Advertisers contract with Internet service providers and search engines to place ad on websites. They pay a fee based on the number of potential customers who click on their ad. Unfortunately, click fraud-the practice of someone clicking on and ad solely for the purpose of driving up advertising revenue-- - has become a problem. Forty percent of advertisers claim they have been a victim of click fraud (business Week, March 13, 2006). Suppose a simple random sample of 380 advertisers will be taken to learn more about how they are affected by this practice. A. What is the probability that the sample proportion will be within + / -.04 of the population experiencing click fraud? B. What is the probability that the sample reapportion will be greater that. 45?

b. P (ρ^>0.45) = 0.0233

Step-by-step explanation:

Hello!

To calculate the asked probabilities you need to first summarize the information given in the problem.

Study variable:

X: "Number of advertisers affected by click fraud"

n = 380 advertisers

ρ = 0.4 (probability of suffering click fraud)

ρ^ (sample proportion)

To calculate the probability of the sample proportion to take any number, you'll have to apply the Central Limit Theorem to approximate it's distribution to normal, that way

ρ^≈N (ρ; ([ρ * (1-ρ) ]/n)

Under this distribution, we can use de statistic Z = (ρ^-ρ) / √[ρ * (1-ρ) ]/n ≈ N (0; 1) to calculate the wanted probabilities.

a.

b.

P (ρ^>0.45) ⇒ 1 - P (ρ^≤0.45) ⇒ 1 - P (Z≤ (0.45 - 0.4) / √[0.4 * (1-0.4) ]/380)

⇒ 1 - P (Z≤ 1.99) = 1 - 0.9767 = 0.0233

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