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 201. What is the probability that the sample proportion will be at least 4 percent more than the population proportion? Answer (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

The distribution of the sample proportion here is obtained as:

$p \sim N(\mu = \hat p , \sigma = \sqrt{\frac{p(1-p)}{n}})$

$p \sim N(\mu = 0.75 , \sigma = \sqrt{\frac{0.75(1-0.75)}{201}})$

$p \sim N(\mu = 0.75 , \sigma = 0.0305)$

Probability that the sample proportion will be at least 4% or 0.04 more than the population proportion is computed here as:

$P(p > 0.75 + 0.04)$

$P(p > 0.79)$

Converting it to a standard normal variable, we get here:

$P(Z > \frac{0.79 - 0.75}{0.0305})$

$P(Z > 1.3097)$

Getting it from the standard normal tables, we get here:

$P(Z > 1.3097) = 0.0952$

Therefore 0.0952 is the required probability here.