Suppose a national survey intends to identify the children experiencing physical bullying in a population of junior high school students in

Question

Suppose a national survey intends to identify the children experiencing physical bullying in a population of junior high school students in Texas. Out of a sample of 6,458, a total of 754 reported that they have experienced physical bullying. Calculate 95% confidence intervals for the proportion of the sample who have been bullied.

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Trung Dũng 5 months 2021-08-22T00:36:55+00:00 1 Answers 2 views 0

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    2021-08-22T00:38:25+00:00

    Answer:

    The 95% confidence intervals is (0.109, 0.125).

    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}.

    Out of a sample of 6,458, a total of 754 reported that they have experienced physical bullying.

    This means that n = 6458, \pi = \frac{754}{6458} = 0.1168

    95% confidence level

    So \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.

    The lower limit of this interval is:

    \pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1168 - 1.96\sqrt{\frac{0.1168*0.8832}{6458}} = 0.109

    The upper limit of this interval is:

    \pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1168 + 1.96\sqrt{\frac{0.1168*0.8832}{6458}} = 0.125

    The 95% confidence intervals is (0.109, 0.125).

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