In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home. Find a 96% confidence interval for the tr

Question

In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home. Find a 96% confidence interval for the true proportion of voters who feel this way. a.Find a point estimate for p, the population proportion who thought US spends too little on fighting hunger at home.b.Find a 96% confidence interval for p.

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Khánh Gia 2 months 2021-07-24T17:27:09+00:00 1 Answers 1 views 0

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    2021-07-24T17:28:44+00:00

    Answer:

    a. 0.48

    b. The 96% confidence interval for p is (0.457, 0.503).

    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}.

    In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home.

    This means that n = 2000, \pi = \frac{960}{2000} = 0.48.

    This is the answer for question a.

    96% confidence level

    So \alpha = 0.04, z is the value of Z that has a pvalue of 1 - \frac{0.04}{2} = 0.98, so Z = 2.056.

    The lower limit of this interval is:

    \pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 - 2.056\sqrt{\frac{0.48*0.52}{2000}} = 0.457

    The upper limit of this interval is:

    \pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 + 2.056\sqrt{\frac{0.48*0.52}{2000}} = 0.503

    The 96% confidence interval for p is (0.457, 0.503).

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