SOLUTION: I need help with my statistics class. Can someone please help me with this? Thanks As an experiment, a self-confessed connoisseur of cheap popcorn carefully counted 773 kernels

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Question 186936: I need help with my statistics class. Can someone please help me with this? Thanks
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(57214) (Show Source):
You can put this solution on YOUR website!
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.
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sample proportion: 86/773 = 0.11 ;Sample standard deviation: sqrt(0.11*0.89/773)
= 0.011
E = 1.645*0.0113 = 0.0186
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90% CI: 0.11-0.0186 < p < 0.11+0.0186
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(b) Check the normality assumption.
pn >5 and qn >5
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(c) Try the Very Quick Rule.
I'm not sure what that is.
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Does it work well here?
Why, or why not?
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(d) Why might this sample not be typical?
The sample might not be representative of the population.
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Cheers,
Stan H.