Question

According to a government agency, 14.9% of the population of a certain country smoked in 2016. In 2018, a random sample of 528 citiz a. Construct a 90% confidence interval to estimate the actual proportion of people who smoked in the country in 2018. The confidence interval has a lower limit of and an upper limit of (Round to three decimal places as needed.) b. What is the margin of error for this sample? The margin of error is (Round to three decimal places as needed.) c. Is there any evidence that this proportion has changed since 2016 based on this sample? This sample evidence that this proportion has changed since 2016, since the Enter your answer in each of the answer boxes.

Answer 1

a) The sample proportion here is computed as:
p = x/n = 72 / 528 = 0.1364

For 90% confidence level, we have from standard normal tables:
P(-1.645 < Z < 1.645) = 0.9

Therefore the lower and upper limits here are computed as:

$L = p - z^*\sqrt{\frac{p(1-p)}{n}} = 0.1364 - 1.645\sqrt{\frac{0.1364(1 - 0.1364)}{528}}$

$L = p - z^*\sqrt{\frac{p(1-p)}{n}} = 0.1364 - 0.0246 = 0.1118$

p(1-P) U =p+*1 0.1364 +1.645 -1.645V 0.1364(1 – 0.1364) 528 n

$U = p + z^*\sqrt{\frac{p(1-p)}{n}} = 0.1364 + 0.0246 = 0.161$

Therefore 0.112 is the lower limit here and 0.161 is the required upper limit here.

b) The margin of error is already computed in the above part as:

MOE = 0.025

Therefore 0.025 is the required margin of error here.

c) As the given convidence interval contains 14.9%, therefore the sample does not suggest that the proportion has changed since 2016.