/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 According to a New York Times/CB... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to a New York Times/CBS News poll conducted during June \(24-28,2011,55 \%\) of the American adults polled said that owning a home is a very important part of the American Dream (The New York Times, June 30,2011 ). Suppose this result was true for the population of all American adults in \(2011 .\) In a recent poll of 1800 American adults, \(61 \%\) said that owning a home is a very important part of the American Dream. Perform a hypothesis test to determine whether it is reasonable to conclude that the percentage of all American adults who currently hold this opinion is higher than \(55 \%\). Use a \(2 \%\) significance level, and use both the \(p\) -value and the critical-value approaches.

Short Answer

Expert verified
First, set up the null and alternative hypotheses. Then perform a hypothesis test using the test statistic (Z-score) formula. Find the p-value for this Z-score and compare it to the given significance level to make a conclusion. Also determine the critical value from the Z-distribution table for your significance level and compare the Z-score to this value to make a conclusion. Following these steps will lead to the accurate interpretation of the task.

Step by step solution

01

Identify Null and Alternative Hypotheses

The null hypothesis is that the population proportion (\( p \)) is equal to \( 0.55 \) or \( 55\% \), i.e., \( H_0: p = 0.55 \). The alternative hypothesis is that the population proportion is greater than \( 0.55 \) or \( 55\% \), i.e., \( H_1: p > 0.55 \).
02

Conduct Hypothesis Test

The test statistic for a hypothesis test for a proportion is a Z-score. This can be calculated using the formula: \( Z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \), where \( \hat{p} = \frac {x}{n} \) is the sample proportion, \( p_0 \) is the proportion under the null hypothesis and \( n \) is the sample size. Plugging in our values, we get \( Z = \frac{(0.61 - 0.55)}{\sqrt{\frac{0.55(1 - 0.55)}{1800}}} \). Now, we can look up this Z-score in the Z-table (or use statistical software) to find the corresponding p-value.
03

Determine the Critical Value

We can find the critical value from the Z-distribution table by looking up the value corresponding to a one-sided test with significance level \( \alpha = 0.02 \). Let's denote the critical value as \( Z_c \). If the calculated Z-score is greater than \( Z_c \), we reject the null hypothesis.
04

Make a Conclusion

There are two ways to make a conclusion. First, compare the p-value with the significance level \( \alpha = 0.02 \), if the p-value is less than \( \alpha \), we reject the null hypothesis in favour of the alternative. Second, compare the calculated Z-score to the critical value, if the Z-score is greater than the critical value, we again reject the null hypothesis. Thus, if either condition is met, it is reasonable to conclude that the percentage of all American adults who currently hold this opinion is higher than \( 55\% \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is a fundamental concept. It's represented by a statement asserting that there is no effect or no difference, and it serves as the starting point for statistical tests. In our exercise, the null hypothesis (\(H_0\)) is that the proportion of American adults who believe that owning a home is very important is exactly 55%.
This means we assume there has been no change in this belief over time. Initially, we accept this hypothesis unless there's strong evidence against it.
  • The null hypothesis is commonly expressed with an equation, such as \(H_0: p = 0.55\).
  • It's like saying, "Let's assume the previous belief holds until proven otherwise."
This assumption helps form a baseline from where any statistical analysis begins, waiting to see if the data significantly deviates from it to warrant alternative conclusions.
Alternative Hypothesis
The alternative hypothesis stands in contrast to the null hypothesis and is what you would believe if your evidence shows strong support against the null. In our context, the alternative hypothesis (\(H_1\)) suggests that the proportion of American adults who think owning a home is very important is more than 55%.
The alternative hypothesis is expressed as \(H_1: p > 0.55\). This representation of more than 55% indicates that we are specifically looking for evidence that exceeds the previous standard.
  • The alternative hypothesis is the change or "something different" we're testing for.
  • It represents a new theory or observation that suggests a deviation from the past norm.
If our data strongly supports this hypothesis, we may reject the null hypothesis in favor of the alternative, suggesting a change in public opinion.
Significance Level
The significance level in hypothesis testing is essential as it dictates the threshold for statistical decision-making. This threshold is denoted by the Greek letter \(\alpha\) and represents the probability of rejecting the null hypothesis when it is actually true.
In our exercise, we use a significance level of 2%, or \(\alpha = 0.02\), which means we are willing to accept a 2% chance of a wrong rejection of \(H_0\).
  • The lower the significance level, the stricter the criterion for rejecting the null hypothesis.
  • It's like setting a rule where you want high confidence in your findings.
The significance level informs us about how robust our claims must be before considering them statistically significant. It helps in ensuring the conclusion is not due to random variation but due to actual evidence.
Z-score
A Z-score is a type of test statistic that helps in determining how far away a sample statistic lies from the null hypothesis value. It measures the number of standard deviations a point is from the mean population value under testing.
In this scenario, the Z-score formula used with proportions is:
\[Z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]
where \(\hat{p}\) is the sample proportion, \(p_0\) is the null hypothesis proportion, and \(n\) is the sample size.
  • A higher magnitude of Z-score indicates more deviation from what was expected under \(H_0\).
  • Statistical software or Z-tables determine how "extreme" this score is by showing us the corresponding probability (p-value).
If the calculated Z-score is higher than the critical value from Z-distribution at our set significance level, it signals strong evidence against the null hypothesis. This helps in deciding whether the real-world observation significantly departs from previous beliefs.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two years ago, \(75 \%\) of the customers of a bank said that they were satisfied with the services provided by the bank. The manager of the bank wants to know if this percentage of satisfied customers has changed since then. She assigns this responsibility to you. Briefly explain how you would conduct such a tes

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? Comment on the results of parts a and \(\mathrm{b}\).

Consider the null hypothesis \(H_{0}=\mu=100\). Suppose that a random sample of 35 observations is taken from this population to perform this test. Using a significance level of \(.01\), show the rejection and nonrejection regions and find the critical value(s) of \(t\) when the alternative hypothesis is as follows. a. \(H_{1}: \mu \neq 100 \quad\) b. \(H_{1}: \mu>100 \quad\) c. \(H_{1}: \mu<100\)

In each of the following cases, do you think the sample size is large enough to use the normal distribution to make a test of hypothesis about the population proportion? Explain why or why not. a. \(n=40\) and \(p=.11\) b. \(n=100\) and \(p=.7\) c. \(n=80 \quad\) and \(\quad p=.05\) d. \(n=50\) and \(p=.14\)

Consider \(H_{0}=p=.70\) versus \(H_{1}: p \neq .70\). a. A random sample of 600 observations produced a sample proportion equal to .68. Using \(\alpha=.01\), would you reject the null hypothesis? b. Another random sample of 600 observations taken from the same population produced a sample proportion equal to. \(76 .\) Using \(\alpha=.01\), would you reject the null hypothesis? Comment on the results of parts a and \(\mathrm{b}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.