Question 197259: Biting an unpopped kernel of corn 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 corn 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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sample proportion = 86/773 = 0.11
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
ME = 1.645*sqrt[0.11*0.89/773] = 0.0185
90% CI: 0.11 - 0.0185 < p < 0.11 + 0.0185
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(b) Check the normality assumption.
I'll leave that to you. Check your textbook.
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(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
Not sure what that means.
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(d) Why might this sample not be typical?
It is only a sample of convenience sample; not a simple random sample.
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Cheers,
Stan H.
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