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1.   In 2011, a random sample of 1345 American adults over the age of 20 was selected and 34.9% of them were found to be overweight.

a.   List and check the assumptions are needed to construct a 90% confidence interval to estimate the true proportion of overweight American adults over the age of 20.

b.    Construct and interpret a 90% confidence interval to estimate the true proportion of overweight American adults over the age of 20.

c.   You would like to conduct a new study to estimate the true proportion of overweight American adults (over the age of 20). Using the percentage from the previous study, how large of a sample would be required to construct a 90% confidence level with a 2% margin of error?

Explanation / Answer

(a) Assumptions are np>=10 and n(1-p)>10
We also assume the normal distribution here, and use the z-value for 90% CI.

(b) For 90% CI, z-value = 1.64
Standard Error, SE = sqrt(0.349*(1-0.349)/1345) = 0.013

lower limit = pcap - z*SE = 0.349 - 1.64*0.013 = 0.3277
upper limit = pcap + z*SE = 0.349 + 1.64*0.013 = 0.3703

(c)
As we know that margin of error is given by
ME = z*SE
0.02 = 1.64 * sqrt(0.349*(1-0.349)/n)
n = (1.64/0.02)^2*0.349*(1-0.349) = 1527.686076

hence required sample size = 1528

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