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13 December, 17:30

You wish to estimate a population proportion with a confidence interval with a margin of error no larger than 0.03. You have a budget of $5000 for collecting sample data with which to determine the interval estimate. Collection c sample data in similar situations has cost $10 per item. A plot sample yielded a sample proportion of 0.75. Using the largest possible sample (given the budget) and p^dot = 0.75, the confidence level that you can have in your estimate is: 0.9394. 0.4394. 0.8788. 0.1212.

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  1. 13 December, 18:03
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    The answer is 0.8788

    Step-by-step explanation:

    From question given, let us recall the following:

    We know that Ƶα / 2 * √p (1-p) / n

    when we use n≤ 5000/10 = 500

    P = 0.75

    The Margin of error = 0.03

    Putting this values together we arrive at

    Ƶα/2 = 0.03/√0.75 * 0.25/500

    = 1.549

    Now,

    Ф (1.549) = 0.9394

    Therefore the confidence level becomes:

    1 - (1-∝) / 2 = 0.9394

    ∝ = 0.8788

    The answer is 0.8788
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