According to a census bureau 125 of the population in a cert

According to a census bureau, 12.5% of the population in a certain country changed addresses from 2008 to 2009. In 2010, 34 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.

a. Construct a 99% confidence interval to estimate the actual proportion of people who changed addresses from 2009 to 2010.A 99% confidence interval to estimate the actual proportion has a lower limit of nothing and an upper limit of nothing.

(Round to three decimal places as needed.)

b. What is the margin of error for this sample? The margin of error is nothing.

(Round to three decimal places as needed.)

c. Is there any evidence that this proportion has changed since 2009 based on this sample? Because the confidence interval found in part a either (includes, or does not include) (pick one) the reported proportion from 2009, this sample either (provieds or does not provide) (pick one) evidence that this proportion has changed since then.

Solution

a)

First, we get the point estimate of the proportion, p^,              
              
p^ = x / n =    0.085          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p(1 - p) / n] =    0.016535946          
              
Now, for the critical z,              
alpha/2 =   0.005          
Thus, z(alpha/2) =    2.575829304          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp   0.042406227          
upper bound = p^ + z(alpha/2) * sp =    0.127593773          
              
Thus, the confidence interval is              
              
(   0.042406227   ,   0.127593773   ) [ANSWER]

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B)

margin of error = (upper bound - lower bound)/2 = 0.042593773 [answer]

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c)

FIRST BLANK: INCLUDES
2ND BLANK: DOES NOT PROVIDE