Page 194 - Statistics for Dummies

178 Part III: Distributions and the Central Limit Theorem

When you plug in the numbers for this example, you get:

And then you find P(Z > 1.44) = 1 – 0.9251 = 0.0749 using Table A-1 in the
appendix. So if it’s true that 0.38 percent of all students taking the exam want
math help, the chance of taking a random sample of 100 students and finding
more than 45 needing math help is approximately 0.0749 (by the CLT).

As noted in the previous section on sample means, you can use sample pro-
portions to check out a claim about a population proportion. (This procedure
is a hypothesis test for a population proportion; all the details are found in
Chapter 15.) In the ACT example, the probability that more than 45% of the
students in a sample of 100 need math help (when you assumed 38% of the
population needed math help) was found to be 0.0749. Because this prob-
ability is higher than 0.05 (the typical cutoff for blowing the whistle on a claim
about a population value), you can’t dispute their claim that the percentage in
the population needing math help is only 38%. Our sample result is just not a
rare enough event. (See Chapter 15 for more on hypothesis testing for a popu-
lation proportion.)

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