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 113. What is the probability that the sample proportion will be at least 4 percentage points higher than the population proportion?

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

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

Answer 1

Solution:

Given:  the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above

Thus p = 0.75

Sample size = n = 113

We have to find:the probability that the sample proportion will be at least 4 percentage points higher than the population proportion.

That is:

Plô-p> 0.04) = ........

Find standard error:

px (1-P) Op = Vn

0.75 x (1 – 0.75) Op = V 113

0.75 x 0.25 Op = V 113

Op = V0.00165929

Op = 0.0407344

Divide 0.04 by standard error to get z value:

Pp -p > 0,04) = P( (P-P 0 0.04 - 0.0407344

Plo-p> 0.04) = P(Z > 0.981971)

Plo-p> 0.04) = 1 - P(Z < 0.981971)

Now use Excel command to get probability:

=1-NORM.S.DIST(z , cumulative)

=1-NORM.S.DIST(0.981971, TRUE)

=0.163057

=0.1631

Thus the sample proportion will be at least 4 percentage points higher than the population proportion is 0.1631