Question 134667: I need some help please, I've searched the website and I'm not finding this answer. NO ISBN as I'm in an online class.
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.
(b) Check the normality assumption.
(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
(d) Why might this sample not be typical?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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