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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. [2.5 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

User LeTex
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1 Answer

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Answer:

A sample of 2017 should be taken.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error is of:


M = z\sqrt{(\pi(1-\pi))/(n)}

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.

This means that
\pi = 0.3

Confidence level:

Not given, so I will use 95%.

This means that
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

a. [2.5 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?

A sample of n should be taken.

n is found fo M = 0.02. So


M = z\sqrt{(\pi(1-\pi))/(n)}


0.02 = 1.96\sqrt{(0.3*0.7)/(n)}


0.02√(n) = 1.96√(0.3*0.7)


√(n) = (1.96√(0.3*0.7))/(0.02)


(√(n))^2 = ((1.96√(0.3*0.7))/(0.02))^2


n = 2016.8

Rounding up:

A sample of 2017 should be taken.

User Hanish Singla
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