Suppose that in a random sample of 300 employed Americans, there are 57 individuals who say that they would fire their boss if they could. C

Question

Suppose that in a random sample of 300 employed Americans, there are 57 individuals who say that they would fire their boss if they could. Calculate a 95% confidence interval for the population proportion. Write a sentence or two that interprets this interval.

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Kinsley 2 weeks 2021-09-15T22:21:41+00:00 1 Answer 0

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    2021-09-15T22:23:07+00:00

    Answer:

    The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

    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 = 300, \pi = \frac{57}{300} = 0.19

    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}} = 0.19 - 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.1456

    The upper limit of this interval is:

    \pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.19 + 1.96\sqrt{\frac{0.19*0.81}{300}} = 0.2344

    The 95% confidence interval for the population proportion is (0.1456, 0.2344). This means that we are 95% sure that the true proportion of employed American who say that they would fire their boss if they could is between 0.1456 and 0.2344.

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