) Beth wants to determine a 99 percent confidence interval for the true proportion p of high school students in the area who attend their ho

Question

) Beth wants to determine a 99 percent confidence interval for the true proportion p of high school students in the area who attend their home basketball games. Out of n randomly selected students she finds that that exactly half attend their home basketball games. About how large would n have to be to get a margin of error less than 0.01 for p

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Ngọc Hoa 3 years 2021-09-04T17:03:23+00:00 1 Answers 33 views 0

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  1. Delwyn
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    2021-09-04T17:05:07+00:00
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    Answer:

    n has to be at least 16577.

    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:

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

    Out of n randomly selected students she finds that that exactly half attend their home basketball games.

    This means that \pi = 0.5

    99% confidence level

    So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

    About how large would n have to be to get a margin of error less than 0.01 for p

    We have to find n for which M = 0.01. So

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

    0.01 = 2.575\sqrt{\frac{0.5*0.5}{n}}

    0.01\sqrt{n} = 2.575*0.5

    \sqrt{n} = \frac{2.575*0.5}{0.01}

    (\sqrt{n})^2 = (\frac{2.575*0.5}{0.01})^2

    n = 16576.6

    Rounding up

    n has to be at least 16577.

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