According to a census bureau, 11.5% of the population in a certain country chang

Question

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.)

Explanation / Answer

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]

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