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Chapter 9: Problem 26

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 adult Americans who think lying is never justified is based on the calculation from step 2.\n b. The \(90 \% \) confidence interval for adult Americans who think it is okay to lie to avoid hurting feelings is based on the calculation from step 3.\n c. The inconsistency lies in the overlapping of the two confidence intervals as analysed in step 4.

Step by step solution

01

Analyse the Given Problem

There are 1000 randomly selected adults who participated in the survey. Out of them, \(52 \%\) think that lying is never justified and 650 individuals think it's okay to lie to avoid hurting someone's feelings. We need to construct a \(90 \%\) confidence interval for both these proportions.
02

Calculate Confidence Interval for Proportion of people who think lying is never justified

The formula to calculate confidence interval is \(p 卤 z \times \sqrt{{p \times (1-p)}/{n}}\). Here, 'n' is total number of participants (1000), 'p' is proportion of participants that think lying is never justified (0.52), and 'z' is the Z-score associated with \(90 \% \) confidence level which is 1.645. By substituting these values into the equation the interval can be calculated.
03

Calculate Confidence Interval for Proportion of people who think it's okay to lie to avoid hurting someone's feelings

We need to use the same confidence interval formula as in the step 2. Here, 'n' is total number of participants (1000) and 'p' is proportion of participants that think it's okay to lie to avoid hurting feelings (650/1000=0.65). By substituting these values into the formula the interval can be calculated.
04

Comment on the apparent inconsistency in the responses

For this step, looking at the two calculated intervals, it can be analysed that there is an overlap between the two intervals. Meaning, there are Americans who think lying is never justified but also believe it is okay to lie to avoid hurting someone's feelings, thus reflecting inconsistency in the responses.

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Most popular questions from this chapter

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