8.64 Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confes

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8.64 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

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iting 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) ============================ (b) Check the normality assumption. pn= (86/773)773>5; nq=(687/773)773>5 ----------- (c) Try the Very Quick Rule. Does it work well here? Why, or why not? (d) Why might this sample not be typical? -------- Comment: Hopefully you know what "c" and "d" are all about.

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