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1 September, 04:26

A newspaper found that, on average, 7.6% of people stay overnight in the hospital with a 1.5% margin of error. Construct a 95% confidence interval.

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  1. 1 September, 04:30
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    Given:

    p = 7.6% = 0.076, the percentage of people who stay overnight at the hospital.

    E = 1.5% = 0.015, margin of error

    95% confidence interval.

    The standard error is

    Es = / sqrt{ / frac{p (1-p) }{n} }

    where

    n = the sample size.

    The margin of error is

    E=z^{*}E_{s}

    where

    z * = 1.96 at the 95% confidence level.

    Because the margin of error is given, there is no need to calculate it.

    The 95% confidence interval is

    p + / - E = 0.076 + / - 0.015 = (0.061, 0.091) = (6.1%, 9.1%)

    Answer:

    The 95% confidence interval is between 6.1% and 9.1%.
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