic.

To calculate the test statistic in our exercise, we use the formula:

\[ Z = \frac{ \hat{p} - p_0 }{ \sqrt{ p_0 * (1 - p_0) / n} } \]

where \( \hat{p} \) is the sample proportion, \( p_0 \) is the proportion under the null hypothesis (in this case 0.75), and \( n \) is the sample size. By substituting the sample data into this formula, we obtain a z-score which tells us how extreme the sample is, assuming the null hypothesis is true. A high absolute value of the test statistic indicates that the observed sample data is unlikely under the null hypothesis and can lead to a rejection of H0.
P-Value and Critical Value
The p-value is a probability that reflects the strength of the evidence against the null hypothesis. Specifically, it is the probability of observing a test statistic as extreme as, or more extreme than, the one computed from the sample data, under the assumption that the null hypothesis is true.

The critical value is a threshold that determines the boundary for rejecting the null hypothesis. It's derived from the level of significance of the test, denoted as \( \alpha \), which is the probability of making a Type I error -- rejecting the null hypothesis when it's actually true.

In the case of the exercise, using the 0.05 level of significance for a one-tailed test indicates that the test statistic should be compared against a critical value from the standard normal distribution. If our calculated test statistic exceeds this critical value or if the p-value is less than 0.05, we reject the null hypothesis, concluding that a higher proportion of apartments exclude children. This pivotal comparison is what determines the statistical validity of our hypotheses.
Test Power Analysis
The power of a test is the probability that it correctly rejects a false null hypothesis (i.e., avoids a Type II error). It's essentially the test's sensitivity to detect an actual effect when there is one.

To perform a power analysis, as mentioned in Part B of the exercise, we calculate the probability of successfully rejecting the null hypothesis when a certain true proportion (in this context, is specified as \( p = 0.8 \)) is present. This calculation requires information on the significance level (\( \alpha \)), the effect size (difference between \( p \) and \( p_0 \)), and the sample size (\( n \)).

High power is desirable and means that there's a good chance of detecting an effect if one exists. If the calculated power is low, the test may not be sensitive enough, which implies that even if apartments significantly exclude children more than the hypothesized proportion, the test could fail to provide enough evidence to take legislative actions. Thus, comprehending the power of a test contributes to the reliability and adequacy of statistical conclusions.

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Most popular questions from this chapter

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 \(=.350\)

A certain university has decided to introduce the use of plus and minus with letter grades, as long as there is evidence that more than \(60 \%\) of the faculty favor the change. A random sample of faculty will be selected, and the resulting data will be used to test the relevant hypotheses. If \(p\) represents the proportion of all faculty that favor a change to plus-minus grading, which of the following pair of hypotheses should the administration test: $$ H_{0}: p=.6 \text { versus } H_{a}: p<.6 $$ or $$ H_{0}: p=.6 \text { versus } H_{a}: p>.6 $$ Explain your choice.

Let \(\mu\) denote the mean diameter for bearings of a certain type. A test of \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) will be based on a sample of \(n\) bearings. The diameter distribution is believed to be normal. Determine the value of \(\beta\) in each of the following cases: a. \(\quad n=15, \alpha=.05, \sigma=0.02, \mu=0.52\) b. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.48\) c. \(\quad n=15, \alpha=.01, \sigma=0.02, \mu=0.52\) d. \(\quad n=15, \alpha=.05, \sigma=0.02, \mu=0.54\) e. \(n=15, \alpha=.05, \sigma=0.04, \mu=0.54\) f. \(\quad n=20, \alpha=.05, \sigma=0.04, \mu=0.54\) g. Is the way in which \(\beta\) changes as \(n, \alpha, \sigma,\) and \(\mu\) vary consistent with your intuition? Explain.

The paper "Playing Active Video Games Increases Energy Expenditure in Children" (Pediatrics [2009]: \(534-539\) ) describes an interesting investigation of the possible cardiovascular benefits of active video games. Mean heart rate for healthy boys age 10 to 13 after walking on a treadmill at \(2.6 \mathrm{~km} /\) hour for 6 minutes is 98 beats per minute (bpm). For each of 14 boys, heart rate was measured after 15 minutes of playing Wii Bowling. The resulting sample mean and standard deviation were 101 bpm and 15 bpm, respectively. For purposes of this exercise, assume that it is reasonable to regard the sample of boys as representative of boys age 10 to 13 and that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. a. Does the sample provide convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is different from the known mean heart rate after 6 minutes walking on the treadmill? Carry out a hypothesis test using \(\alpha=.01\). b. The known resting mean heart rate for boys in this age group is \(66 \mathrm{bpm}\). Is there convincing evidence that the mean heart rate after Wii Bowling for 15 minutes is higher than the known mean resting heart rate for boys of this age? Use \(\alpha=.01\). c. Based on the outcomes of the tests in Parts (a) and (b), write a paragraph comparing the benefits of treadmill walking and Wii Bowling in terms of raising heart rate over the resting heart rate.

"Most Like it Hot" is the title of a press release issued by the Pew Research Center (March 18, 2009 , www.pewsocialtrends.org). The press release states that "by an overwhelming margin, Americans want to live in a sunny place." This statement is based on data from a nationally representative sample of 2260 adult Americans. Of those surveyed, 1288 indicated that they would prefer to live in a hot climate rather than a cold climate. Do the sample data provide convincing evidence that a majority of all adult Americans prefer a hot climate over a cold climate? Use the nine-step hypothesis testing process with \(\alpha=.01\) to answer this question.

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