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 196. What is the probability that the sample proportion will be 4 or more percent below 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

Sample size: n=196

Population proportion: p'=0.75

Let sample proportion is p

For large n(>30), we approximate the distribution of the data by standard normal distribution by central limit theorem.

sample proportion will be within percentage points of the population proportion mean

p = p^′ ± 0.04 = 0.75 ± 0.04 = ( 0.71 , 0.79)

here standars deviation of sampling distribution σ =√((0.75 × 0.25)/196) = 0.030

Thus, ℙ( 0.71 < p < 0.79)

Step 2:

Since μ=0.75 and σ=0.030 we have:

P ( 0.71<X<0.79 )=P ( 0.71−0.75< X−μ<0.79−0.75 )=P ( 0.71−0.750.030<X−μσ<0.79−0.750.030)

Since Z=x−μσ , 0.71−0.750.030=−1.33 and 0.79−0.750.030=1.33 we have:

P ( 0.71<X<0.79 )=P ( −1.33<Z<1.33 )

Step 3: Use the standard normal table to conclude that:

P ( −1.33<Z<1.33 )=0.8164