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. (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.
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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.
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