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One thousand randomly selected adult Americans participated in a survey conducted by the Associated 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 OK. 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 OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the re-

Short Answer

Expert verified
The first confidence interval is for the proportion of adults who think lying is never justified and the second confidence interval is for the proportion who think lying to avoid hurting someone is justified. Part c would involve interpreting the overlap or disjointedness of these intervals, and any apparent inconsistency therein.

Step by step solution

01

Part A Step 1: Identify Given Values

Several values can be obtained from the problem. The number of randomly selected adult Americans included in the survey is 1000. The count or number of successes for the first question is $52\%$ of 1000, which is 520.
02

Part A Step 2: Calculate the Margin of Error

The formula to calculate the margin of error is \(E = Z \cdot \sqrt{\frac{{p(1-p)}}{n}}\). Using the z-value for a \(90 \%\) confidence interval (which is approximately 1.645), \(p = 0.52\) (the proportion of success) and \(n = 1000\) (total number of trials), we can find E.
03

Part A Step 3: Construct the Confidence Interval

The confidence interval is given by (p - E, p + E). Calculate this to obtain the interval.
04

Part B Step 1: Identify the New Given Values

For the second part of the exercise, 650 out of 1000 adults think it's okay to lie to avoid hurting someone's feelings. So, the new \(p = \frac{650}{1000} = 0.65\).
05

Part B Step 2: Calculate the Margin of Error Again

Use the same formula and process from Part A Step 2 but use this new \(p\) value to calculate the new margin of error.
06

Part B Step 3: Construct the Confidence Interval

Use the same step as Part A Step 3 to construct the confidence interval for the second part of the problem.
07

Part C: Compare the Confidence Intervals and Offer Explanation

Look at the two calculated confidence intervals and compare them. Offer an explanation based on the relationship, difference and contradictions between them.

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Key Concepts

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

Margin of Error
When analyzing survey data, the margin of error is crucial as it provides an estimation of the uncertainty in the sampling process. The margin of error measures how much the survey's results can vary and still be considered accurate within a specific confidence level. For example, in our exercise, we aim for a 90% confidence level. This confidence level indicates that if we conducted the survey 100 times, 90 of those times, the results would fall within the calculated interval.

To calculate the margin of error (\(E\)), we use the formula: \[E = Z \cdot \sqrt{\frac{{p(1-p)}}{n}}\] where \(Z\) is the Z-score corresponding to our confidence level, \(p\) is the sample proportion, and \(n\) is the sample size. In our example, with \(p = 0.52\) and \(n = 1000\), we calculate \(E\) for part A using a Z-score of approximately 1.645.

The margin of error gives us the "wiggle room" around the percentage error, helping in creating confidence intervals that provide a range within which the true proportion lies.
Proportion
The concept of proportion in survey analysis refers to the fraction or percentage of the population that exhibits a certain characteristic. In our exercise, there are two proportions. The first is the proportion of adult Americans who believe that lying is never justified. This proportion is given as 0.52, meaning 52% of respondents think lying is never justified.

The second proportion is about those who find it acceptable to lie to avoid hurting feelings, which is 0.65. From a sample of 1000 respondents, 650 agreed to this, providing us with the proportion. Calculating proportions helps determine how a certain belief or behavior is spread among a population.

In statistical terms, once we've identified the proportions, they serve as the basis for constructing confidence intervals. These intervals allow us to estimate the proportion's range within the entire population while considering the calculated margin of error.

Proportions are critical in giving us an initial snapshot of survey outcomes and are fundamental in broader survey analysis processes.
Survey Analysis
Survey analysis is the process of interpreting the data collected from a survey to understand the views, beliefs, or behaviors of a group. The key to effective survey analysis is not just collecting data, but also understanding what this data tells us about the population.

In our exercise, we start with responses to two separate questions dealing with the justifiability of lying. By analyzing these survey results, we calculate proportions, margins of error, and construct confidence intervals:
  • The consistency in responses can be tested using these statistical tools.
  • An apparent inconsistency might arise if overlapping between confidence intervals occurs or if one interval is wholly contained within another.
This analysis sheds light on potential contradictions or insights within public opinions, such as why some people find lying acceptable to spare feelings while another group holds a stricter viewpoint.

Ultimately, survey analysis allows us to summarize the data effectively. It gives us a more profound understanding of underlying trends and helps identify significant patterns or puzzling inconsistencies in the data.

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Most popular questions from this chapter

The formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

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 percent 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.

would result in a wider large-sample cont?dence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

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Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were as follows: \(\begin{array}{lllll}6 & 17 & 11 & 22 & 29\end{array}\) Assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a \(95 \%\) confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program.

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