Question

According to a census? bureau,

13.513.5?%

of the population in a certain country changed addresses from 2008 to 2009. In? 2010,

3131

out of a random sample of

400400

citizens of this country said they changed addresses during the previous year? (in 2009). Complete parts a through c below.

a. Construct a

9595?%

confidence interval to estimate the actual proportion of people who changed addresses from 2009 to 2010.A

9595?%

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

?

does not include

includes

the reported proportion from? 2009, this sample

?

does not provide

provides

evidence that this proportion has changed since then.

Answer 1

Solution:-

p = 0.135

x = 31, n = 400

a) 95?% confidence interval to estimate the actual proportion of people who changed addresses from 2009 to 2010 is C.I = (0.044, 0.111).

C.I = 0.0775 + 1.96 × 0.01709

C.I = 0.0775 + 0.0335

C.I = (0.044, 0.111)

b) Margin of error for this sample is 0.0335.

M.E = 1.96 × 0.01709

M.E = 0.0335

c) Because the confidence interval found in part a, does not include the reported proportion from? 2009, this sample provide evidence that this proportion has changed since then.