Question 398352
Please help with part (a). I understand that I need to use a students t bar but my knowledge on this subject is very limited. Please disregard the text references, I am just trying to find starting points for solving this problem. 
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. 
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p-hat = 86/773 = 0.111255
ME = (1.645)sqrt[0.111*0.889/773] = 0.018605
90% CI: 0.111-0.0186 < p < 0.111+0.0186 
90%CI: 0.0923 < p < 0.1296
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You could run a 1-PropZInt on x = 86, n= 773 and get this answer.
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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? 
p+-(1/sqrt(n) = 0.111+-(1/sqrt(773)) = 0.111+-0.035968
gives 90%CI:  0.111-0.036 < p < 0.111+0.036
0.075 < p < 0.147
Not good because p-hat = 0.111 is not near 0.5

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(d) Why might this sample not be typical? 
p = x/n = 86/773 = 0.1113
The z-value = 0.10 = 1.645 
According to text on page 315, 90% = .10, 99% = .01 etc. 
a) We use a Student's T test. 
b) Page 374 
c) The Very Quick Rule can be found on page 325 of the text. Test once solve (a) and (b)
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d) This sample is not typical because the connoisseur only tested one bag of popcorn. Like in class with the M & M's, we would have to have a large sample size to make accurate predictions.
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Cheers,
Stan H.