Question

 

The population proportion is 0.26. 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 = 1,000

(e)

What is the advantage of a larger sample size?

There is a higher probability

σ_p

will be within ±0.04 of the population standard deviation.We can guarantee

p

will be within ±0.04 of the population proportion p.    As sample size increases,

E(p)

approaches p.There is a higher probability

p

will be within ±0.04 of the population proportion p.

Answer 1

a)

Here, μ = 0.26, σ = 0.0439, x1 = 0.22 and x2 = 0.3. We need to compute P(0.22<= X <= 0.3). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.22 - 0.26)/0.0439 = -0.91
z2 = (0.3 - 0.26)/0.0439 = 0.91

Therefore, we get
P(0.22 <= X <= 0.3) = P((0.3 - 0.26)/0.0439) <= z <= (0.3 - 0.26)/0.0439)
= P(-0.91 <= z <= 0.91) = P(z <= 0.91) - P(z <= -0.91)
= 0.8186 - 0.1814
= 0.6372

b)

Here, μ = 0.26, σ = 0.031, x1 = 0.22 and x2 = 0.3. We need to compute P(0.22<= X <= 0.3). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.22 - 0.26)/0.031 = -1.29
z2 = (0.3 - 0.26)/0.031 = 1.29

Therefore, we get
P(0.22 <= X <= 0.3) = P((0.3 - 0.26)/0.031) <= z <= (0.3 - 0.26)/0.031)
= P(-1.29 <= z <= 1.29) = P(z <= 1.29) - P(z <= -1.29)
= 0.9015 - 0.0985
= 0.803

c)

Here, μ = 0.26, σ = 0.0196, x1 = 0.22 and x2 = 0.3. We need to compute P(0.22<= X <= 0.3). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.22 - 0.26)/0.0196 = -2.04
z2 = (0.3 - 0.26)/0.0196 = 2.04

Therefore, we get
P(0.22 <= X <= 0.3) = P((0.3 - 0.26)/0.0196) <= z <= (0.3 - 0.26)/0.0196)
= P(-2.04 <= z <= 2.04) = P(z <= 2.04) - P(z <= -2.04)
= 0.9793 - 0.0207
= 0.9586

d)
Here, μ = 0.26, σ = 0.0139, x1 = 0.22 and x2 = 0.3. We need to compute P(0.22<= X <= 0.3). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.22 - 0.26)/0.0139 = -2.88
z2 = (0.3 - 0.26)/0.0139 = 2.88

Therefore, we get
P(0.22 <= X <= 0.3) = P((0.3 - 0.26)/0.0139) <= z <= (0.3 - 0.26)/0.0139)
= P(-2.88 <= z <= 2.88) = P(z <= 2.88) - P(z <= -2.88)
= 0.998 - 0.002
= 0.996

e)

There is a higher probability

σp

will be within ±0.04 of the population standard deviation