In a recent international poll, nine of 48 respondents rated the likelihood of a

Question

In a recent international poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as "likely" or "very likely." Use the "Plus Four" method to create a 96% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. (Round your answers to two decimal places.) 0.07x 03035x Explain what this confidence interval means in the context of the problem. ( There is a 96% chance that an American adult believes that a terrorist attack in their community is likely or very likely We are 96% confident that the proportion of American adults in the sample of 48 adults that believe that a terrorist attack in their community is likely or very likely lies within this interval we are 96% confident that the true proportion of American adults that believe that a terrorist attack in their community is likely or very likely lies within this interval. o we are 96% confident that an American adult believes that a terrorist attack in their community is likely or very likely

Explanation / Answer

Proportion of American adults who believe that a terrorist attack is likely or very likely = (Y + 2)/(n + 4)

where Y = 9 and n = 48

Proportion of American adults who believe that a terrorist attack is likely or very likely p^ = (9 + 2)/ (48 + 4) = 0.2115

96% confidence interval = p^ +- Z96% sqrt [p^ * (1-p^)/ (N+4)]

= 0.2115 +- 2.055 * sqrt [ 0.2115 * 0.7885 /52]

= 0.2115 +- 2.055 * 0.0566

= (0.095 ,0.328)

Option C is correct that we are 96% confident that the true proportion of American adults that a terrorist attack in their community is likely and very likely.

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