Question

QUESTION 10

1. What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

   QUESTION 8 Question 8-10 are based on the following information: The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .35. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)? Use 95% confidence. 2000 2312 2185 2285 0.075 points    QUESTION 9 Assume that the study uses your sample size recommendation in Question 8 and finds 520 smokers. What is the point estimate of the proportion of smokers in the population (to 4 decimals)? 0.2380 0.2250 0.2423 0.2678 0.075 points    QUESTION 10 What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)? 0.2280, 0.2673 0.2103, 0.2549 0.2211, 0.5509 0.2201, 0.2558

Answer 1

8)

Solution :

Given that,

= 0.35

1 - = 1 - 0.35 = 0.65

margin of error = E = 0.02

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.02)² * 0.35 * 0.65

= 2184.91 = 2185

sample size = 2185

9)

n = 2185

x = 520

Point estimate = sample proportion = = x / n = 520 / 2185 = 0.2380

10)

1 - = 1 - 0.2380 = 0.7620

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.2380 * 0.7620) / 2185)

= 0.0179

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.2380 - 0.0179 < p < 0.2380 + 0.0179

0.2201 < p < 0.2558

(0.2201 , 0.2358)