Fifteen percent of Australians smoke. By introducing tough laws banning brand la
ID: 3134405 • Letter: F
Question
Fifteen percent of Australians smoke. By introducing tough laws banning brand labels on cigarette packages, Australia hopes to reduce the percentage of people smoking to 10% by 2018 (Reuters website, October 23, 2012). Answer the following questions based on a sample of 240 Australians. a. Show the sampling distribution of p, the sample proportion of Australians who are smokers.
A) Show the sampling distribution of P, the sample proportion of Australians that are smokers.
B) What is the probability the sample proportion will be within +/-.04 of the population proportion?
C) What is the probability the sample proportion will be within +/-.02 of the population proportion?
Explanation / Answer
Fifteen percent of Australians smoke. By introducing tough laws banning brand labels on cigarette packages, Australia hopes to reduce the percentage of people smoking to 10% by 2018 (Reuters website, October 23, 2012). Answer the following questions based on a sample of 240 Australians. a. Show the sampling distribution of p, the sample proportion of Australians who are smokers.
A) Show the sampling distribution of P, the sample proportion of Australians that are smokers.
Sample mean p=0.10
Standard deviation = sqrt(p*(1-p)/n) = sqrt(0.10*0.90/240) = 0.0194
B) What is the probability the sample proportion will be within +/-.04 of the population proportion?
Z value for +/-.04 of the population proportion, =(-2.06,2.06)
P( -2.06<z<2.06)
=0.9803-0.0197
=0.9606
C) What is the probability the sample proportion will be within +/-.02 of the population proportion?
Z value for +/-.02 of the population proportion, =(-1.03,1.03)
P( -1.03<z<1.03)
= 0.8485 - 0.1515
=0.697
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