Question

The population proportion is 0.36. What is the probability that a sample proportion will be within ±0.04 of the population proportion for each of the following sample sizes? (Round your answers to 4 decimal places.)

A)n=100

B)n=200

C)n=500

D)n=1000

What is the advantage of a larger sample size?

Answer 1

a)
Here, μ = 0.36, σ = 0.048, x1 = 0.32 and x2 = 0.4. We need to compute P(0.32<= X <= 0.4). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.32 - 0.36)/0.048 = -0.83
z2 = (0.4 - 0.36)/0.048 = 0.83

Therefore, we get
P(0.32 <= X <= 0.4) = P((0.4 - 0.36)/0.048) <= z <= (0.4 - 0.36)/0.048)
= P(-0.83 <= z <= 0.83) = P(z <= 0.83) - P(z <= -0.83)
= 0.7967 - 0.2033
= 0.5934

b)

Here, μ = 0.36, σ = 0.0339, x1 = 0.32 and x2 = 0.4. We need to compute P(0.32<= X <= 0.4). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.32 - 0.36)/0.0339 = -1.18
z2 = (0.4 - 0.36)/0.0339 = 1.18

Therefore, we get
P(0.32 <= X <= 0.4) = P((0.4 - 0.36)/0.0339) <= z <= (0.4 - 0.36)/0.0339)
= P(-1.18 <= z <= 1.18) = P(z <= 1.18) - P(z <= -1.18)
= 0.881 - 0.119
= 0.7620

c)

Here, μ = 0.36, σ = 0.0215, x1 = 0.32 and x2 = 0.4. We need to compute P(0.32<= X <= 0.4). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.32 - 0.36)/0.0215 = -1.86
z2 = (0.4 - 0.36)/0.0215 = 1.86

Therefore, we get
P(0.32 <= X <= 0.4) = P((0.4 - 0.36)/0.0215) <= z <= (0.4 - 0.36)/0.0215)
= P(-1.86 <= z <= 1.86) = P(z <= 1.86) - P(z <= -1.86)
= 0.9686 - 0.0314
= 0.9372

d)

Here, μ = 0.36, σ = 0.0152, x1 = 0.32 and x2 = 0.4. We need to compute P(0.32<= X <= 0.4). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.32 - 0.36)/0.0152 = -2.63
z2 = (0.4 - 0.36)/0.0152 = 2.63

Therefore, we get
P(0.32 <= X <= 0.4) = P((0.4 - 0.36)/0.0152) <= z <= (0.4 - 0.36)/0.0152)
= P(-2.63 <= z <= 2.63) = P(z <= 2.63) - P(z <= -2.63)
= 0.9957 - 0.0043
= 0.9914

as the sampl esize increases the standard error decreases and probability increases