Question

The Grocery Manufactures of America reported that 76% of consumers read the ingredients listed on a product's label. Assume the a sample of 400 consumers is selected from the population.

a. Calculate the standard error of the proportion.

b. What is the probability that the sample proportion will be more than 81% of the population proportion?

c. What is the probability that the sample proportion will be within ± .03 of the population proportion?

d. Answer part (c) for a sample of 750 consumers.

Answer 1

a)

standard error = sqrt(p *(1-p)/n)
= sqrt(0.76 *(1-0.76)/400)
= 0.0214

b)

Here, μ = 0.76, σ = 0.0214 and x = 0.81. We need to compute P(X >= 0.81). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (0.81 - 0.76)/0.0214 = 2.34

Therefore,
P(X >= 0.81) = P(z <= (0.81 - 0.76)/0.0214)
= P(z >= 2.34)
= 1 - 0.9904 = 0.0096

c)

Here, μ = 0.76, σ = 0.0214, x1 = 0.73 and x2 = 0.79. We need to compute P(0.73<= X <= 0.79). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.73 - 0.76)/0.0214 = -1.4
z2 = (0.79 - 0.76)/0.0214 = 1.4

Therefore, we get
P(0.73 <= X <= 0.79) = P((0.79 - 0.76)/0.0214) <= z <= (0.79 - 0.76)/0.0214)
= P(-1.4 <= z <= 1.4) = P(z <= 1.4) - P(z <= -1.4)
= 0.9192 - 0.0808
= 0.8384

d)

Here, μ = 0.76, σ = 0.0156, x1 = 0.73 and x2 = 0.79. We need to compute P(0.73<= X <= 0.79). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.73 - 0.76)/0.0156 = -1.92
z2 = (0.79 - 0.76)/0.0156 = 1.92

Therefore, we get
P(0.73 <= X <= 0.79) = P((0.79 - 0.76)/0.0156) <= z <= (0.79 - 0.76)/0.0156)
= P(-1.92 <= z <= 1.92) = P(z <= 1.92) - P(z <= -1.92)
= 0.9726 - 0.0274
= 0.9452