Sampling Distributions

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 278. 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

We are interested in finding

P(\hat p ≥ 0.75 + 0.05) = P(\hat p ≥ 0.80)

Since the mean of the sample proportion is the population proportion, which is 0.75, and the sample size is 278 we have that the standard deviation of the sample proportion is

sqrt{\frac{0.75*(1-0.75)}{278}} = sqrt{0.1875/278}

We can now use these facts about the sample proportion and the sample size as well as the first equation to find that

P(\hat p ≥ 0.80) = P(\hat p - 0.75 ≥ 0.80 - 0.75)

= P(\hat p - 0.75 ≥ 0.05)

= P(\frac{\hat p - 0.75}{sqrt{0.1875/278}} ≥ \frac{0.05}{sqrt{0.1875/278}})

= P(z ≥ 1.9253)

= P(z > 1.9253)

= 0.0271