Question

In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 276. What is the probability that the sample proportion will be at least 5 percent more than the population proportion?

Note: You should carefully round any z-values you calculate to at least 4 decimal places to match wamap's approach and calculations.

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Solution

Given that,

p = 0.75

1 - p = 1 - 0.75 = 0.25

n = 276

= p = 0.75

=  [p( 1 - p ) / n] = [(0.75 * 0.25) / 276 ] = 0.0261

= 0.75 + 0.05 = 0.80

P( > 0.80) = 1 - P( < 0.80)

= 1 - P(( - ) / < (0.80 - 0.75) / 0.0261 )

= 1 - P(z < 1.9157)

Using z table

= 1 - 0.9723

= 0.0277