Respuesta :
Answer:
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.
