Question

# In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA...

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 161. What is the probability that the sample proportion will be 4 or more percent below the population proportion?

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

proportion of complaint p = 0.75

sample size n = 161

The normal distribution is appropriate because np = 120.75 and n(1−p) = 40.25 are both greater than 5

Standard deviation = σ = √ (p*(1-p) / n) = √(0.75*0.25 / 161) = 0.03412618

We want the sample proportion to be within 4% of the population proportion, that is we want your population proportion to be between 0.71 and 0.79

When the sample proportion is 0.71, the z-score associated with that would be:

z = (x - p) /σ becomes z = (0.71 - 0.75) / 0.03412618 = -1.1721

When the sample proportion is 0.79, the z-score associated with that would be:

z = (x - p) /σ becomes z = (0.79 - 0.75) / 0.03412618 = 1.1721

So, the area in a standard normal curve between -1.1721 to 1.1721 is 0.7588

Hence, the probability that the sample proportion will be 4 or more percent below the population proportion = 0.7588

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