population proportion?

construct the 95% and the 99% confidence intervals for the population proportion p using the sample statistics below

p=0.733

q=0.267

n=4408

the 95% confidence interval for the population proportion p is ___________ _____________

the 99% confidence interval for the population proportion p is ___________ _____________

2 Answers

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  • M
    Lv 7
    1 decade ago
    Favorite Answer

    ANSWER: 95% Resulting Confidence Interval for 'true mean': = [0.72, 0.746]

    Why???

    POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION

    n = Number of samples 4408

    Number of Successes 3231.064

    p = POPULATION PROPORTION 0.733

    significant digits 3

    Confidence Level 95

    "Look-up" Table 'z-critical value' 1.960

    from Cumulative Normal Distribution Table "Look-up" of 'z-critical value'.

    Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical.

    95% Resulting Confidence Interval for 'true mean':

    p +/- (z critical value) * SQRT[p * (1 - p)/n] = 0.733 +/- 1.96 * SQRT[0.733 * (1 - 0.733)/4408] = [0.72, 0.746]

    ANSWER: 99% Resulting Confidence Interval for 'true mean': = [0.716, 0.75]

    Why???

    POPULATION PROPORTION, CONFIDENCE INTERVAL, NORMAL DISTRIBUTION

    n = Number of samples 4408

    Number of Successes 3231.064

    p = POPULATION PROPORTION 0.733

    significant digits 3

    Confidence Level 99

    "Look-up" Table 'z-critical value' 2.576

    from Cumulative Normal Distribution Table "Look-up" of 'z-critical value'.

    Excel function: NORMSINV(probability) Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero, a standard deviation of one and is symmetrical.

    99% Resulting Confidence Interval for 'true mean':

    p +/- (z critical value) * SQRT[p * (1 - p)/n] = 0.733 +/- 2.576 * SQRT[0.733 * (1 - 0.733)/4408] = [0.716, 0.75]

  • gains
    Lv 4
    4 years ago

    because you recognize the classic deviation, a z-try will suffice. on the ninety% aspect of self assurance, your z-score is z = a million.645. enable s be the classic deviation and n the pattern length. Then the margin of mistakes is defined as: E = zs / n^(a million/2) = (a million.645)(21.9) / 22^(a million/2) = 16.375.

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