EXAMPLE 5.24

Shopping online. A survey by the Consumer Reports National Research Center revealed that 84% of all respondents were very satisfied with their online shopping experience.¹⁵ It was also reported, however, that people over the age of 40 were generally more satisfied than younger respondents. You decide to take a nationwide random sample of 2500 college students and ask if they agree or disagree that “I am very satisfied with my online shopping experience.” Suppose that 60% of all college students would agree if asked this question. What is the probability that the sample proportion who agree is at least 58%?

The count X who agree has the binomial distribution B(2500, 0.6). The sample proportion does not have a binomial distribution because it is not a count. But we can translate any question about a sample proportion into a question about the count X. Because 58% of 2500 is 1450,

P( ≥ 0.58) = P(X ≥ 1450)

= P(X = 1450) + P(X = 1451) + · · · + P(X = 2500)

This is a rather elaborate calculation. We must add more than 1000 binomial probabilities. Software tells us that . But what do we do if we don’t have access to software?