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

remolatg

remolatg

Answered question

2021-02-21

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. 


a.) 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. 


b.) Assume that the study uses your sample size recommendation in part(a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population(to 4 decimals)? 


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

Answer & Explanation

crocolylec

crocolylec

Skilled2021-02-22Added 100 answers

a.) p = 0.30, the confidence level is (1α)=95%, the level of significance is α=0.05
So, E = 0.02, the confidence level is 95%. 
The critical value corresponding Zα2=1.96 from the calculation:
1α=10.95 
α=0.05 
α2=0.052 =0.025 
Area to the left = 1 - Area to the right = 1 - 0.025 = 0.975 
Zα2=1.96[Excelf or μla:NORMINV(0.975,0,1)] 

zα2=1.96,p=0.3,E=0.02
The required sample size is obtained as 2,017: 
n=p^(1p^)×(Zα2E)2 =(0.3×(10.3))×(1.960.02)2 =0.21×9,604 =2,016.84 2,017 (answer)

b.) Among a sample of 2,017 people of age 18 years and older, 520 are identified as smokers, the number of people sampled is n = 2,017, the number of smokers among the sampled people is x = 520, the point estimate of the proportion of smokers in the population is obtained as 0.2578: 
p^=xn =5202,017 =0.2578 (answer)

c.)  p=^0.2578,n=2,017 and zα2=1.96
CI(p^Zα2×p^(1p^)npp^+Zα2×p^(1p^)n  
=(0.25781.96×0.2578(10.2578)2,017p0.2578+1.96×0.2578(10.2578)2,017) 
=(0.25780.0191p0.2578+0.0191) =(0.2387p0.2769)  (answer)

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