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12 November, 01:24

The population proportion is. 30. What is the probability that a sample proportion will be within ±.04 of the population proportion for each of the following sample sizes? a. n=100b. n=200c. n=500d. n=1000e. What is the advantage of a larger sample size?

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  1. 12 November, 01:37
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    a) 0.6180

    b) 0.7817

    c) 0.9484

    d) 0.9941

    e) The advantage of a larger sample size is that the estimate covers more ground and is more reliable.

    Step-by-step explanation:

    z = (x - xbar) / σ

    Standard deviation = σ = √variance = √[ (p) (1-p) / n]

    p = population proportion = 0.3

    n = sample size

    Sample proportion = (x - xbar) = ± 0.04

    z = (sample proportion) / (standard deviation)

    a) sample proportion = ± 0.04

    n = 100

    Standard deviation = √[ (0.3) (1 - 0.3) / 100] = 0.0458

    z = ± 0.04/0.0458 = ± 0.873

    Using normal probability distribution tables

    P (|x - xbar| < 0.04) = P (-0.873 < z < 0.873) = P (z < 0.873) - P (z < - 0.873) = 0.809 - 0.191 = 0.6180

    b) sample proportion = ± 0.04

    n = 200

    Standard deviation = √[ (0.3) (1 - 0.3) / 200] = 0.0324

    z = ± 0.04/0.0324 = ± 1.23

    Using normal probability distribution tables

    P (|x - xbar| < 0.04) = P (-1.23 < z < 1.23) = P (z < 1.23) - P (z < - 1.23) = 0.891 - 0.1093 = 0.7817

    c) sample proportion = ± 0.04

    n = 500

    Standard deviation = √[ (0.3) (1 - 0.3) / 500] = 0.0205

    z = ± 0.04/0.0205 = ± 1.95

    Using normal probability distribution tables

    P (|x - xbar| < 0.04) = P (-1.95 < z < 1.95) = P (z < 1.95) - P (z < - 1.95) = 0.974 - 0.0256 = 0.9484

    d) sample proportion = ± 0.04

    n = 1000

    Standard deviation = √[ (0.3) (1 - 0.3) / 1000] = 0.0145

    z = ± 0.04/0.0145 = ± 2.76

    Using normal probability distribution tables

    P (|x - xbar| < 0.04) = P (-2.76 < z < 2.76) = P (z < 2.76) - P (z < - 2.76) = 0.997 - 0.0029 = 0.9941

    e) the advantage of a larger sample size is that the estimate covers more ground and is more reliable.
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