Please provide a clear, easy to follow, legible answer. Thanks. 6. The 2003 Stat

Question

Please provide a clear, easy to follow, legible answer.
Thanks. 6. The 2003 Statistical Abstract of the United States reported that 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) [2 PTs] 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 90% confidence. (b) [2 PTS] How large a sample should be taken to estimate the proportion of smokers in the population with a margin of of 0.02? Use 95% confidence.

Explanation / Answer

Solution:- p = 0.3 , q = 1-p = 0.7

a) 90% conifdence interval for Z = 1.645 , E = 0.02
n = (Z/E)^2*p*q  
= (1.645/0.02)^2*0.3*0.7
= 1420.6631
n = 1421
b) 95% conifdence interval for Z = 1.96 , E = 0.02
n = (Z/E)^2*p*q  
= (1.96/0.02)^2*0.3*0.7
= 2016.84
n = 2017

7)

Solution:- p = 0.74 , q = 1-p = 1-0.74 = 0.26

a) 95% confidence interval for the proportion of the young tech: p +/- Z*sqrt(pq/n)
: 0.74 +/- 1.96*sqrt(0.74*0.26/1677)
:(0.7190 , 0.7610)

b) 99% confidence interval for the proportion of the young tech: p +/- Z*sqrt(pq/n)
: 0.74 +/- 2.575*sqrt(0.74*0.26/1677)
: (0.7124 , 0.7676)

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