ccording to Time CNN poll, 58% ofphone interviews responded that they were happi

Question

ccording to Time CNN poll, 58% ofphone interviews responded that they were happier than their -nB% 4.T parents. Suppose that this percentage is based o n a random sample of 450 U.S, adults. (a) Construct a 95% confidence interval for the proportion of all U.S. adults who feel that they are happier than their parents (b) Find the sample size that would limit the maximum error to be within 0.025 of the population proportion for a 95% confidence interval. what will be the sample size for the most conservative case? (ie, this is the largest sample size)

Explanation / Answer

a)

CI for 95%

n = 450

p = 0.58

z-value of 95% CI 1.9600

SE = sqrt(p*(1-p)/n)

= sqrt(0.58*0.42/450)

= 0.02327

ME = z*SE

= 1.96*0.02327

= 0.04560

Lower Limit = p - ME = 0.58 - 0.0456 = 0.53440

Upper Limit = p + ME = 0.58 - 0.0456 = 0.62560

95% CI (0.5344 , 0.6256 )

b)

Given CI Level 95%

p = 0.58

ME = 0.025

z-value of 95% CI = 1.9600

n = (z/ME)^2*p*(1-p)

= (1.96/0.025)^2*0.58*0.42

= 1497

c)

Given CI Level 95%

For conservative approach, p = 0.5

ME = 0.025

z-value of 95% CI = 1.9600

n = (z/ME)^2*p*(1-p)

= (1.96/0.025)^2*0.5*0.5

= 1537

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