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?

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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