/*! 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 26 One thousand randomly selected a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

One thousand randomly selected adult Americans participated in a survey conducted by the Assodated Press (June 2006 ). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes okay. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes okay to lie to avoid hurting someone's feelings. c. Comment on the apparent inconsistency in the responses given by the individuals in this sample.

Short Answer

Expert verified
a. The 90% confidence interval for the proportion of adult Americans who think lying is never justified is (0.49, 0.55). b. The 90% confidence interval for the proportion of adult Americans who think that it is often or sometimes okay to lie to avoid hurting someone's feelings is (0.62, 0.68). c. There is an apparent inconsistency in the responses given by the individuals in this sample as at least 130 individuals hold inconsistent views.

Step by step solution

01

Construct the 90% confidence interval (a)

Step 1: Identify the sample size (n) and the sample proportion (\(\hat{p}\)). Here, n = 1000 and \(\hat{p}\) = 0.52. Step 2: Compute the z-value for the 90% confidence level. The z-value for 90% confidence is approximately 1.645. Step 3: Insert n, \(\hat{p}\), and z-value (1.645) into the confidence interval equation. The result is \(0.52 \pm 1.645 * \sqrt{\frac{0.52(1-0.52)}{1000}}\) which equals to \(0.52 \pm 0.03\) or the interval is (0.49, 0.55) this means we are 90% confident the population proportion lies in this interval.
02

Construct the 90% confidence interval (b)

Step 1: Identify the sample size (n) and sample proportion (\(\hat{p}\)). n = 1000 and \(\hat{p}\) = 650/1000 = 0.65. Step 2: Compute the z-value for the 90% confidence level. The z-value for 90% confidence is usually around 1.645. Step 3: Insert n, \(\hat{p}\), and z-value (1.645) into the confidence interval equation: \(0.65 \pm 1.645 * \sqrt{\frac{0.65(1-0.65)}{1000}}\) which simplifies to \(0.65 \pm 0.03\) or the interval is (0.62, 0.68). This means that we are 90% confident the population proportion lies in this interval.
03

Analyzing the apparent inconsistency in the responses

It's apparent that the responses are inconsistent. While 52% people believe that lying is never justified which is about \(0.52*1000 = 520\) people, 65% believe lying to avoid hurting someone's feelings is often or sometimes okay, which represents 650 people. This discrepancy shows that at least 170 individuals (650 - 520 = 130) hold inconsistent views. That is, they either did not consider lying to protect someone's feelings as 'lying' or they, inexplicably, changed their opinion within the period of the survey.

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.

Proportion of Adult Americans
The concept of a proportion in statistics represents the part of a whole. Specifically, when we refer to the proportion of adult Americans in this context, we're discussing the fraction or percentage of the adult American population that holds a certain opinion or behavior pattern. A survey that queries a sample can estimate this proportion, but it's important to remember that the exact proportion in the entire population is typically unknown. This is where confidence intervals come into play.

To illustrate, consider a survey result saying '52% of the sample believes lying is never justified.' This 52% is a sample proportion, denoted as \(\hat{p}\). It is calculated by dividing the number of individuals in the sample who have a certain trait by the total number of sampled individuals. In our case, if 1000 adult Americans were surveyed, and 520 said lying is never justified, \(\hat{p} = 0.52\). It's a piece of the puzzle, hinting at the potential stance of the broader population.
Survey Data Analysis
Survey data analysis is a process that involves collecting, processing, and interpreting information gathered from a group of individuals to generalize their responses to a larger population. This analysis can often include the computation of proportions and confidence intervals to make inferences about the population.

For example, when conducting a survey, one must consider the randomness of the sample selection to ensure that it's representative of the entire population. The accuracy of the analysis is affected by factors such as sample size and question phrasing. After data collection, statisticians interpret the results, such as calculating the proportion of respondents who hold a certain view, and then use statistical techniques, like confidence intervals, to estimate the range within which the true population proportion is likely to fall. A 90% confidence interval suggests that if we were to take 100 similar samples and compute an interval from each sample, we would expect about 90 of those intervals to contain the true population proportion.
Statistics in Decision Making
Statistics plays a crucial role in decision making in various fields, including policy, business, healthcare, and beyond. When decision-makers face uncertainty, statistical methods help quantify and manage risks. In the context of survey data, statistical findings inform not only the public but also those creating strategies, laws, or campaigns.

In the exercise provided, decision making could be based around the attitudes of adult Americans toward the morality of lying. The confidence intervals for the two related survey questions provide numerical ranges which help to understand the variability and potential error in the estimates. Decision-makers need to weigh these statistics along with other factors to make informed choices. The discrepancy observed between the percentages of participants who condone lying outright and those who accept lying to protect someone's feelings offers a nuanced understanding of public opinion that could influence how decisions are made regarding this topic.

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

USA Today (October 14, 2002) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Assuming a \(95 \%\) confidence level, do you agree with the reported bound on the error? Explain.

Given a variable that has a \(t\) distribution with the specified degrees of freedom, what percentage of the time will its value fall in the indicated region? a. \(10 \mathrm{df}\), between -1.81 and 1.81 b. \(10 \mathrm{df}\), between -2.23 and 2.23 c. 24 df, between -2.06 and 2.06 d. \(24 \mathrm{df}\), between -2.80 and 2.80 e. 24 df, outside the interval from -2.80 to 2.80 f. \(24 \mathrm{df}\), to the right of 2.80 g. \(10 \mathrm{df}\), to the left of -1.81

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006\()\) reported that \(37 \%\) of college freshmen and \(48 \%\) of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a \(90 \%\) confidence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two \(90 \%\) confidence intervals from Parts (a) and (b) are not the same width.

The article "Viewers Speak Out Against Reality TV (Associated Press, September 12. 2005\()\) included the following statement: "Few people believe there's much reality in reality TV: a total of \(82 \%\) said the shows are either "totally made up' or 'mostly distorted'." This statement was based on a survey of 1002 randomly selected adults. Compute and interpret a bound on the error of estimation for the reported percentage.

The article "Most Canadians Plan to Buy Treats, Many Will Buy Pumpkins, Decorations and/or Costumes" (Ipsos-Reid, October 24, 2005 ) summarized results from a survey of 1000 randomly selected Canadian residents. Each individual in the sample was asked how much he or she anticipated spending on Halloween during \(2005 .\) The resulting sample mean and standard deviation were \(\$ 46.65\) and \(\$ 83.70,\) respectively. a. Explain how it could be possible for the standard deviation of the anticipated Halloween expense to be larger than the mean anticipated expense. b. Is it reasonable to think that the distribution of the yariable anticipated Halloween expense is approximately normal? Explain why or why not. c. Is it appropriate to use the \(t\) confidence interval to estimate the mean anticipated Halloween expense for Canadian residents? Explain why or why not. d. If appropriate, construct and interpret a \(99 \%\) confidence interval for the mean anticipated Halloween exnense for Canadian residents.

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.