Question

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 .30.

1. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02? Use 95% confidence.
2. Assume that the study uses your sample size recommendation in part (a) and find 520 smokers. What is the point estimate of the proportion of smokers in the population?
3. What is the 95 % confidence interval for the proportion of smokers in the population?

PLEASE WRITE CLEAR!

Answer 1

Saution: (a) Given that. proportion of the probablutey p= 0.30. i alven that. Margin of error for 95% confidence intervat = 0.02 a= 1-c. 1-951100 1-0.95 = 0.05 Z2 = 20.05 2412 = 2005/ = 20.025. - 1.96 from t table. C.I = 72/2 0.09. - 1.96 0.30 (1 -0.30) . 196 10.30 x 0.70 = 196 10:21 = 1.96 0.458257) 0.02.2 0.898184 0-898184 vn - m = 0.09

44.9092418 (n) = (44.90924189V n = 2016.84. i na 2016) (6). Given that n= 2016 X- 520 point estimate CP -6292016 0.2579365079 C. LV n : point estimate (P)-10.2579] 95vo Cill for proportion of Smoker (P) is C.I = P & 242 SP (1-P) = 0.2579 † 196 0.2579 x(1-0.2573) 2016 = 0.2579 & 1.96 0.2579 (0.2421) 2016 0.19138755 2579 +696 2016 0.25797 196 (0.00974 34244) = 0.2579 +0.019097 = 0.2579-0.019097 , 0.2579 +0.019090 < (0.238803, 0.276997) 0.238803 = P = 0.276997 ::