Statistics and Probability

Question

In a recent year, the Better Business Bureau settled 75% of complaints they received. 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. Suppose your sample size is 117. What is the probability that the sample proportion will be at least 2 percent more than the population proportion?

Answer 1

The given information and wording makes this question a bit ambiguous; this will become more clear in a bit. I'm NOT going to assume that the population percentage for the current year is equal to 75%.

Under the given assumptions, it's reasonable to assume that the sample proportions (for all samples of size 117) follow a Normal distribution with:

Mean = p (the true population proportion for this year)

Standard Deviation = sqrt(p*(1-p)/117)

"2 percent more than the population proportion" is given by the expression (p + .02). We can (almost) compute the z-score for this value by the usual formula:

z = ([Actual Value] - [Mean]) / (Standard Deviation)

The numerator is [p + .02] - [p] = .02, so it doesn't matter that the true value of p is unknown.

To compute the denominator, we need an actual numerical value of p. The best we can do is approximate it using last year's value of p = .75, giving an approximate Standard Deviation of .0400 using the formula above.

Thus the z-score for "2 percent more than the population proportion" is

z = (.02)/(.0400) = .4996

You can use technology or a standard Normal table to find the proportion of values with a z-score *less than* this (if you're still using tables, round to z = 0.50 first); I get a proportion of .6913 using Geogebra with the unrounded z-score.

The probability of getting a z-score greater than this will be 1 - .6913 = .3087; this is what the question is asking.