Construct a 90% confidence interval of the population proportion using the given information. X= 175 n= 250

Question

Construct a 90% confidence interval of the population proportion using the given information.

X= 175

n= 250

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Arianna 2 hours 2021-10-14T03:00:11+00:00 1 Answer 0

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    2021-10-14T03:01:30+00:00

    Answer:

    The 90% confidence interval of the population proportion is (0.6523, 0.7477).

    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}.

    For this problem, we have that:

    n = 250, \pi = \frac{X}{n} = \frac{175}{250} = 0.7

    90% confidence level

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

    The lower limit of this interval is:

    \pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7 - 1.645\sqrt{\frac{0.7*0.3}{250}} = 0.6523

    The upper limit of this interval is:

    \pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.7 + 1.645\sqrt{\frac{0.7*0.3}{250}} = 0.7477

    The 90% confidence interval of the population proportion is (0.6523, 0.7477).

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