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 139. What is the probability that the sample proportion will be within 5 percentage points of the population proportion? (Tip: If you compute the standard error [i.e. the standard deviation of the sampling distribution], round it to 4 digits if you have to use it in later computations.) Answer = (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

np=	104.3
n(1-p)=	34.8

since np and n(1-p) both are greater than 10, we can use normal approximation of binomial distribution:

for normal distribution z score =(p̂-p)/σp
here population proportion=     μp=	0.7500
sample size       =n=	139

std error of proportion=σp=√(p*(1-p)/n)=

0.0367

probability that the sample proportion will be within 5 percentage points of the population proportion:

probability =P(0.7<X<0.8)=P((0.7-0.75)/0.037)<Z<(0.8-0.75)/0.037)=P(-1.36<Z<1.36)=0.9131-0.0869=0.8262