In: Statistics and Probability
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.
(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.