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Chapter 10: Problem 62

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
a. No, based on a one-sample Z-test with level .05, it would not be statistically significant to conclude that more than 75% of all apartments exclude children. b. The power of this test is about 0.89 when if the real proportion is 0.8.

Step by step solution

01

Specify the Null and Alternative Hypothesis

The null hypothesis, \(H_0: p = 0.75\), suggests that 75% or less apartments exclude children. The alternative hypothesis, \(H_1: p > 0.75\), implies that more than 75% of the apartments exclude children.
02

Conduct Level .05 Z-test

A Z-test is conducted. The sample proportion \(\hat{p}\) equals 102 out of 125, or 0.816. The standard error of the proportion is calculated by \(SE = \sqrt{p(1-p)/n} = \sqrt{0.75 * (1 - 0.75) / 125} = 0.0408\). The Z-score is calculated by ( 0.816 - 0.75) / 0.0408 = 1.6172. From the Z-table, the one-tailed probability associated with a Z-score of 1.6172 is approximately 0.053, which is greater than the significance level of 0.05. Therefore, we fail to reject the null hypothesis. The evidence suggests that 75% or less apartments exclude children.
03

Calculate the power of test

The power of the test is the probability that the test correctly rejects the Null hypothesis if the real proportion is indeed 0.8. The power can be calculated by finding the Z-score using the alternate proportion and the standard error, the same way we calculated for our test. Here, the Z-score for the alternate proportion (0.8) is \(Z_{0.8} = (0.8 - 0.75) / 0.0408 = 1.2255\). The power of the test is then the probability that we observe a Z-score greater than this score. From the Z-table, the one-tailed probability associated with a Z-score of 1.2255 is about 0.1103. This means that there is approximately an 11% chance that we would reject the Null hypothesis even if the true proportion is 0.8. Therefore, the power of the test is 1 - 0.1103 = 0.8897 or approximately 0.89.

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Key Concepts

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

Proportion Test
A proportion test is a statistical method used to determine if there is a significant difference between a sample proportion and a known or hypothesized population proportion. It's a common tool in hypothesis testing.
In the example exercise, the city council aims to determine if more than 75% of apartments exclude children. The test focuses on comparing the sample proportion, which in this case is 102 out of 125 units, or 0.816, to the hypothesized population proportion of 0.75.

Key points of a Proportion Test:
  • Used for categorical data divided into two outcomes (e.g., includes children vs. excludes children).
  • Helps in understanding if the observed difference is due to a random chance or if it's statistically significant.
  • Involves calculating a test statistic to measure the extent of deviation from the hypothesized proportion.
Z-Test
The Z-test is a statistical tool used to determine if there is a significant difference between a sample mean or proportion and a population mean or proportion. In this case, it's used to test proportions.
For the exercise at hand, a Z-test is conducted to examine whether the proportion of apartments excluding children significantly exceeds 75%. The Z-test involves calculating a Z-score, which indicates how many standard deviations an element is from the mean.

The Z-score calculation involves:
  • Calculating the sample proportion (\( \hat{p} = \frac{102}{125} = 0.816 \)).
  • Determining the standard error (\( SE = \sqrt{0.75 \times (1 - 0.75) / 125} = 0.0408 \)).
  • Computing the Z-score (\( Z = \frac{0.816 - 0.75}{0.0408} = 1.6172 \)).
By comparing this Z-score with a critical value from the Z-table, one can decide whether to reject or fail to reject the null hypothesis.
Power of a Test
The power of a test is the probability that it correctly rejects a false null hypothesis. It's an essential aspect of hypothesis testing, as it indicates the test's ability to detect an effect when there is one.
In the given exercise, we evaluate the power of the test at a true proportion (\( p = 0.8 \)) and significance level (\( \alpha = 0.05 \)). A higher power means a greater chance of detecting a true effect.
  • The Z-score for the hypothesized higher proportion (\( Z_{0.8} = \frac{0.8 - 0.75}{0.0408} = 1.2255 \)).
  • The probability associated with this Z-score is approximately 0.1103, representing the chance of Type II error (failing to reject the null when false).
  • The power of the test is calculated as 1 minus this probability (1 - 0.1103 = 0.8897), implying a roughly 89% likelihood of correctly identifying that more than 75% of apartments exclude children if that's the true state.
Understanding power helps in planning studies to ensure that they are capable of detecting meaningful effects.

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Most popular questions from this chapter

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