According to a census bureau 115 of the population in a cert

According to a census bureau, 11.5% of the population in a certain country changed addresses from 2008 to 2009. In 2010, 30 out of a random sample of 400 citizens of this country said they changed addresses during the previous year (in 2009). Complete parts a through c below. Construct a 95% confidence interval to estimate the actual proportion of people who changed addresses from 2009 to 2010. A 95% 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 for this sample The margin of error is (Round to three decimal places as needed.)

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.075          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.013169567          
              
Now, for the critical z,              
alpha/2 =   0.025          
Thus, z(alpha/2) =    1.959963985          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.049188123          
upper bound = p^ + z(alpha/2) * sp =    0.100811877          
              
Thus, the confidence interval is              
              
(   0.049188123   ,   0.100811877   ) [ANSWER]

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Margin of Error E = (upper bound - lower bound)/2 = 0.025811877 [ANSWER]