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 129. What is the probability that the sample proportion will be within 10 percentage points of the population proportion?

Answer 1

It is given that in a recent year, the Better Business Bureau settled 75% of complaints they received. Given that, the population proportion of complaints settled for the computer stores is 0.75. The sample size is 129. We have to find the probability that the sample proportion will be within 10 percent of the population proportion.

Thys, we need sample proportion between 0.75-0.10=0.65 and 0.75+0.10=0.85.

It is given that, p=0.75 and n=129

So, z-score for sample proportion =0.65 is

z= (-p)/√[p(1-p)/n]​​ = (0.65-0.75)/√[0.75(1-0.75)/129]

= -0.10/√(0.1875/129) = -0.10/0.038 = -2.623

And, z-score for sample proportion =0.85 is

z= (-p)/√[p(1-p)/n] = (0.85-0.75)/√[0.75(1-0.75)/129]

= 0.10/√(0.1875/129) = 0.10/0.038 = 2.623

Thus, the probability that the sample proportion will be within 10 percent of the population proportion is

P(0.65<<0.85)= P(-2.623<z<2.623)= P(z<2.623) - P(z< -2.623)

= 0.9956 - 0.0044 [from z-score table]

= 0.9912

Thus, the required probability is 0.9912.