Question 175907: 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. (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 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.
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
p-hat = 86/773 = 0.111..
E = 1.645*sqrt[(0.111)(0.889)/773]=0.0186
90% CI: 0.111-0.0186 < p < 0.111+0.0186
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(b) Check the normality assumption
I'll let you do that; it differs with every text.
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(c) Try the Very Quick Rule.
Does it work well here? Why, or why not?
I'm not sure what that is; it depends on your text.
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(d) Why might this sample not be typical?
The 773 is probably not a simple-random-sample
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Cheers,
Stan H.
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