Find a sample size for a confidence interval, SD not known : 41% of Jacksonville residents said that they had been in a hurricane. How many adults should be surveyed to estimate the true proportion of adults who have been in a hurricane, with a 95% confidence interval 6% wide?
Answer
Figure out the following variables:
z a/2 = z-score(CI/2 .95 / 2 = .475) = z-score for .475 is 1.96
E (margin of error) = width/2 = 6% / 2 = .06 / 2 = .03
p = 41% = .41
q = 1-p = 1 - .41 = .59
Multiply p by q = .41 × .59 = .2419
Sample size = (z a/2 : E)^2 x (p x q)
Question
Find a sample size for a confidence interval, SD not known : 41% of Jacksonville residents said that they had been in a hurricane. How many adults should be surveyed to estimate the true proportion of adults who have been in a hurricane, with a 95% confidence interval 6% wide?
Answer
?
Question
Find a sample size for a confidence interval, SD not known : 41% of Jacksonville residents said that they had been in a hurricane. How many adults should be surveyed to estimate the true proportion of adults who have been in a hurricane, with a 95% confidence interval 6% wide?
Answer
Figure out the following variables:
z a/2 = z-score(CI/2 .95 / 2 = .475) = z-score for .475 is 1.96
E (margin of error) = width/2 = 6% / 2 = .06 / 2 = .03
p = 41% = .41
q = 1-p = 1 - .41 = .59
Multiply p by q = .41 × .59 = .2419
Sample size = (z a/2 : E)^2 x (p x q)
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owner: naxplast06 - (no access) - Deviant S. - The Practically Cheating Statistics Handbook (2010, CreateSpace Independent Publishing Platform).pdf, p108
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