A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him

Question

A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed

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Calantha 2 months 2021-07-23T03:58:39+00:00 1 Answers 1 views 0

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    2021-07-23T04:00:13+00:00

    Answer:

    A sample size of 752 is needed.

    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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

    The margin of error is of:

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

    90% confidence level

    So \alpha = 0.1, z is the value of Z that has a p-value of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

    If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed?

    We have no estimate of the proportion, so we use \pi = 0.5.

    The sample size is n for which M = 0.03. So

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

    0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}

    0.03\sqrt{n} = 1.645*0.5

    \sqrt{n} = \frac{1.645*0.5}{0.03}

    (\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2

    n = 751.67

    Rounding up:

    A sample size of 752 is needed.

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