Question 148687
Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 86.
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p-hat = 86/773 = 0.111
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(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
E = 1.645*sqrt[(0.111)(0.889)/773] = 1.645*0.0113 = 0.018586...
90% CI: (0.111-0.018586 < p < 0.111+0.018586)
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(b) Check the normality assumption.
pn = 0.111*773=85 ; qn = 687.197
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(c) Try the Very Quick Rule. Does it work well here? Why, or why not? 
I don't know what the Quick Rule is.
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(d) Why might this sample not be typical?
One person counting his unpopped kernels does not a random sample make.
Cheers,
Stan H.