Question 428135: : Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 739 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 82.
(a) Construct a 95 % 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 739 kernels and put them in a popper. After popping, the unpopped kernels were counted. There were 82.
(a) Construct a 95 % 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?
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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.
E = z*(p-hat/sqrt(n))
E = 1.645*sqrt[(86/773)(687/773)/773) = 0.01131
90% C.I. = ((86/773)-0.01131 , (86/773)+0.01131)
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(b) Check the normality assumption.
pn= (86/773)773>5; nq=(687/773)773>5
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(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
(d) Why might this sample not be typical?
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Comment: Hopefully you know what "c" and "d" are all about.
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Cheers,
Stan H.
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