Question

Advertisers contract with Internet service providers and search engines to place ads 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 an ad solely for the purpose of driving up advertising revenue—has become a problem. According to BusinessWeek 43% of advertisers claim they have been a victim of click fraud. Suppose a simple random sample of 300 advertisers will be taken to learn more about how they are affected by this practice. Use z-table.

a. What is the probability that the sample proportion will be within +- 0.03 of the population proportion experiencing click fraud?

(to 4 decimals)

b. What is the probability that the sample proportion will be greater than 0.49?

(to 4 decimals)

Answer 1

Solution

Given that,

p = 0.43

1 - p = 1 - 0.43 = 0.57

n = 300

= p = 0.43

  [p ( 1 - p ) / n] = [(0.43 * 0.57) / 300 ] = 0.0286

a) P( 0.40 < < 0.46 )

= P[(0.40 - 0.43) /0.0286 < ( - ) / < (0.46 - 0.43) / 0.0286]

= P(-1.05 < z < 1.05)

= P(z < 1.05) - P(z < -1.05)

Using z table,   

= 0.8531 - 0.1469

= 0.7062

b) P( > 0.49) = 1 - P( < 0.49 )

= 1 - P(( - ) / < (0.49 - 0.43) / 0.0286)

= 1 - P(z < 2.10)

Using z table

= 1 - 0.9821

= 0.0179