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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 and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would say that lying is never justified. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Using the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses.

Short Answer

Expert verified
a. The 90% confidence interval for the proportion who think lying is never justified is approximately (\(0.49, 0.55\)). b. The 90% confidence interval for the proportion who believe it's sometimes okay to lie to preserve feelings is approximately (\(0.62, 0.68\)). c. Both intervals indicate the majority of respondents, but they do not overlap, suggesting some inconsistency in views on lying. However, this apparent inconsistency could be due to the different contexts of the questions.

Step by step solution

01

Calculate the Confidence Interval for those who think lying is never justified

First, calculate the standard error for the observed proportion (0.52) using the formula \(\sqrt{p(1-p)/n}\). Then, to calculate the 90% confidence interval, take the observed proportion (0.52) and add/subtract the product of z-score (1.645) and the standard error.
02

Calculate the Confidence Interval for those who think it's okay to lie to avoid hurting feelings

Similar to step 1, calculate the proportion who believe it's okay to lie to avoid hurting feelings, which can be found by dividing the number who responded this way (650) by the total number of respondents (1000). Then calculate the standard error for this proportion and calculate the confidence interval using this proportion, the z-score, and the standard error.
03

Compare and comment on the two confidence intervals

Once both confidence intervals are calculated, they can be compared for any inconsistency. For example, if the confidence intervals overlap, it could suggest that the same people who think lying is never justified are also saying it's okay to lie to preserve feelings. Discussion of this possible inconsistency between views would form the final part of this problem.

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Key Concepts

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

Proportion Estimation
When conducting a survey or an experiment involving a large population, it's impossible to reach every individual. That's where proportion estimation comes into play. This statistical method helps researchers estimate the fraction of a population that holds a certain view or characteristic based on a sample. For instance, in our exercise, a sample of 1,000 Americans was used to estimate the proportion of the entire adult population who believe lying is never justified.

Proportion estimation is particularly effective when the sample is random and representative of the whole population. It uses the formula \( \hat{p} = \frac{x}{n} \), where \( \hat{p} \) is the estimated proportion, \( x \) is the number of individuals in the sample with the characteristic, and \( n \) is the total sample size. With this method, we estimated that 52% of the sample, and by extension of adult Americans, believe that lying is never justified.

In our exercise, the proportion of people thinking lying is sometimes justified was also calculated using the responses from the sample.
Standard Error Calculation
In order to assess the precision of our proportion estimates, the concept of standard error is utilized. It quantifies the variability of the proportion estimate due to sampling error and indicates how far the estimate might be from the true population proportion. We use the formula \( SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \) to calculate the standard error, where \( \hat{p} \) is the estimated proportion and \( n \) is the sample size.

This calculation acknowledges that a sample may not perfectly reflect the population. When applying this to our example, we can compute the standard error of the proportion of American adults who believe lying is never justified, providing a measure of confidence in the estimate made from the survey data.
Z-Score Application
A z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. In the context of confidence intervals, a z-score enables us to define the margin of error around a sample proportion estimate. Using standard z-scores from the normal distribution, we can create confidence intervals of different levels — 90%, 95%, or 99% are common.

For a 90% confidence interval, the z-score is typically 1.645, meaning that we are confident that the true population proportion falls within our calculated range 90% of the time. In our survey analysis example, by multiplying this z-score with the standard error, we can determine the range in which we expect the true population proportion will fall with 90% confidence.
Survey Data Analysis
When analyzing survey data, we aim to draw conclusions about a population based on a sample. This process involves several steps, including collecting data, estimating proportions, calculating standard errors, and applying z-scores to generate confidence intervals. The analysis also often requires reconciling seemingly contradictory results.

In our exercise, once the data was collected from 1,000 Americans about their views on lying, we estimated proportions of those who think lying is never justified and those who think it's okay in certain circumstances. By creating confidence intervals for these proportions, we were able to interpret the range within which the true values for the entire population likely fall. Finally, we were tasked to comment on any inconsistencies found between the two viewpoints, suggesting nuances in how individuals consider the moral implications of lying.

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Most popular questions from this chapter

An Associated Press article on potential violent behavior reported the results of a survey of 750 workers who were employed full time (San Luis Obispo Tribune, September 7 , 1999). Of those surveyed, 125 indicated that they were so angered by a coworker during the past year that they felt like hitting the coworker (but didn't). Assuming that it is reasonable to regard this sample as representative of the population of full-time workers, use this information to construct and interpret a \(90 \%\) confidence interval estimate of \(p,\) the proportion of all full-time workers so angered in the last year that they wanted to hit a coworker.

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It probably wouldn't surprise you to know that Valentine's Day means big business for florists, jewelry stores, and restaurants. But did you know that it is also a big day for pet stores? In January \(2010,\) the National Retail Federation conducted a survey of consumers in a representative sample of adult Americans ("This Valentine's Day, Couples Cut Back on Gifts to Each Other, According to NRF Survey," www.nrf.com). One of the questions in the survey asked if the respondent planned to spend money on a Valentine's Day gift for his or her pet. a. The proportion who responded that they did plan to purchase a gift for their pet was 0.173 . Suppose that the sample size for this survey was \(n=200\). Construct and interpret a \(95 \%\) confidence interval for the proportion of all adult Americans who planned to purchase a Valentine's Day gift for their pet. b. The actual sample size for the survey was much larger than 200\. Would a \(95 \%\) confidence interval calculated using the actual sample size have been narrower or wider than the confidence interval calculated in Part (a)?

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