The 2003 Statistical Abstract of the United States reported that the percentage of people 18 years of age and older who smoke. Suppose that

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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

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Diễm Kiều 6 months 2021-08-21T22:43:50+00:00 1 Answers 1 views 0

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    2021-08-21T22:44:57+00:00

    Answer:

    A sample of 2017 should be taken.

    Step-by-step 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{\frac{\pi(1-\pi)}{n}}

    In which

    z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

    The margin of error is of:

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

    0.02 = 1.96\sqrt{\frac{0.3*0.7}{n}}

    0.02\sqrt{n} = 1.96\sqrt{0.3*0.7}

    \sqrt{n} = \frac{1.96\sqrt{0.3*0.7}}{0.02}

    (\sqrt{n})^2 = (\frac{1.96\sqrt{0.3*0.7}}{0.02})^2

    n = 2016.8

    Rounding up:

    A sample of 2017 should be taken.

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