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91Ó°ÊÓ

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 \(\mathrm{OK}\) 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
The 90% confidence interval for the proportion of adult Americans who think lying is never justified is approximately (0.4937, 0.5463), while the confidence interval for those who think it is often or sometimes OK to lie to avoid hurting someone's feelings is about (0.6231, 0.6769). These intervals do not overlap, indicating an inconsistency in responses. This suggests that some participants might believe lying is never justified, but in specific situations, like avoiding hurting someone's feelings, they think lying can be justified, or that the questions were interpreted differently.

Step by step solution

01

Confidence interval formula for proportions

The confidence interval for a proportion is given by the formula: \[p \pm Z * \sqrt{\frac{p(1 - p)}{n}}\] Here, - \(p\) is the sample proportion - \(n\) is the number of observations (sample size) - \(Z\) is the Z-score corresponding to the desired level of confidence We will first find the proportion for each situation, and then calculate the 90% confidence interval for each. #a.# Construct and interpret a 90% confidence interval for the proportion of adult Americans who would say that lying is never justified.
02

Find the proportion

We are given that 52% of the survey respondents said that lying is never justified. In our sample of 1000 participants, we have: \(p = 0.52\)
03

Find the Z-score

For a 90% confidence interval, the Z-value is \(1.645\).
04

Calculate the confidence interval

Now, using the confidence interval formula, we have the following: \[CI = 0.52 \pm 1.645 * \sqrt{\frac{0.52(1 - 0.52)}{1000}}\] Calculating the margin of error: \[1.645 * \sqrt{\frac{0.52(0.48)}{1000}} \approx 0.0263\] Subtracting and adding the margin of error to the sample proportion: \[0.52 - 0.0263 = 0.4937\] \[0.52 + 0.0263 = 0.5463\] Therefore, the 90% confidence interval for the proportion of adult Americans who would say that lying is never justified is approximately \((0.4937, 0.5463)\). #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.
05

Find the proportion

We are given that 650 respondents said that it is often or sometimes OK to lie to avoid hurting someone's feelings. In our sample of 1000 participants, we have: \(p = 0.65\)
06

Find the Z-score

For a 90% confidence interval, the Z-values stay the same, and is \(1.645\).
07

Calculate the confidence interval

Now, using the confidence interval formula, we have the following: \[CI = 0.65 \pm 1.645 * \sqrt{\frac{0.65(1 - 0.65)}{1000}}\] Calculating the margin of error: \[1.645 * \sqrt{\frac{0.65(0.35)}{1000}} \approx 0.0269\] Subtracting and adding the margin of error to the sample proportion: \[0.65 - 0.0269 = 0.6231\] \[0.65 + 0.0269 = 0.6769\] Therefore, the 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 is approximately \((0.6231, 0.6769)\). #c.# Using the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses.
08

Compare the confidence intervals

The 90% confidence interval for those who think lying is never justified is \((0.4937, 0.5463)\), while the 90% confidence interval for those who think it is often or sometimes OK to lie to avoid hurting someone's feelings is \((0.6231, 0.6769)\). The two confidence intervals do not overlap, indicating that there is an apparent inconsistency in the responses. This suggests that some respondents might believe lying is never justified in general, but in specific situations, such as avoiding hurting someone's feelings, they believe lying can be justified. Another interpretation could be that respondents interpreted the questions differently, leading to different proportions for these two situations.

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Key Concepts

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

Proportion
The concept of proportion in statistics is a fundamental building block when we're dealing with data involving fractions or percentages of a total. A proportion reflects the part of the whole that shares a certain characteristic. For example, in the original exercise, one proportion is the participants who think lying is never justified. If we have a group of 1000 people, and 520 of them believe that lying is never justified, then the proportion, represented as a decimal or percentage, is 0.52 or 52%, respectively. Proportions help us to understand the composition of a group and make predictions about what might hold true for the entire population. They are a snapshot of our sample's opinion and can also be used to make inferences about larger populations from which the sample is drawn. Always remember, the larger the sample size, the more accurate the proportion is likely to represent the population.
Margin of Error
The margin of error is an essential concept when constructing confidence intervals. It gives us an idea of how much we can expect the sample proportion to differ from the true population proportion. It signifies the range within which the true proportion is likely to fall, given a certain level of confidence (like 90%, in this case).How is it calculated? It's derived from the standard error of the proportion, which accounts for sample variability. The formula for margin of error when dealing with proportions is:\[ MOE = Z * \sqrt{\frac{p(1-p)}{n}} \]Where:
  • Z: is the Z-score that corresponds to our desired confidence level.
  • p: is the sample proportion.
  • n: refers to the sample size.
The larger the margin of error, the wider our confidence interval, indicating less certainty about the true population parameter. It can be decreased by increasing the sample size or by choosing a lower confidence level, although doing so might sacrifice the reliability of our inferences.
Z-score
The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. In the context of confidence intervals, the Z-score is critical as it helps determine the margin of error. It is associated with the desired confidence level of an interval. For example, when constructing a 90% confidence interval, the Z-score is approximately 1.645. This score is derived from the standard normal distribution, which supplies us with values for different confidence levels:
  • 1.645: for 90% confidence
  • 1.96: for 95% confidence
  • 2.576: for 99% confidence
The Z-score adjusts the margin of error. A higher Z-score for a higher confidence level results in a broader confidence interval. Using the correct Z-score ensures that we accurately reflect the certainty or uncertainty of our confidence interval about the true population parameter.

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Most popular questions from this chapter

The formula used to calculate a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) b. \(98 \%\) c. \(85 \%\)

Will \(\hat{p}\) from a random sample from a population with \(60 \%\) successes tend to be closer to 0.6 for a sample size of \(n=400\) or a sample size of \(n=800 ?\) Provide an explanation for your choice.

Consider taking a random sample from a population with \(p=0.40\). a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be greater for samples of size 100 or samples of size \(200 ?\) c. If the sample size were doubled from 100 to 200 , by what factor would the standard error of \(\hat{p}\) decrease?

The report "Parents, Teens and Digital Monitoring" (Pew Research Center, January \(7,2016,\) www.pewinternet .org/2016/01/07/parents-teens-and-digital- monitoring, retrieved May 5,2017 ) reported that \(61 \%\) of parents of teens aged 13 to 17 said that they had checked which web sites their teens had visited. The \(61 \%\) figure was based on a representative sample of 1060 parents of teens in this age group. a. Use the given information to estimate the proportion of parents of teens age 13 to 17 who have checked which web sites their teen has visited. What statistic did you use? b. Use the sample data to estimate the standard error of \(\hat{p}\). c. Calculate and interpret the margin of error associated with the estimate in Part (a).

Consider taking a random sample from a population with \(p=0.70\) a. What is the standard error of \(\hat{p}\) for random samples of size \(100 ?\) b. Would the standard error of \(\hat{p}\) be smaller for samples of size 100 or samples of size \(400 ?\) c. Does decreasing the sample size by a factor of \(4,\) from 400 to \(100,\) result in a standard error of \(\hat{p}\) that is four times as large?

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