Question 134667
Biting an un-popped 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 un-popped kernels were counted. There were 86. 
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
p-hat = 86/773 = 0.1113
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E = 1.96*sqrt[0.1113*0.8887/773] = 0.0221
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CI = (0.1113-0.0221 , 0.1113+0.0221)
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(b) Check the normality assumption.
pn = 0.113*773 = 86+
qn = even higher
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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?
The geographic distribution of kernels is enormous. The sample
was gathered by one person in one place so it probably is not
a simple-random-sample becasue not every kernel had an equal
chance of being selected.
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Cheers,
Stan H.