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  1. #1
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    need help stats

    Confidence interval for a population proportion.
    In a random sample of 400 students at a university, 332 stated that they were nonsmokers. Based on this sample, compute a 95% confidence interval for the proportion of all students at the university who are nonsmokers. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places
    (a)What is the lower limit of the confidence interval?
    (b) What is the upper limit of the confidence interval?
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  2. #2
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    Confidence interval for the population proportion: \hat{P} - E < p < \hat{P} + E, where E = z_c \sqrt{\frac{\hat{p} \cdot \hat{q}}{n}}.

    Make a list of the given information for easier reference.
    n=400
    x=332
    c=95\%=0.95
    z_c=1.96 by looking at the standard normal distribution table
    \hat{p} = \frac{x}{n} = \frac{332}{400} = 0.83
    \hat{q} = 1 - \hat{p} = 1 - 0.83 = 0.17

    Solving for E, we get:
    E = 1.96 \sqrt{\frac{0.83 \cdot 0.17}{400}} \approx 0.0368

    And substituting back into the confidence interval formula:
    0.83 - 0.0368 < p < 0.83 + 0.0368
    \boxed{0.793 < p < 0.867}
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by harry View Post
    Confidence interval for a population proportion.
    In a random sample of 400 students at a university, 332 stated that they were nonsmokers. Based on this sample, compute a 95% confidence interval for the proportion of all students at the university who are nonsmokers. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places
    (a)What is the lower limit of the confidence interval?
    (b) What is the upper limit of the confidence interval?
    Just to be arkward I will observe that you cannot get such a confidence
    interval using this data. A significant number of respondents lied, giving
    the answer that they thought would make them look good in the eyes
    of the survey organisers.

    All you can do is obtain a 95% confidence interval for the proportion of
    students at the university who would have claimed to be non-smokers
    on a survey worded and conducted like the one in question.

    RonL
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