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 5 or more percent below the population proportion?
Given: p = 0.75, the population proportion n = 129, the sample size
We want the sample proportion to be 5% or more below the population proportion.
The standard error is [tex]SE_{p}= \sqrt{ \frac{p(1-p)}{n} } = \sqrt{ \frac{0.75*0.25}{129} } =0.0381[/tex]
5% of p = 0.05*0.75 = 0.0375
The unknown value of z* determines the confidence interval (or probability). [tex]z^{*}SE_{p}=0.0375\\ z^{*}= \frac{0.0375}{0.0381} =0.9843[/tex] This is a probability of about 98%
From standard tables, z* of 2.15 yields an area of 0.9843 at about 80% confidence level (Not required).
