Question

In 2014, the Centers for Disease Control reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 0.30. (a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02? Use 95% confidence. (Round your answer up to the nearest integer.) 2017 Correct: Your answer is correct. (b) Assume that the study uses your sample size recommendation in part (a) and finds "680" smokers. What is the point estimate of the proportion of smokers in the population?

Answer 1

Answer:

a) 2017

b) 0.3371

Step-by-step explanation:

Given:

Population proportion, P= 0.30

a) To find the sample size.

Let's use the expression:

$n = \frac{(Z_\alpha/2)^2 * P(1-P)}{(M.E)^2}$

Where,

Margin of error, M.E = 0.02

At confidence interval of 95%, = 0.95

$\alpha = 1 - 0.95 = 0.05,$ $\alpha /2 = 0.025, 1 - 0.025 = 0.975$

Using Z table, the Z value for probability of 0.975 is 1.96.

Therefore, Substituting values in the formula, we have:

$n = \frac{(1.96/2)^2 * 0.30(1 - 0.30)}{(0.02)^2}$

n = 2016.84

≈ 2017

The sample size is approximately 2017.

b) With sample size, n = 2017

Sample mean, X = 680

The point estimate of the proportion of smokers in the population will be calculated using the formula :

$P' = \frac{X}{n}$

$P' = \frac{680}{2017} = 0.3371$

The point estimate of the proportion of smokers in the population is 0.3371