According to a census bureau, 12.5% of the population in a certain country chang

Question

According to a census bureau, 12.5% of the population in a certain country changed addresses from 2008 to 2009, In 2010. 16 out of a random sample of 200 citizens of this country said they changed addresses previous year (in 2009). Complete parts a through C below. Construct a 99% confidence interval to estimate the actual proportion of people who changed address from 2009 to 2010. A 99% confidence interval to estimate the actual proportion has a lower limit of and an upper limit of. (Round to three decimal places as needed.) What is the margin of error of this sample? The margin of error is. (Round to three decimal places as needed.) Is there any evidence that this proportion has changed since 2009 based on this sample? Because the confidence interval found in part a the reported proportion from 2009, this sample evidence that this proportion has changed since then.

Explanation / Answer

here estimated proportion phat =16/200=0.08

hence std error =(p(1-p)/n)1/2 =0.0192

also for 99% CI;z= 2.5758

hence confidence interval =estimated proportion -/+ z*std error =0.0306 ; 0.1294

b) margin of error =z*std error =0.0494

c)in part a ; contains the.........do not have sufficient evidence

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