Question 256946
The U.S. Customs Service conducted a random check of Miami longshoremen and found that 36 of 50 had arrest records. 
(a) Construct a 90 percent confidence interval for the true proportion.
sample proportion: 36/50 = 0.72
standard error: 1.645*sqrt[0.72*0.18/50] = 0.0837...
90% CI: 0.72-0.0837 < p < 0.72+0.0837
90% CI: 0.6363 < p < 0.8037 
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(b) Is the sample size large enough to be convincing? Explain.
I'll leave that up to you.
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(c) How would you select a random sample of longshoremen?
I'll leave that up to you.
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Cheers,
Stan H.