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 105. What is the probability that the sample proportion will be within 10 percent of the population proportion?

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

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Population proportion =p =0.75

Within 10% means within 0.75 - 0.10 and 0.75 + 0.10 =0.65 and 0.85

Sample proportion =

Sample size =n =105

The sample proportion, is normally distributed with a mean of =p and a standard deviation of = as long as np 10 and n(1 - p) 10.

np =105(0.75) =78.75 > 10

n(1 - p) =105(1 - 0.75) =26.25 > 10

So, is normally distributed.

Mean = =p =0.75

Standard deviation =_{​​​​​​​​​​​​}​​​​= = =0.042257713

Z =(X - ​​​​​​)/ =(0.65 - 0.75)/0.042257713 = -2.3664

Z =(X - ​​​​​​)/ =(0.85 - 0.75)/0.042257713 =2.3664

The probability that the sample proportion will be within 10 percent of the population proportion =P(0.65 0.85) =P(-2.3664 Z 2.3664) =0.9820