Question

In: Statistics and Probability

According to a census​ bureau, 12.5​% of the population in a certain country changed addresses from...

According to a census​ bureau, 12.5​% of the population in a certain country changed addresses from 2008 to 2009. In​ 2010, 14 out of a random sample of 200 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 ▼ includes does not include the reported proportion from​ 2009, this sample ▼ does not provide provides evidence that this proportion has changed since then.

Solutions

Expert Solution

(a)

P = Population Proportion = 0.125

Q = 1 - P = 0.875

n = Sample Size =200

SE =

= 0.01

From Table, critical values of Z = 2.576

p = Sample Proportion = 14/200 = 0.07

Confidence Interval:

0.07 (2.576 X 0.0234)

= 0.07 0.0603

= ( 0.010 ,0.130)

So,

Answer is:

A 99% confidence interval to estimate the actual proportion has a lower limit of 0.010 and upper limit of 0.130.

(b)

The margin of error = 2.576 X 0.0234

                             = 0.060

So,

Answer is:

The margin of error is 0.060

(c)

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


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