Question 129135
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 are 86. 
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p-hat = 86/773=0.1113
std = sqrt[0.1113*0.8887/773]= 0.113
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(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not pop.
E = 1.645*0.0113 = 0.0186
C.I.: (p-hat-E , p-hat+E) or (0.0927 , 0.1299)
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(b) Check the normality assumption.
Check pn>5 and qn>5
0.1113*773 = 86-plus ; q*p-hat is even larger
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(c) Try the Very Quick Rule. Does it work well here?Why or why not?
I am not familiar with that Rule.
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(d) Why might this sample not be typical?
Since the C.I. is a statement about the whole population of popcorn
kernnels, the sample here might not be representative.
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Cheers,
Stan H.