/*! 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 57 Do You Know Your Neighbors? A su... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Do You Know Your Neighbors? A survey of 2255 randomly selected US adults found that \(51 \%\) said they know all or most of their neighbors. \({ }^{17}\) Does this provide evidence that more than half of US adults know most or all of their neighbors?

Short Answer

Expert verified
The determination of this will be based on whether the z-score of the statistic is greater than the critical z-value, to decide whether to reject or fail to reject the null hypothesis. The actual answer will depend on the computed z-score from the mathematical calculation.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) is that half (or 0.50) of US adults know most or all of their neighbors. This can be expressed as \(P_0 = 0.50\). The alternative hypothesis \(H_a\) is that more than half (or greater than 0.50) of US adults know most or all of their neighbors. This can be written as \(P > 0.50\) or \(!H_0\).
02

Compute the Test Statistic

We'll use the formula for the test statistic \(Z = \frac{P - P_0}{\sqrt{\frac{P_0(1- P_0)}{n}}}\). Here, \(P = 0.51\) is the sample proportion, \(P_0 = 0.50\) is the hypothesized population proportion, and \(n = 2255\) is the sample size. By substituting these values into the formula, we'll get the value of the test statistic.
03

Determine the Critical Value and Make a Decision

We have to decide our significance level. Assuming a significance level (α) of 0.05 and a one-tailed test, the critical z-value is 1.645. This is found in the z-table or using a calculator. If the test statistic is greater than this critical value, we will reject the null hypothesis.
04

Interpret the Results

If the null hypothesis is rejected, it suggests that there is enough evidence to say that more than 50 percent of US adults know most or all of their neighbors. This needs to be determined after the calculations are made.

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
The null hypothesis, often denoted as \(H_0\), is a fundamental concept in hypothesis testing. It's a statement that there is no effect or no difference, and it's what you aim to test against. In many cases, it represents a baseline or default position. In statistical terms, the null hypothesis is a statement that we assume to be true unless we have evidence to prove otherwise.

In the context of the exercise, the null hypothesis proposed is that exactly half, or 50%, of US adults know most or all of their neighbors. This can be mathematically expressed as \( P_0 = 0.50 \). The challenge here is to determine whether the data provides sufficient evidence to reject this assumption.

Formulating the null hypothesis is usually the first step in any hypothesis testing process. Once the null hypothesis is established, the researcher proceeds to collect and analyze data to see if there is enough evidence to reject \(H_0\). If rejected, it brings into question the assumption and suggests that an alternative hypothesis might hold true instead.
Alternative Hypothesis
The alternative hypothesis, represented as \(H_a\), is opposed to the null hypothesis. It suggests that there is a statistically significant effect or difference. In hypothesis testing, if the evidence strongly suggests that the null hypothesis does not hold, the alternative hypothesis is considered.

For the survey about US adults knowing their neighbors, the alternative hypothesis is that more than 50% of them know most or all of their neighbors. This is indicated by \(P > 0.50\). Essentially, \(H_a\) posits that the true proportion of adults knowing their neighbors is greater than the hypothesized 50%.

Alternative hypotheses can be of different types:
  • Two-tailed: tests for any significant difference, either higher or lower.
  • One-tailed: tests only for an increase or only for a decrease.
In this case, a one-tailed test is appropriate as the focus is on whether more than half of adults know their neighbors. The goal is to gather enough statistical proof against the null hypothesis so that the alternative hypothesis can be accepted.
Z-test
The Z-test is a statistical method used to determine whether there is a significant difference between sample and population means, proportions, or other such measures. It is applicable when the sample size is large, typically over 30.

For this exercise, we use a Z-test to compare the sample proportion of people knowing their neighbors to the population proportion. The formula for the Z-test is:\[Z = \frac{P - P_0}{\sqrt{\frac{P_0(1- P_0)}{n}}} \]Where:
  • \(P\) is the sample proportion.
  • \(P_0\) is the hypothesized population proportion (null hypothesis).
  • \(n\) is the sample size.
In this example, the formula evaluates whether the observed proportion (51%) of US adults knowing their neighbors is statistically significantly different from the assumed population proportion (50%).

After calculating the Z-value, we compare it with a critical value from the standard normal distribution table to make decisions. If the calculated Z-value exceeds the critical value, the null hypothesis is rejected, suggesting the sample provides enough evidence to support the alternative hypothesis.

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

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.05 .

As part of the same study described in Exercise 6.254 , the researchers also were interested in whether babies preferred singing or speech. Forty-eight of the original fifty infants were exposed to both singing and speech by the same woman. Interest was again measured by the amount of time the baby looked at the woman while she made noise. In this case the mean time while speaking was 66.97 with a standard deviation of \(43.42,\) and the mean for singing was 56.58 with a standard deviation of 31.57 seconds. The mean of the differences was 10.39 more seconds for the speaking treatment with a standard deviation of 55.37 seconds. Perform the appropriate test to determine if this is sufficient evidence to conclude that babies have a preference (either way) between speaking and singing.

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

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.